# A Ramsey theorem for the reals

Tanmay C. Inamdar  Department of Mathematics, Ben-Gurion University of the
Negev, P.O.B. 653, Be’er Sheva, 84105 Israel<EMAIL_ADDRESS>
(Date: Preliminary version as of )

###### Abstract.

We prove that for every colouring of pairs of reals with finitely-many
colours, there is a set homeomorphic to the rationals which takes no more than
two colours. This was conjectured by Galvin in 1970, and a colouring of
Sierpiński from 1933 witnesses that the number of colours cannot be reduced to
one. Previously in 1985 Shelah had shown that a stronger statement is
consistent with a forcing construction assuming the existence of large
cardinals. Then in 2018 Raghavan and Todorčević had proved it assuming the
existence of large cardinals. We prove it in ZFC. In fact Raghavan and
Todorčević proved, assuming more large cardinals, a similar result for a large
class of topological spaces. We prove this also, again in ZFC.

###### Key words and phrases:

Ramsey theory, partition relations, PCF theory.

###### 2020 Mathematics Subject Classification:

Primary 03E02; Secondary 03E04, 03E55, 05D10, 05C55

##  1\. Introduction

We prove here the following:

######  Theorem A.

Let $X$ be an uncountable set of reals. Let $K$ be a positive natural number
and let $c:[X]^{2}\rightarrow\\{0,1,\ldots,K-1\\}$ be a colouring of pairs of
elements of $X$ into $K$-many colours. Then for some $Y\subseteq X$
homeomorphic to $\mathbb{Q}$, $c$ takes at most $2$ colours on pairs of
elements of $Y$; that is, $|c``[Y]^{2}|\leq 2$.111For a set $Z$, we denote by
$[Z]^{2}$ the collection of all two element subsets of $Z$, and $|Z|$ denotes
the cardinality of the set $Z$. Additional notation can be found in Section 2.

Theorem A is the most interesting case of a more general result, Theorem D,
that we will discuss in Section 1.1. An immediate consequence of Theorem A is
the following Ramsey theorem for the reals.

######  Corollary B.

For every positive natural number $K$ and
$c:[\mathbb{R}]^{2}\rightarrow\\{0,1,\ldots,K-1\\}$ a colouring of pairs of
reals, there is a set of reals $Y$, homeomorphic to $\mathbb{Q}$, such that
$|c``[Y]^{2}|\leq 2$.

We now commence on a short survey of infinite Ramsey theory aimed towards
placing Theorem A and Corollary B in their appropriate context. We will
deliberately avoid using the Erdős-Hajnal-Rado arrow [8, § 18] in order to
avoid overloading the reader with notation. Below, by standard set-theoretic
convention, an ordinal will also represent the set of its predecessors. In
particular, for $K$ a natural number, $K=\\{0,1,\ldots,K-1\\}$.

We start in 1928 with Ramsey’s Theorem.

######  Theorem (Ramsey, [21]).

Let $K$ be a positive natural number. For any colouring
$c:[\mathbb{N}]^{2}\rightarrow K$, there is an infinite $Y\subseteq\mathbb{N}$
such that $|c``[Y]^{2}|=1$.

Interpreting the above as a result about the ordertype (as opposed to the
size222For a perspective centred on cardinalities, starting from Erdős and
Tarski’s isolation of the notion of _weakly compact cardinals_ , [10], see
[16].) of the natural numbers, our focus here will be attempts to generalise
Ramsey’s Theorem to the most basic infinite structures.

Sierpiński, answering a question of Knaster, constructed in 1933 the following
colouring ([31]). Let $<^{*}$ be a well-ordering of $\mathbb{R}$ and $<$ the
usual order on $\mathbb{R}$. Let $c_{\mathrm{s}}:[\mathbb{R}]^{2}\rightarrow
2$ be defined via: for $x<y$ from $\mathbb{R}$,

$c_{\mathrm{s}}(x,y):=\begin{cases}0&\text{ if }x<^{*}y;\\\ 1&\text{ if
}y<^{*}x.\end{cases}$

Then for any uncountable set $Y\subseteq\mathbb{R}$,
$c_{\mathrm{s}}``[Y]^{2}=2$. This is because there cannot be an order-
preserving embedding of the first uncountable ordinal $\omega_{1}$ or its
reverse into $\mathbb{R}$, since $\mathbb{R}$ has a countable dense set. It
follows that there is no direct analogue of Ramsey’s Theorem for
$\mathbb{R}$.333Curiously, in [31] Sierpiński seems to be unaware of Ramsey’s
work, and indeed proves Ramsey’s Theorem himself in the paper. That is, the
infinite Ramsey theorem for pairs, as stated above. He also mentions that
Lindenbaum informed him that he was aware of this fact.

Examining Sierpiński’s colouring $c_{\mathrm{s}}$, since $<^{*}$ is a well-
ordering, we also have that on any set $Y\subseteq\mathbb{R}$ order-isomorphic
to $\mathbb{Z}$, $c_{\mathrm{s}}``[Y]^{2}=2$. Galvin in a letter to Laver,
[12], observed that on any non-empty $Y\subseteq\mathbb{R}$ which is dense-in-
itself, that is, has no isolated points, $c_{\mathrm{s}}``[Y]^{2}=2$. This
uses the fact that any non-empty $Y\subseteq\mathbb{R}$ which is dense-in-
itself contains a set order-isomorphic to $\mathbb{Q}$, and hence also
contains a set order-isomorphic to $\mathbb{Z}$.444Some details are provided
in Section 2.

Since any set of reals homeomorphic to $\mathbb{Q}$ is dense-in-itself, it
follows that the number of colours in Theorem A and Corollary B cannot be
reduced to $1$, that is, we cannot obtain a monochromatic set.

Sierpiński’s idea can also be applied to the rational numbers $\mathbb{Q}$.
That is, starting from a well-ordering of $\mathbb{Q}$ instead, we obtain
$c:[\mathbb{Q}]^{2}\rightarrow 2$ such that for any set $Y\subseteq\mathbb{Q}$
order-isomorphic to $\mathbb{Z}$, $c``[Y]^{2}=2$. It follows that there is no
direct analogue of Ramsey’s Theorem for $\mathbb{Q}$ as well.

Galvin was then able to show in an unpublished result (from 1969 per [2, pg.
71]) that the Ramsey theorem holds for $\mathbb{Q}$ _modulo this failure_ (see
discussion of [7, Problem 18] and [2, Chapter 4] and for a proof see [35,
Section 6.3]):

######  Theorem (Galvin).

Let $K$ be a positive natural number. For any colouring
$c:[\mathbb{Q}]^{2}\rightarrow K$, there is a set $Y\subseteq\mathbb{Q}$
order-isomorphic to $\mathbb{Q}$ such that $|c``[Y]^{2}|\leq 2$.

It follows that in Theorem A and Corollary B if we only desire an order-
isomorphic copy of $\mathbb{Q}$, then one can simply colour any countably
infinite subset of $\mathbb{R}$ which is dense as a linearly-ordered set. In
order to see that the conclusion of Theorem A and Corollary B is stronger,
recall the aforementioned fact that any set which is homeomorphic to
$\mathbb{Q}$ contains a set order-isomorphic to $\mathbb{Q}$.

Van Douwen asked the natural question whether Galvin’s Theorem can be
strengthened to obtain in fact a homeomorphic copy of $\mathbb{Q}$ on which at
most $2$ colours are realised. This was answered negatively by Baumgartner in
1982.555For a positive result when the colourings are required to be
continuous, see [38, Theorem 4.1].

######  Theorem (Baumgartner, [1]).

Let $X$ be a countable Hausdorff topological space. Then there is
$c:[X]^{2}\rightarrow\mathbb{N}$ such that for every $Y\subseteq X$
homeomorphic to $\mathbb{Q}$, $c``[Y]^{2}=\mathbb{N}$.

It follows that in Theorem A the set of reals $X$ necessarily must be
uncountable.

Mentioned explicitly in Baumgartner’s paper [1] from 1982 as “the most
interesting open problem in this area”, and credited to the same letter, [12],
from Galvin to Laver in 1970 by Raghavan and Todorčević in [19] is the
conjecture of Galvin which Corollary B verifies.

######  Conjecture (Galvin).

For every positive natural number $K$ and $c:[\mathbb{R}]^{2}\rightarrow K$ a
colouring of pairs of reals, there is a set of reals $Y$ homeomorphic to
$\mathbb{Q}$ such that $|c``[Y]^{2}|\leq 2$.

This appears also in Weiss’ survey on partition relations for topological
spaces, [40, Question 4.2]. We will refer to the conjectured statement, that
is, the statement of Corollary B, as _Galvin’s Conjecture_ in the rest of this
introduction.

Related to this is the following question which Galvin also considered, but
which appears already in the Erdős-Hajnal problem list from 1967 as [6,
Problem 15].

######  Question (Erdős and Hajnal, [6]).

Without assuming the continuum hypothesis, $2^{\aleph_{0}}=\aleph_{1}$, can
one prove that there is a colouring $c:[\mathbb{R}]^{2}\rightarrow 3$ of pairs
of reals into $3$ colours, such that for every uncountable set of reals $Y$,
$|c``[Y]^{2}|=3$?

The assumption of the negation of the continuum hypothesis is justified by the
Erdős-Hajnal-Rado result ([8, Theorem 17]) that assuming
$|\mathbb{R}|=\aleph_{1}$, there is a colouring
$c:[\aleph_{1}]^{2}\rightarrow\aleph_{1}$ such that on any uncountable
$Y\subseteq\aleph_{1}$, $c``[Y]^{2}=\aleph_{1}$. Later in 1984, [37],
Todorčević using his theory of ‘walks on ordinals’ (see the monograph [34])
proved the spectacular result that the same is true just in ZFC, the most
radical failure possible for the Ramsey Theorem at $\aleph_{1}$.666For various
extensions of Todorčević’s result at $\aleph_{1}$ and at other cardinals, see
the survey [23].

These results show that the strengthening of Corollary B where the set $Y$
(according to the notation there) is asked to be uncountable does not hold in
any model satisfying the continuum hypothesis.

Returning to the Erdős-Hajnal question, Shelah in 1985 proved that the answer
to it is negative, by giving a forcing construction assuming the existence of
suitable large cardinals.

######  Theorem (Shelah, [25]).

Assuming the existence of an $\omega_{1}$-Erdős cardinal, the following is
consistent: let $K$ be a positive natural number and let
$c:[\mathbb{R}]^{2}\rightarrow K$ be a colouring of pairs of reals into
$K$-many colours; then for some uncountable $Y\subseteq\mathbb{R}$,
$|c``[Y]^{2}|\leq 2$.777Shelah later improved the large cardinal assumption in
[29, Theorem 1.21] to a strongly Mahlo cardinal. See also [26] for a similar
result for all finite dimensions.

The exact definition of an $\omega_{1}$-Erdős cardinal, or more generally, any
other large cardinal we mention will not be required in the paper so we refer
the reader to [16] for details.

To connect the Erdős-Hajnal question to Galvin’s Conjecture, note that in
Shelah’s model, for every colouring $c:[\mathbb{R}]^{2}\rightarrow K$ of pairs
of reals into finitely-many colours, for some uncountable
$Y\subseteq\mathbb{R}$, $|c``[Y]^{2}|\leq 2$. Then we can shrink $Y$ to a
countably infinite set $Z$ which is dense-in-itself, and hence $Z$ is
homeomorphic to $\mathbb{Q}$888Sierpiński [32] proved that any countable non-
empty metric space which is dense-in-itself is homeomorphic to $\mathbb{Q}$.
In other words, Shelah’s forcing construction establishes that, assuming large
cardinals, Galvin’s Conjecture is consistent.

More recently in 2018 Raghavan and Todorčević made a breakthrough by verifying
the conjecture axiomatically: assuming large cardinals, or hypotheses which
have large cardinal strength.

######  Theorem (Raghavan and Todorčević, [19]).

Suppose that there is a Woodin cardinal or a strongly compact cardinal or a
precipitous ideal on $\aleph_{1}$. Then Galvin’s Conjecture holds.

The advantage of the Raghavan-Todorčević result over the result of Shelah is
that by the Lévy-Solovay phenomenon (see [16, pg. 125]) it follows that
assuming there is a proper class of Woodin or strongly compact cardinals, not
only does Galvin’s Conjecture hold, but furthermore in any set forcing
extension of the universe, Galvin’s Conjecture continues to hold. In
particular, a set forcing extension of an arbitrary model of set theory cannot
produce a model where Galvin’s Conjecture fails, which is strong evidence that
Galvin’s conjecture holds outright.999In contrast, for the Erdős-Hajnal
question, Baumgartner observed (see [7, pg. 274]) that it is consistent that
$|\mathbb{R}|$ is anything reasonable and there is
$c:[\mathbb{R}]^{2}\rightarrow\mathbb{N}$ such that on any uncountable
$Y\subseteq\mathbb{R}$, $c``[Y]^{2}=\mathbb{N}$.

Further evidence is given by Eisworth [5] who improved the hypotheses to a
Ramsey cardinal or a weakly precipitous ideal on $\aleph_{1}$ and in
particular showed that Galvin’s Conjecture may hold in a generic extension of
Gödel’s constructible universe as well.

This brings us to our main result, Corollary B, that Galvin’s Conjecture is
true in ZFC. This also answers the $2$-dimensional case of [4, Open Problem
8.2] as well as [5, Questions 1,2,3]. Our inspiration is a technique of Shelah
from [30] which uses an idea from his work on PCF theory (PCF stands for
‘possible cofinalities’, see the monograph [27]). After we discuss the
structure of the paper we will discuss our proof and compare it with those in
[19] and [5].

As for the matter of whether Theorem A and Corollary B can be strengthened, it
is easy to construct using the standard metric on $\mathbb{R}$ a colouring
$c:[\mathbb{R}]^{2}\rightarrow\mathbb{N}$ which takes infinitely-many values
on $[Y]^{2}$ when $Y$ is not discrete.101010There are some details in Lemma
1.3. This cannot be strengthened to realising all the colours on such a $Y$ by
Corollary B; on the other hand Galvin and Shelah [14] constructed a
$c:[\mathbb{R}]^{2}\rightarrow\mathbb{N}$ such that for any
$Y\subseteq\mathbb{R}$ with $|Y|=|\mathbb{R}|$, $c``[Y]^{2}=\mathbb{N}$.

More interesting is the case of considering higher-dimensional colourings such
as colouring triples of reals. Here Raghavan and Todorčević have proved the
following, generalising Baumgartner’s Theorem above which would correspond to
$n=0$ below.

######  Theorem (Raghavan and Todorčević, [20]).

For every positive natural number $n$, and a set of reals $X$ of size
$\aleph_{n}$, there is $c:[X]^{n+2}\rightarrow\mathbb{N}$ such that for any
$Y\subseteq X$ which is homeomorphic to $\mathbb{Q}$,
$c``[Y]^{n+2}=\mathbb{N}$.

In the simplest case, it follows that we cannot improve Corollary B to
colouring triples of reals unless we also assume that
$|\mathbb{R}|\geq\aleph_{2}$, that is, unless we assume the failure of the
continuum hypothesis.

Before moving on to a discussion of the more general Theorem D, we mention a
corollary which improves on [19, Corollary 8] since our result is in ZFC. A
motivation for this result comes from the discussion of the _expansion
problem_ for a topological space (see [19, § 2] for details) and the following
results connects to Ramsey’s original application of the Ramsey Theorem in
[21] and a related result of Erdős-Rado [9] (see the discussion of [19,
Corollary 8]).

######  Corollary C.

Let $<$ denote the usual order on $\mathbb{R}$ and let $<^{*}$ be a well-
ordering of $\mathbb{R}$. Let $X\subseteq\mathbb{R}$ be an uncountable set,
and let $E\subseteq X^{2}$ be a binary relation (not necessarily symmetric).
Then there is $Y\subseteq X$ which is homeomorphic to $\mathbb{Q}$ such that
$E\cap Y^{2}$ is one of the following restricted to $Y$:
$\top,\perp,=,\neq,<,>,\leq,\geq,<^{*},>^{*},\leq^{*},\geq^{*},<\cap<^{*},<\cap>^{*},>\cap<^{*},>\cap>^{*},\leq\cap<^{*},\leq\cap>^{*},\geq\cap<^{*},\geq\cap>^{*}$.

###### Proof.

The proof is similar to that of [35, Theorem 1.7] using Theorem A instead of
Ramsey’s Theorem. ∎

Closely related to the expansion problem and more generally to the topic of
this paper is the area of _Ramsey degrees_ which has recently seen a lot of
progress, and we refer the interested reader to Dobrinen’s ICM survey [3] for
this, connections with topological dynamics etc. Our Corollary B or the
forthcoming Theorem D can both be refomulated using this language. Another
survey where many related results are collected is [4].

We end by mentioning that much of our knowledge of the history of this subject
comes from Todorčević’s survey [36].

###  1.1. A Ramsey theorem for a class of topological spaces

All topological spaces are assumed to be Hausdorff. As we have mentioned
earlier, Theorem A is a consequence of a more general result that we prove. In
this we are emulating Raghavan and Todorčević [19] who also proved Theorem D
below assuming additionally a class of Woodin cardinals or a strongly compact
cardinal111111This is the main theorem of their paper., and then use it to
draw a corollary about Galvin’s Conjecture, assuming the relevant large
cardinal hypotheses. Before we state Theorem D, let us recall the definitions.
The first is standard, the second is a property intermediate between being
first countable and being second countable, and the last is due to Hajnal and
Juhász, [15].

######  Definition 1.1.

Let $(X,\tau)$ be a topological space.

  1. (i)

We say that $(X,\tau)$ is _regular_ if for every closed $C\subseteq X$ and
$x\in X\setminus C$, we can find pairwise disjoint open sets $U,V$ with
$C\subseteq U$ and $x\in V$.

  2. (ii)

A base $\mathcal{B}\subseteq\tau$ is said to be _point-countable_ if for every
$x\in X$, $\\{U\in\mathcal{B}\mid x\in U\\}$ is countable.

  3. (iii)

We say that $(X,\tau)$ is _left-separated_ if there is $<^{*}$ a well-ordering
of $X$ such that for every $x\in X$, $\\{y\in X\mid y<^{*}x\\}$ is a closed
set.

The property of left-separation is a notion of smallness among topological
spaces, for example a countable Hausdorff space is always left-separated.

######  Theorem D.

Let $(X,\tau)$ be a regular topological space which is not left-separated and
has a point-countable basis. Let $K$ be a positive natural number and let
$c:[X]^{2}\rightarrow K$ a colouring of pairs of elements of $X$. Then there
is a $Y\subseteq X$ homeomorphic to $\mathbb{Q}$ such that $|c``[Y]^{2}|\leq
2$.121212The reader unfamiliar with these topological notions will find in
Section 4 the set-theoretic consequences of theirs which we use in the proof.

###### Proof of Theorem A.

In order to see that Theorem A follows from Theorem D, note that the reals
have a countable basis, and that any uncountable set of reals has a countable
subset which is dense in it. So, any uncountable set of reals satisfies the
hypotheses of Theorem D. ∎

The motivation for considering this specific class of topological spaces is
laid out by Raghavan and Todorčević in [19, § 3]. As they argue there, this is
“an essentially optimal class of topological spaces” for which a result such
as Theorem A holds. As for the matter of consistency strength, on [19, pg. 22]
they say “it would not be surprising if [Theorem D] turns out to be
equiconsistent with some large-cardinal axiom.” An example is given from [33]
to justify this statement. Eisworth [5] improved the large cardinal bounds by
obtaining Theorem D from a class of Ramsey cardinals. As our result shows, ZFC
is enough.

Rather than merely repeating [19, § 3] here to justify the class of
topological spaces in Theorem D, we refer the reader to it and focus here on
the class of metrisable spaces where Theorem D provides a provably optimal
result. This too can be found in more detail in [19, § 2,§ 3] so our treatment
will be brief and we refer the reader to [19] for a more comprehensive
treatment.131313According to [19, pg. 9] special stationary Aronszajn lines
with the order topology are another example of spaces to which Theorem D
applies.

It is clear that a metric space is regular Hausdorff; by the Nagata-Bing-
Smirnov Metrisation Theorem (see [18, § 6]), it also has a point-countable
basis. So Theorem D combined with the following lemma due to Fleissner yields
a proof of Corollary E below.

######  Lemma 1.2 (Fleissner, [11]).

Let $X$ be a metric space. The following are equivalent:

  1. (i)

$X$ is left-separated;

  2. (ii)

$X$ is $\sigma$-discrete: there is a decomposition $X=\bigcup_{n<\omega}Y_{n}$
where each $Y_{n}$ is a discrete subspace with the relative topology.

###### Proof.

For (i) implies (ii) see [11, pg. 666] and for (ii) implies (i) see [11,
Theorem 2.2]. ∎

######  Corollary E.

Let $X$ be a metric space which is not $\sigma$-discrete. Then for every
positive natural number $K$ and colouring $c:[X]^{2}\rightarrow K$, there is a
$Y\subseteq X$ homeomorphic to $\mathbb{Q}$ such that $|c``[Y]^{2}|\leq 2$.

The claim that Corollary E is optimal in the class of metric spaces is
justified by the following three lemmas. The first is trivial, showing that we
cannot increase the number of colours.

######  Lemma 1.3.

Let $X$ be a metric space. Then there is $c:[X]^{2}\rightarrow\mathbb{N}$ such
that for every $Y\subseteq X$ which is not discrete, $c``[Y]^{2}$ is infinite.

###### Proof sketch.

Let $d$ denote the metric on $X$. Let $c:[X]^{2}\rightarrow\mathbb{N}$ be
given by: for $x\neq y$ from $X$,

$c(x,y):=\begin{cases}\sup\\{k\in\mathbb{N}\mid\frac{1}{k}\geq
d(x,y)\\}&\text{ if }d(x,y)\leq 1;\\\ 0&\text{ otherwise}.\end{cases}$

Then it is easy to see that $c$ is as required. ∎

The second, showing that an analogue of Sierpiński’s colouring holds for
metric spaces, we learnt from [19, Theorem 12].

######  Lemma 1.4.

Let $X$ be a metric space. Then there is $c:[X]^{2}\rightarrow\\{0,1\\}$ such
that for every $Y\subseteq X$ which is dense-in-itself,
$c``[Y]^{2}=\\{0,1\\}$.

###### Proof sketch.

Every metric space can be embedded into a space of the form $[0,1]^{\kappa}$
for some cardinal $\kappa$. Considering $X$ to be a subset of $[0,1]^{\kappa}$
for some cardinal $\kappa$, recall the _lexicographic order_
$<_{\mathrm{lex}}$ on $[0,1]^{\kappa}$ is defined via: for distinct
$f,g\in[0,1]^{\kappa}$,

$f<_{\mathrm{lex}}g\iff f(\alpha)<g(\alpha)\text{ where
}\alpha:=\min\\{\beta<\kappa\mid f(\beta)\neq g(\beta)\\}.$

Then we can imitate Sierpiński’s colouring by defining a function
$c:[X]^{2}\rightarrow 2$ using $<_{\mathrm{lex}}$ and $<^{*}$ where $<^{*}$ is
some well-ordering of $X$. See [19, Lemma 11] and [19, Theorem 12] for
details. ∎

The last is the following unpublished result of Todorčević and Weiss which
provides a negative result for any $\sigma$-discrete metric space.

######  Lemma 1.5 (Todorčević-Weiss, [39]).

Let $X$ be a $\sigma$-discrete metric space. Then there is
$c:[X]^{2}\rightarrow\mathbb{N}$ such that for every $Y\subseteq X$
homeomorphic to $\mathbb{Q}$, $c``[Y]^{2}=\mathbb{N}$.

###  1.2. Structure of the paper

We now describe the layout of the paper.

In Section 2 we collect in one place general notation and other preliminaries.

In Section 3 we recall standard notation, terminology, and facts about ideals
on countable sets. In Subsection 3.1 we generalise to this context an operator
on ideals introduced by Shelah in his work on PCF theory. We will use this in
the proof of the main lemma.

In Section 4 we start the proof of Theorem D. In particular we fix here
various objects we shall use throughout the proof. We also explain here the
role of the topological hypotheses of Theorem D. Beyond this point the proof
of Theorem D proceeds in three steps.

In Section 5 we recall the idea of Raghavan and Todorčević which provides us
with a sufficient condition to construct the homeomorphic copy of the
rationals as required. The main result here is Theorem 5.5. This is step one.
The main lemma, which is step three, is to show that this sufficient condition
is met.

In Section 6 we perform step two, which is to find the appropriate input for
our main lemma. The main result here is Lemma 6.7.

In Section 7 we prove the main lemma, Lemma 7.1 and conclude the proof of
Theorem D.

###  1.3. Discussion of the proof

As we have mentioned, our main result is the same as that of [19] with the
difference being that we work only in ZFC. We now compare our result with
[19], the reader can look there for undefined notions.

There are two places where large cardinals are required in [19]. One is simply
for the definition of Woodin’s stationary tower forcing (see [41] or [17]),
this needs a class of strongly inaccessible cardinals. The second is for the
precipitousness of the stationary tower, this needs considerably more large
cardinals: a class of Woodin cardinals or a strongly compact cardinal.
Eliminating large cardinals from the first step is somewhat straightforward,
the reader may compare the results of Section 4 with [19, Lemma 26].

The key argument is in the proof of the main lemma, Lemma 7.1, in Section 7.
This replaces the use of games related to precipitousness in [19] and [5] by
an argument using ranks inspired by [30]. 141414The argument can be translated
into the language of precipitousness-type games however we do not do so over
here. In particular, this step achieves the more substantial reduction of
large cardinals to a ZFC argument.

The necessary machinery for ranks is developed in Subsection 3.1. To make this
machinery applicable we need to generalise the basic idea from [19] by working
not just with the fixed ideal of non-stationary sets, but allowing the ideal
to vary. The idea of varying the ideals is present in [5, § 3.1] as well,
though our treatment is rather different. In this greater generality we then
repeat some arguments from [19]. This is done in Section 6.

The last piece is to restrict ourselves to the class of fine normal ideals,
see Lemma 6.5.

It should be clear by now that our paper heavily builds on the ideas of [19].
However, there are enough minor differences that we have taken the decision to
make the paper self-contained.

##  2\. Notation and preliminaries

For a set $X$ we denote its powerset by $\operatorname{\mathcal{P}}(X)$ and
its cardinality by $|X|$ and the set of its ordered pairs by $X^{2}$ and the
set of its two-element subsets by $[X]^{2}$. Note that this will also apply to
natural numbers, so for $K$ a natural number, by $K^{2}$ we will designate the
set of all ordered pairs $(i,j)$ for $i,j<K$ not necessarily distinct. For a
function $f:X\rightarrow Y$ and $Z\subseteq X$, $f``[Z]:=\\{f(x)\mid x\in
Z\\}$.

By $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$ we denote the collection
of countable subsets of $X$,

$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X):=\\{M\subseteq
X\mid|M|<\aleph_{1}\\}.$

Ideals on sets of the form $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$
play a central role in this paper. Notation and some important facts on them
are collected in Section 3.

We will consider colourings $c:[X]^{2}\rightarrow K$ for $X$ some set and $K$
a natural number. We will use the convention that if $x,y\in X$ and $c(x,y)$
has been defined, has some specific value etc., then there is an implicit
assumption that $x\neq y$. On the other hand if we only know that $c(x,y)\neq
i$ for some $i<K$ then this may also be because in fact $x=y$.

We will talk about trees of finite sequences at some points, so let us refresh
the reader’s memory and introduce some relevant terminology we will use as
well. For $Y$ some set, ${}^{<\omega}Y$ denotes the set of finite sequences of
elements of $Y$. For $\sigma,\rho$ finite sequences of elements of some set
$Y$, $\sigma\sqsubset\rho$ shall denote that $\sigma$ is a proper initial
segment of $\rho$. The empty sequence we denote by $\langle\rangle$. If
$\sigma,\rho$ are two elements of ${}^{<\omega}Y$, then
$\sigma{}^{\smallfrown}\rho$ denotes the concatenation of $\sigma$ and $\rho$.

We say that $\mathcal{T}\subseteq{}^{<\omega}Y$ is a _tree_ if it is closed
under taking initial segments. For $Y$ some set, and
$\mathcal{T}\subseteq{}^{<\omega}Y$ a tree, it is _well-founded_ if for every
$b\in{}^{\omega}Y$, for some $k<\omega$, $b\restriction k\notin\mathcal{T}$.
In this case we can define a ordinal-valued rank function. The details will be
provided.

For subtrees of ${}^{<\omega}\omega$, we use $<_{\mathrm{lex}}$ to refer to
the lexicographic order, so that for $\sigma,\rho\in{}^{<\omega}\omega$
distinct and such that $\sigma\not\sqsubset\rho$ and
$\rho\not\sqsubset\sigma$, $\sigma<_{\mathrm{lex}}\rho$ if for the least $i$
such that $\sigma(i)\neq\rho(i))$, we have that $\sigma(i)<\rho(i)$.

For a cardinal $\Omega$, $H(\Omega)$ denotes the collection of all sets of
hereditary cardinality less than $\Omega$. In the context of taking elementary
submodels by $H(\Omega)$ we shall mean the structure
$(H(\Omega),\in,\triangleleft)$ where $\triangleleft$ will denote some fixed
well-ordering of $H(\Omega)$. We will denote that $M$ is an elementary
submodel of this structure by $M\prec H(\Omega)$.

Lastly we discuss some topological preliminaries. All spaces are assumed to be
Hausdorff. The main class of spaces we study has been defined in Definition
1.1.

For $(X,\tau)$ be a topological space and $M\subseteq X$, $\overline{M}$ will
denote the closure of $M$. If $\mathcal{B}\subseteq\tau$ is a base for the
topology, then for $Y\subseteq X$, we let

$\mathcal{B}_{Y}:=\\{U\in\mathcal{B}\mid U\cap Y\neq\emptyset\\}.$

Given a set $X$ and a point $x\notin X$, we will say that $X$ is _dense around
$x$_ if every open neighbourhood of $x$ contains a point of $X$. We will say
that $X$ is _dense-in-itself_ if for every $x\in X$, $X\setminus\\{x\\}$ is
dense around $x$, in other words, $x$ is not an isolated point of $X$.

An old theorem of Sierpiński [32] which we have mentioned before is that any
countable non-empty metric space which is dense-in-itself is homeomorphic to
$\mathbb{Q}$. Recall also the Urysohn Metrisation Theorem which states that
any second countable regular Hausdorff space is metrisable.

Finally we justify some statements from the introduction. Let
$X\subseteq\mathbb{R}$ be uncountable subset of the reals, then there is
$Y\subseteq X$ also uncountable such that $Y$ is dense in itself. Indeed, let
$\mathcal{U}$ denote the set of all rational intervals $U$ such that $U\cap X$
is countable, and then let $Y:=X\setminus(\bigcup\mathcal{U})$. Given such an
$Y$, there is $Z\subseteq Y$ countably infinite such that $Z$ is also dense-
in-itself.

We also mentioned that any $X\subseteq\mathbb{R}$ which is homeomorphic to
$\mathbb{Q}$ contains a set order-isomorphic to $\mathbb{Q}$, so let us
quickly check the details. First we define an equivalence relation $\sim$ on
$X$ via: $x\sim y$ if $(x,y)\cap X=\emptyset$. Then each equivalence class has
size at most $2$. So let $Y\subseteq X$ be such that $Y$ contains exactly one
element of each equivalence class of $\sim$. So $Y$ must be countably infinite
too. Then for any $x<y$ from $Y$, $(x,y)\cap Y\neq\emptyset$. In other words,
$Y$ is dense as a linear order. Now discarding the end points of $Y$, of which
there are at most two, we obtain a set $Z$ which is a countable dense linear
order with no endpoints, so by Cantor’s Theorem $Z$ is order-isomorphic to
$\mathbb{Q}$, and hence every countable linear order can be order-embedded
into $Z$.

##  3\. Ideals of countable sets

Recall that for $H$ a non-empty set,
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ denotes the set
$\\{M\subseteq H\mid|M|<\aleph_{1}\\}$. We start with standard notation,
terminology, and facts about $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$
which can be found, for example, in [16]. We finish with Subsection 3.1 which
generalises some ideas of Shelah [24] to this context.

The following can be found in [16, § 25].

######  Definition 3.1.

Let $H$ be a non-empty set.

  1. (i)

A set $C\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ is _club in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$_ if the following holds:

     * •

for every $N\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ there is an
$M\in C$ with $N\subseteq M$.

     * •

for every $0<\xi<\omega_{1}$ and a $\subseteq$-increasing sequence $\langle
M_{\nu}\mid\nu<\xi\rangle$ of elements of $C$, $\bigcup_{\nu<\xi}M_{\xi}\in
C$.

  2. (ii)

A set $S\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ is
_stationary in $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$_ if for every
$C\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ club in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, $S\cap C\neq\emptyset$.

######  Fact 3.2.

Let $X\subseteq H$ be non-empty sets and let
$S\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$ and
$T\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$.

  1. (i)

If $S$ is stationary in $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$ then
$S^{\uparrow H}$ is stationary in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ where

$S^{\uparrow H}:=\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid
M\cap X\in S\\}.$

  2. (ii)

If $T$ is stationary in $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ then
$T^{\downarrow X}$ is stationary in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$ where

$T^{\downarrow X}:=\\{M\cap X\mid M\in T\\}.$

The following can be found in [16, pg. 301 and § 25].

######  Definition 3.3.

Let $H$ a non-empty set. Let
$\mathcal{I}\subseteq\operatorname{\mathcal{P}}(\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H))$.

  1. (i)

$\mathcal{I}$ is a _ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$_ if

     * •

for every $A,B\in\mathcal{I}$, $A\cup B\in\mathcal{I}$,

     * •

for every $A\in\mathcal{I}$ and $B\subseteq A$, $B\in\mathcal{I}$, and

     * •

$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\notin\mathcal{I}$.

  2. (ii)

An ideal $\mathcal{I}$ is _$\aleph_{1}$ -complete_ if given a sequence
$\langle A_{n}\mid n<\omega\rangle$ of elements of $\mathcal{I}$,
$\bigcup_{n<\omega}A_{n}\in\mathcal{I}$.

  3. (iii)

An $\aleph_{1}$-complete ideal $\mathcal{I}$ is a _fine ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$_ if for every $x\in H$,
$\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid x\not\in
M\\}\in\mathcal{I}$.

  4. (iv)

For a sequence $\langle A_{x}\mid x\in H\rangle$ of elements of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, the _diagonal union_ of
$\langle A_{x}\mid x\in H\rangle$, denoted
$\operatorname{\bigtriangledown}_{\\!x\in H}A_{x}$, is the set
$\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid M\in\bigcup_{x\in
M}A_{x}\\}$.

  5. (v)

For a sequence $\langle A_{x}\mid x\in H\rangle$ of elements of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, the _diagonal intersection_
of $\langle A_{x}\mid x\in H\rangle$, denoted $\bigtriangleup_{x\in H}A_{x}$,
is the set $\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid
M\in\bigcap_{x\in M}A_{x}\\}$.

  6. (vi)

$\mathcal{I}$ is _normal_ if for every sequence $\langle A_{x}\mid x\in
H\rangle$ of elements of $\mathcal{I}$,
$\operatorname{\bigtriangledown}_{\\!x\in H}A_{x}\in\mathcal{I}$.

  7. (vii)

For an ideal $\mathcal{I}$ the dual filter, denoted $\mathcal{I}^{*}$, is the
set

$\mathcal{I}^{*}:=\\{A\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\setminus
A\in\mathcal{I}\\}.$

  8. (viii)

For an ideal $\mathcal{I}$ the collection of all $\mathcal{I}$-positive sets,
denoted $\mathcal{I}^{+}$, is the set

$\mathcal{I}^{+}:=\\{A\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid
A\notin\mathcal{I}\\}.$

So by default we assume all ideals are proper ideals and that fine ideals are
$\aleph_{1}$-complete. An example of a fine normal ideal is the collection of
non-stationary subsets of $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$
for $H$ an uncountable set.

Given $H^{\prime}\subseteq H$ and a sequence $\langle A_{x}\mid x\in
H^{\prime}\rangle$ of elements of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, we will let
$\operatorname{\bigtriangledown}_{\\!x\in H^{\prime}}A_{x}$ denote the set
$\operatorname{\bigtriangledown}_{\\!x\in H}B_{x}$ where for $x\in H^{\prime}$
we take $B_{x}:=A_{x}$ and for $x\in H\setminus H^{\prime}$ we take
$B_{x}:=\emptyset$. We can similarly take diagonal intersections of a sequence
indexed by a subset of $H$ by adding
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ at the missing indices. Note
that given a sequence $\langle A_{x}\mid x\in H\rangle$ if $H=H_{0}\cup
H_{1}$, then

$(\operatorname{\bigtriangledown}_{\\!x\in
H_{0}}A_{x})\cup(\operatorname{\bigtriangledown}_{\\!x\in
H_{1}}A_{x})=\operatorname{\bigtriangledown}_{\\!x\in H}A_{x}.$

For $A\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, we will call
a function $\Phi:A\rightarrow H$ a _choice function_ if for every $M\in A$,
$\Phi(M)\in M$.

The following are some standard facts which serve also to introduce some
notation we shall use throughout the paper.

######  Fact 3.4.

Let $H$ be a non-empty set and $\mathcal{I}$ an ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$.

  1. (i)

If $\mathcal{I}$ is a fine normal ideal, then $\mathcal{I}$ extends the ideal
of non-stationary subsets of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$.

  2. (ii)

$\mathcal{I}$ is normal if and only if $\mathcal{I}^{*}$ is closed under
diagonal intersections.

  3. (iii)

Let $A\in\mathcal{I}^{+}$ and let $\Phi:A\rightarrow H$ be a choice function.
Then there is $B\subseteq A$, $B\in\mathcal{I}^{+}$, and an $x\in H$ such that
$\Phi``[B]=\\{x\\}$.

  4. (iv)

For $A\in\mathcal{I}^{+}$, there is a smallest ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ containing
$\mathcal{I}\cup\\{\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\setminus
A\\}$ which we denote by $\mathcal{I}+A$, and it is described explicitly by
the following: for
$B\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$,

$B\in\mathcal{I}+A\iff B\cap A\in\mathcal{I}.$

  5. (v)

For $A\in\mathcal{I}^{+}$, the dual filter to $\mathcal{I}+A$,
$(\mathcal{I}+A)^{*}$, is described explicitly by the following: for
$B\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$,

$B\in(\mathcal{I}+A)^{*}\iff
B\cup(\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\setminus
A)\in\mathcal{I}^{*}.$

  6. (vi)

If $\mathcal{I}$ is a fine normal ideal and $A\in\mathcal{I}^{+}$, then
$\mathcal{I}+A$ is a fine normal ideal as well.

  7. (vii)

Suppose that $\mathcal{I}$ is a fine ideal. Let $\langle A_{x}\mid x\in
H\rangle$ be a sequence of elements of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ such that
$A\in\mathcal{I}^{+}$, where $A:=\bigtriangleup_{x\in H}A_{x}$. Then, for
every $x\in H$, $A_{x}\in(\mathcal{I}+A)^{*}$.

For $H$ a non-empty set and $\mathcal{I}$ an ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ and $A\in\mathcal{I}^{+}$ we
will use $\exists^{\mathcal{I}}M\in A\,\phi(M)$ to abbreviate that

$\\{M\in A\mid\phi(M)\\}\in(\mathcal{I}+A)^{+},$

and $\forall^{\mathcal{I}}M\in A\,\phi(M)$ to abbreviate that

$\\{M\in A\mid\phi(M)\\}\in(\mathcal{I}+A)^{*}.$

###  3.1. An operator on ideals following Shelah

Our purpose in this subsection is to generalise to ideals on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ some ideas implicit in [24],
one of Shelah’s first papers on PCF theory. The definitions of the ranks below
are variants of the ranks first used by Galvin and Hajnal [13]. This was
continued by Shelah in [24] and more. Definition 3.6 can be compared with [24,
Remark 20(B)] and Lemma 3.7 with [24, Lemma 19]. More explicitly, what follows
is an analogue of the $\mathbf{J}[f,D]$ operator of Shelah, see [28, § 5].

For the rest of this section, fix $H$ a non-empty set. For $\mathcal{I}$ an
ideal on $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ and functions
$f:\operatorname{\mathcal{P}}_{\\!\omega_{1}}(H)\rightarrow\operatorname{Ord}$
and
$g:\operatorname{\mathcal{P}}_{\\!\omega_{1}}(H)\rightarrow\operatorname{Ord}$,
$f<_{\mathcal{I}}g$ denotes that

$\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\mid\neg(f(M)<g(M))\\}\in\mathcal{I}.$

We similarly define the relations $\leq_{\mathcal{I}}$ and $=_{\mathcal{I}}$.

When $\mathcal{I}$ is $\aleph_{1}$-complete we have that the relation
$<_{\mathcal{I}}$ is well-founded on
${}^{\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)}\operatorname{Ord}$.
Hence we can define an ordinal-valued rank function
$\operatorname{rk}_{\mathcal{I}}$ on functions
$f:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
recursively as follows: $\operatorname{rk}_{\mathcal{I}}(f)$ is the least
ordinal $\alpha$ such that for every
$g:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$,

$g<_{\mathcal{I}}f\implies\operatorname{rk}_{\mathcal{I}}(g)<\alpha.$

It should be clear that for $\aleph_{1}$-complete ideals
$\mathcal{I}\subseteq\mathcal{J}$ on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ and
$f:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$,

$\operatorname{rk}_{\mathcal{I}}(f)\leq\operatorname{rk}_{\mathcal{J}}(f).$

In particular, for $A\subseteq B$ both elements of $\mathcal{I}^{+}$ for
$\mathcal{I}$ an $\aleph_{1}$-complete ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, as
$\mathcal{I}+B\subseteq\mathcal{I}+A$,

$\operatorname{rk}_{\mathcal{I}+B}(f)\leq\operatorname{rk}_{\mathcal{I}+A}(f).$

For the rest of this section, fix $\mathcal{I}$ an $\aleph_{1}$-complete ideal
on $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$. If $Z\in\mathcal{I}^{*}$
and
$f,g:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
are such that $f\restriction Z=g\restriction Z$, then we have that
$\operatorname{rk}_{\mathcal{I}}(f)=\operatorname{rk}_{\mathcal{I}}(g)$, and
for this reason we can talk about $\operatorname{rk}_{\mathcal{I}}(h)$ for any
ordinal-valued function $h$ so long as
$\operatorname{dom}(h)\in\mathcal{I}^{*}$.

For completeness we point out the following simple fact.

######  Fact 3.5.

Let
$f:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
and let $\alpha:=\operatorname{rk}_{\mathcal{I}}(f)$.

Then for every $\beta<\alpha$, there is a
$g:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
with $g<_{\mathcal{I}}f$ and $\operatorname{rk}_{\mathcal{I}}(g)=\beta$.

######  Definition 3.6.

Let
$f:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$.
Then

$\mathbf{I}[f,\mathcal{I}]:=\mathcal{I}\cup\\{A\in\mathcal{I}^{+}\mid\operatorname{rk}_{\mathcal{I}+A}(f)>\operatorname{rk}_{\mathcal{I}}(f)\\}.$

######  Lemma 3.7.

Let
$f:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
and let $\mathcal{I}$ be a fine ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$.

  1. (i)

$\mathbf{I}[f,\mathcal{I}]$ is a fine ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$ extending $\mathcal{I}$.

  2. (ii)

If $\mathcal{I}$ is a normal ideal then $\mathbf{I}[f,\mathcal{I}]$ is a
normal ideal.

###### Proof.

It is clear that $\mathbf{I}[f,\mathcal{I}]\supseteq\mathcal{I}$. Since
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\in\mathcal{I}^{*}$,
$\mathcal{I}+\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)=\mathcal{I}$, and
so it is clear that
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\notin\mathbf{I}[f,\mathcal{I}]$.
For $A\subseteq B$ both in $\mathcal{I}^{+}$, we have that
$\operatorname{rk}_{\mathcal{I}+B}(f)\leq\operatorname{rk}_{\mathcal{I}+A}(f)$,
so it follows that $\mathbf{I}[f,\mathcal{I}]$ is closed under taking subsets.
For $A\in\mathcal{I}^{+}$ and $B\in\mathcal{I}$, $\mathcal{I}+(A\cup
B)=\mathcal{I}+A$. This makes it clear that for
$A\in\mathbf{I}[f,\mathcal{I}]$ and $B\in\mathcal{I}$, $A\cup
B\in\mathbf{I}[f,\mathcal{I}]$.

Given a sequence $\langle A_{n}\mid n<\omega\rangle$ of elements of
$\mathbf{I}[f,\mathcal{I}]$, we can find a pairwise disjoint sequence $\langle
B_{n}\mid n<\omega\rangle$ such that for every $n<\omega$, $B_{n}\subseteq
A_{n}$, and $\bigcup_{n<\omega}A_{n}=\bigcup_{n<\omega}B_{n}$. It may be that
some of the $B_{n}$ are elements of $\mathcal{I}$ and some others not, but
combined with the previous observations and the fact that $\mathcal{I}$ is
$\aleph_{1}$-complete the following claim proves that
$\mathbf{I}[f,\mathcal{I}]$ is an ideal and $\aleph_{1}$-complete, hence a
fine ideal.

######  Claim 3.7.1.

Let $\langle B_{n}\mid n<\omega\rangle$ be a sequence of pairwise disjoint
elements of $\mathcal{I}^{+}$, and let $B:=\bigcup_{n<\omega}B_{n}$. Let
$\langle f_{n}\mid n<\omega\rangle$ be a sequence of functions in
${}^{\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)}\operatorname{Ord}$ such
that for every $n<\omega$, $f_{n}\leq_{\mathcal{I}+B_{n}}f$.

Then

  1. (i)

for every ordinal $\zeta$, if
$\zeta\leq\min\\{\operatorname{rk}_{\mathcal{I}+B_{n}}(f_{n})\mid
n<\omega\\}$, then $\zeta\leq\operatorname{rk}_{\mathcal{I}+B}(f)$;

  2. (ii)

$\operatorname{rk}_{\mathcal{I}+B}(f)\geq\min\\{\operatorname{rk}_{\mathcal{I}+B_{n}}(f_{n})\mid
n<\omega\\}$.

###### Proof.

It is clear that (i) implies (ii). We will prove (i) by induction on $\zeta$.

Since the rank is always at least $0$, it is clear for $\zeta=0$, and it is
also clear for $\zeta$ limit. Now, suppose that $\zeta=\eta+1$. Then by Fact
3.5 there are $\langle g_{n}\mid n<\omega\rangle$ such that for every
$n<\omega$, $g_{n}<_{\mathcal{I}+B_{n}}f_{n}$ and
$\operatorname{rk}_{\mathcal{I}+B_{n}}(g_{n})\geq\eta$. Let
$g:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
be a function such that for $n<\omega$, $g\restriction B_{n}=g_{n}\restriction
B_{n}$. Applying the induction hypothesis to $\langle g_{n}\mid
n<\omega\rangle$ and $g$ and $\eta$, we see that
$\operatorname{rk}_{\mathcal{I}+B}(g)\geq\eta$. So, if we show that
$g<_{\mathcal{I}+B}f$ then we will be done. Suppose that this is not so, so
that

$\\{M\in B\mid g(M)\geq f(M)\\}\in\mathcal{I}^{+}.$

As $\mathcal{I}$ is $\aleph_{1}$-complete, it follows that there is some
$n<\omega$ such that

$\\{M\in B_{n}\mid g(M)\geq f(M)\\}\in\mathcal{I}^{+}.$

This contradicts that

$g=_{\mathcal{I}+B_{n}}g_{n}<_{\mathcal{I}+B_{n}}f_{n}\leq_{\mathcal{I}+B_{n}}f,$

and we finish. ∎

We now wish to show that if $\mathcal{I}$ is normal, then so is
$\mathbf{I}[f,\mathcal{I}]$. So, suppose that $\mathcal{I}$ is normal, and let
$\langle A_{x}\mid x\in H\rangle$ be a sequence of elements of
$\mathbf{I}[f,\mathcal{I}]$ and let
$A:=\operatorname{\bigtriangledown}_{\\!x\in H}A_{x}$. We wish to show that
$A\in\mathbf{I}[f,\mathcal{I}]$ too.

Fix $<^{*}$ a well-order of $H$. Now, by recursion along $<^{*}$, for $x\in H$
define a set $B_{x}\subseteq A_{x}$ as follows: for
$M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, $M\in B_{x}$ if and
only if $x$ is the $<^{*}$-least element such that

  * •

$M\in A_{x}$, and

  * •

$x\in M$.

Note that for $M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$,
$M\in\bigcup_{x\in M}A_{x}$ if and only if $M\in\bigcup_{x\in M}B_{x}$. In
other words,

$\operatorname{\bigtriangledown}_{\\!x\in
H}B_{x}=\operatorname{\bigtriangledown}_{\\!x\in H}A_{x}=A,$

and for every $M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, $\\{x\in
M\mid M\in B_{x}\\}$ has cardinality at most $1$.

Furthermore, for every $x\in H$, $B_{x}\subseteq A_{x}$, and as
$\mathbf{I}[f,\mathcal{I}]$ is closed under taking subsets we have that for
every $x\in H$, $B_{x}\in\mathbf{I}[f,\mathcal{I}]$ too. However, some of the
$B_{x}$ may be in $\mathcal{I}$ and some others not. That motivates the
following claim which combined with previous observations proves that
$\mathbf{I}[f,\mathcal{I}]$ is normal.

######  Claim 3.7.2.

Let $H^{\prime}\subseteq H$ and let $\langle B_{x}\mid x\in H^{\prime}\rangle$
be a sequence of elements of $\mathcal{I}^{+}$ such that for every
$M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, $\\{x\in M\cap
H^{\prime}\mid M\in B_{x}\\}$ has cardinality at most $1$. Let
$B:=\operatorname{\bigtriangledown}_{\\!x\in H^{\prime}}B_{x}$. Let $\langle
f_{x}\mid x\in H^{\prime}\rangle$ be a sequence of functions in
${}^{\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)}\operatorname{Ord}$ such
that for every $x\in H^{\prime}$, $f_{x}\leq_{\mathcal{I}+B_{x}}f$.

Then

  1. (i)

for every ordinal $\zeta$, if
$\zeta\leq\min\\{\operatorname{rk}_{\mathcal{I}+B_{x}}(f_{x})\mid x\in
H^{\prime}\\}$, then $\zeta\leq\operatorname{rk}_{\mathcal{I}+B}(f)$;

  2. (ii)

$\operatorname{rk}_{\mathcal{I}+B}(f)\geq\min\\{\operatorname{rk}_{\mathcal{I}+B_{x}}(f_{x})\mid
x\in H^{\prime}\\}$.

###### Proof.

It is clear that (i) implies (ii). We will prove (i) by induction on $\zeta$.

Since the rank is always at least $0$, it is clear for $\zeta=0$, and it is
also clear for $\zeta$ limit. Now, suppose that $\zeta=\eta+1$. Then by Fact
3.5 there are $\langle g_{x}\mid x\in H^{\prime}\rangle$ such that for every
$x\in H^{\prime}$, $g_{x}<_{\mathcal{I}+B_{x}}f_{x}$ and
$\operatorname{rk}_{\mathcal{I}+B_{x}}(g_{x})\geq\eta$.

Let
$g:\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)\rightarrow\operatorname{Ord}$
be a function such that for every
$M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H)$, if there is a
(necessarily unique) $x\in M\cap H^{\prime}$ such that $M\in B_{x}$, then
$g(M)=g_{x}(M)$. We first observe that for $x\in H^{\prime}$,
$g=_{\mathcal{I}+B_{x}}g_{x}$. Indeed, for $x\in H^{\prime}$, as
$B_{x}\in\mathcal{I}^{+}$ and $\mathcal{I}$ is a fine ideal, $\\{M\in
B_{x}\mid x\notin M\\}\in\mathcal{I}+B_{x}$.

Now, applying the induction hypothesis to $\langle g_{x}\mid x\in
H^{\prime}\rangle$ and $g$ and $\eta$, we see that
$\operatorname{rk}_{\mathcal{I}+B}(g)\geq\eta$. So, if we show that
$g<_{\mathcal{I}+B}f$ then we will be done. So, let

$B^{\prime}:=\\{M\in B\mid g(M)\geq f(M)\\},$

and suppose towards a contradiction that $B^{\prime}\in\mathcal{I}^{+}$. So,
let $\Phi:B^{\prime}\rightarrow H$ be the choice function defined via: for
$M\in B^{\prime}$, $\Phi(M)$ is the unique $x\in M\cap H^{\prime}$ such that
$M\in B_{x}$. Then $\Phi$ takes a constant value (which we will denote by $x$,
which is an element of $H^{\prime}$) on a set in $\mathcal{I}^{+}$. So, the
set $B^{\prime}\cap B_{x}\in\mathcal{I}^{+}$, which means that
$g\not<_{\mathcal{I}+B_{x}}f$

However, we also have that

$g=_{\mathcal{I}+B_{x}}g_{x}<_{\mathcal{I}+B_{x}}f_{x}\leq_{\mathcal{I}+B_{x}}f,$

which means that we have reached a contradiction. ∎

With that the proof is completed. ∎

##  4\. Start of the proof

We start the proof of Theorem D in this section and mainly fix a number of
objects which we shall need in our proof. We start by explaining the role of
the topological hypotheses. The objects are different, but the ideas are
imported from Raghavan and Todorčević, [19], to this context.

The role of left-separation is explained by the following result of Fleissner.

######  Theorem 4.1 (Fleissner, [11]).

Let $(X,\tau)$ be a regular space with point-countable basis. Then $(X,\tau)$
is not left-separated if and only if
$\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)\mid\overline{M}\setminus
M\neq\emptyset\\}$ is stationary in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$.151515Note that the result
is true even for merely $T_{1}$ spaces instead of regular Hasudorff, however
our extra hypotheses are for the purpose of finding homeomorphic copies of
$\mathbb{Q}$. See the use of Urysohn’s Metrisation Theorem in Theorem 5.5.

We will also need the following consequence of a point-countable basis of a
topological space.

######  Lemma 4.2.

Let $(X,\tau)$ be a topological space with $\mathcal{B}\subseteq\tau$ a point-
countable basis. Let $\Omega$ be an uncountable regular cardinal such that
$X,\tau,\mathcal{B}\in H(\Omega)$. Let $M\prec H(\Omega)$ be a countable
elementary submodel with $X,\tau,\mathcal{B}\in M$.

If $x\in\overline{M\cap X}$, then $\mathcal{B}_{\\{x\\}}\subseteq M$.

###### Proof.

Let $U\in\mathcal{B}$ be such that $x\in U$. As $x\in\overline{M\cap X}$,
there is a $y\in M\cap X$ such that $y\in U$. Now as $\mathcal{B}_{\\{y\\}}$
is countable, $\mathcal{B}_{\\{y\\}}\subseteq M$, so that $U\in M$. ∎

We now start the proof of Theorem D. First we fix some objects which we will
use throughout the paper.

  * •

Let $(X,\tau)$ be a regular topological space which is not left-separated and
let $\mathcal{B}\subseteq\tau$ be a point-countable basis for it.

  * •

Let $\Omega$ be an uncountable regular cardinal such that
$X,\tau,\mathcal{B}\in H(\Omega)$.

  * •

Fix $\langle U_{x,k}\mid x\in X,k<\omega\rangle$ such that for every $x\in X$,
$\langle U_{x,k}\mid k<\omega\rangle$ enumerates $\mathcal{B}_{\\{x\\}}$.

By Theorem 4.1 there is
$\Lambda_{0}\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(X)$ a
stationary set such that for every $M\in\Lambda_{0}$, $\overline{M}\setminus
M\neq\emptyset$.

The set of countable elementary $M\prec H(\Omega)$ containing
$X,\tau,\mathcal{B}$ is club in
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$. By Fact 3.2(i), the
set

$\\{M\in\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))\mid
X,\tau,\mathcal{B}\in M,\,M\cap X\in\Lambda_{0}\\}$

is stationary in $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$. So
fix
$\Lambda_{1}\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$
a stationary set such that $M\in\Lambda_{1}$ implies that

  * •

$M\prec H(\Omega)$, $|M|=\aleph_{0}$, and $X,\tau,\mathcal{B}\in M$, and

  * •

$M\cap X\in\Lambda_{0}$ so that $\overline{M\cap X}\setminus M\neq\emptyset$.

Fix also a function $F:\Lambda_{1}\rightarrow X$ such that for every
$M\in\Lambda_{1}$, $F(M)\in\overline{M\cap X}\setminus M$.

######  Definition 4.3.

The set $\mathcal{C}_{1}$ consists of pairs $(A,\mathcal{I})$ where

  * •

$\mathcal{I}$ is a fine normal ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$, and

  * •

$A\subseteq\Lambda_{1}$, and $A\in\mathcal{I}^{+}$.

Given $(A,\mathcal{I})$ and $(B,\mathcal{J})$ from $\mathcal{C}_{1}$, we let
$(B,\mathcal{J})\leq_{1}(A,\mathcal{I})$ denote that

  * •

$B\subseteq A$, and

  * •

$\mathcal{J}\supseteq\mathcal{I}$.

It is clear that $\leq_{1}$ partially orders $\mathcal{C}_{1}$. It is also
clear that $\mathcal{C}_{1}$ is non-empty since taking $\mathcal{I}$ to be the
ideal of non-stationary subsets of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$,
$(\Lambda_{1},\mathcal{I})\in\mathcal{C}_{1}$. The requirement that
$A\subseteq\Lambda_{1}$ for every $(A,\mathcal{I})\in\mathcal{C}_{1}$ is
needed in the following lemma, which we will need in the proof of the main
lemma, Lemma 7.1.

######  Lemma 4.4.

Let $(A,\mathcal{I})\in\mathcal{C}_{1}$ and let $k<\omega$. Then there is a
$(B,\mathcal{I})\leq_{1}(A,\mathcal{I})$ and $U\in\mathcal{B}$ such that for
every $M\in B$, $U_{F(M),k}=U$.

###### Proof.

Let $\Phi:A\rightarrow\mathcal{B}$ be the function $M\mapsto U_{F(M),k}$. The
nature of $F$ and Lemma 4.2 ensure that $\Phi$ is a choice function on $A$,
and of course $A\in\mathcal{I}^{+}$ and $\mathcal{I}$ is a normal ideal. So
let $B\subseteq A$ in $\mathcal{I}^{+}$ and $U\in\mathcal{B}$ be such that for
every $M\in B$, $\Phi(M)=U$. ∎

The requirement that $F(M)\notin M$ for $M\in\Lambda_{1}$ is needed for the
following, which in turn will be needed in Lemma 6.2.

######  Lemma 4.5.

Let $(A,\mathcal{I})\in\mathcal{C}_{1}$ and let $x\in X$.

Then $\\{M\in A\mid F(M)=x\\}\in\mathcal{I}$.

###### Proof.

Let $B:=\\{M\in A\mid F(M)=x\\}$ and suppose towards a contradiction that
$B\in\mathcal{I}^{+}$. As $\mathcal{I}$ is a fine ideal, $\\{M\in B\mid x\in
M\\}\in\mathcal{I}^{+}$. But if $M\in B$ then $F(M)=x$ so that
$x\in\overline{M\cap X}\setminus M$ whereas also $x\in M$, a contradiction. ∎

##  5\. The construction of Raghavan and Todorčević

We recall some objects we have fixed in the previous section.

  * •

$(X,\tau)$ is a regular topological space which is not left-separated and has
a point-countable basis $\mathcal{B}$.

  * •

$\Omega$ is an uncountable regular cardinal such that $X,\tau,\mathcal{B}\in
H(\Omega)$.

  * •

$\langle U_{x,k}\mid x\in X,k<\omega\rangle$ is such that for every $x\in X$,
$\langle U_{x,k}\mid k<\omega\rangle$ enumerates $\mathcal{B}_{\\{x\\}}$.

  * •

$\Lambda_{1}\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$
is a fixed stationary subset of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$ such that for every
$M\in\Lambda_{1}$, $M\prec H(\Omega)$ and $X,\tau,\mathcal{B}\in M$, and
$\overline{M\cap X}\setminus M\neq\emptyset$.

  * •

$F:\Lambda_{1}\rightarrow X$ is such that for every $M\in\Lambda_{1}$,
$F(M)\in\overline{M\cap X}\setminus M$.

Additionally, fix $K$ a positive natural number and $c:[X]^{2}\rightarrow K$ a
colouring of pairs from $X$. Also, let $\mathbb{I}$ be the collection of fine
normal ideals on $\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$.

The main result of this section is Theorem 5.5 which gives a sufficient
condition for the existence of a subset of $X$ homeomorphic to $\mathbb{Q}$ on
which $c$ takes at most $2$ colours. Again, the ideas here are all from
Raghavan and Todorčević’s [19] only the objects are different.

Recall that $\mathcal{C}_{1}$ has been defined earlier in Definition 4.3. The
following is a variant.

######  Definition 5.1.

Let $\mathcal{C}_{2}$ denote the set of all triples $(A,B,\mathcal{I})$ where

  * •

$\mathcal{I}\in\mathbb{I}$ and

  * •

$A,B\subseteq\Lambda_{1}$ and $A,B\in\mathcal{I}^{+}$.

We define a relation $\leq_{2}$ on $\mathcal{C}_{2}$ as follows: for
$(A,B,\mathcal{I}),(C,D,\mathcal{J})\in\mathcal{C}_{2}$,

$(C,D,\mathcal{J})\leq_{2}(A,B,\mathcal{I})\text{ if }C\subseteq A\ \&\
D\subseteq B\ \&\ \mathcal{J}\supseteq\mathcal{I}.$

Just as $\leq_{1}$ partially orders $\mathcal{C}_{1}$, $\leq_{2}$ partially
orders $\mathcal{C}_{2}$. And just as $\mathcal{C}_{1}$ is non-empty,
$\mathcal{C}_{2}$ is also non-empty.

We remind the reader of our convention that if $c(x,y)$ is defined, that is,
has a specific value, then we implicitly assume that also $x\neq y$.

######  Definition 5.2.

Let $(A,B,\mathcal{I})\in\mathcal{C}_{2}$ and let $(i,j)\in K^{2}$.

  1. (i)

We say that $(A,B)$ is _weakly $(i,j)$-saturated over $\mathcal{I}$_ if

$\exists^{\mathcal{I}}M\in A\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i]$

and

$\exists^{\mathcal{I}}N\in B\exists^{\mathcal{I}}M\in A\,[c(F(M),F(N))=j].$

  2. (ii)

We say that $(A,B)$ is _$(i,j)$ -saturated over $\mathcal{I}$_ if for every
$C\subseteq A$ and every $D\subseteq B$ with $C,D\in\mathcal{I}^{+}$, $(C,D)$
is weakly $(i,j)$-saturated over $\mathcal{I}$.

The following are the only properties of these notions that we will need in
this section. In Secion 6 we will study them in more detail.

######  Lemma 5.3.

Let $(A,B,\mathcal{I})\in\mathcal{C}_{2}$ and let $(i,j)\in K^{2}$. Suppose
that $(A,B)$ is $(i,j)$-saturated over $\mathcal{I}$.

  1. (i)

Then,

$\forall^{\mathcal{I}}M\in A\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i]$

and

$\forall^{\mathcal{I}}N\in B\exists^{\mathcal{I}}M\in A\,[c(F(M),F(N))=j].$

  2. (ii)

For every $C\subseteq A$ and $D\subseteq B$ with $C,D\in\mathcal{I}^{+}$, also
$(C,D)$ is $(i,j)$-saturated over $\mathcal{I}$.

###### Proof.

  1. (i)

Suppose that the conclusion does not hold. Since the other case is symmetric,
suppose that this is because it is not the case that

$\forall^{\mathcal{I}}M\in A\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i].$

So, let $C:=\\{M\in A\mid\neg(\exists^{\mathcal{I}}N\in
B\,[c(F(M),F(N))=i])\\}$, and then $C\in\mathcal{I}^{+}$. Now we have that
$(C,B)$ is weakly $(i,j)$-saturated over $\mathcal{I}$. However, clearly

$\\{M\in C\mid\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i]\\}=\emptyset,$

and so we reach a contradiction.

  2. (ii)

Easy.

So the lemma is proved. ∎

Compare the following with [19, Lemma 45].

######  Definition 5.4.

Let $(A,\mathcal{I})\in\mathcal{C}_{1}$.

  1. (i)

Let $B\in\mathcal{I}^{+}$ with $B\subseteq A$. Let $x\in X$ and $(i,j)\in
K^{2}$. We say that $x$ is an _$(i,j)$ -winner below $B$ over $\mathcal{I}$_
if there is an $M\in B$ with $F(M)=x$ and there is a sequence $\langle
T_{k}\mid k<\omega\rangle$ such that

     * •

for every $k<\omega$, $T_{k}\subseteq B$ and $T_{k}\in\mathcal{I}^{+}$;

     * •

for every $N\in\bigcup_{k<\omega}T_{k}$, $F(N)\neq x$ and $c(x,F(N))=i$;

     * •

for every $k<l<\omega$, $(T_{l},T_{k})$ is $(i,j)$-saturated over
$\mathcal{I}$;

     * •

for every $k<\omega$, for every $N\in T_{k}$, $F(N)\in U_{x,k}$.

  2. (ii)

For $(i,j)\in K^{2}$, we say that $(A,\mathcal{I})$ is an _$(i,j)$ -winning
pair_ if for every $B\subseteq A$ with $B\in\mathcal{I}^{+}$, there is an
$(i,j)$-winner below $B$ over $\mathcal{I}$.

  3. (iii)

We say that $(A,\mathcal{I})$ is a _winning pair for $c$_ if $(A,\mathcal{I})$
is an $(i,j)$-winning pair for some $(i,j)\in K^{2}$.

######  Theorem 5.5 (Raghavan-Todorčević).

Let $(A,\mathcal{I})\in\mathcal{C}_{1}$ and suppose that $(A,\mathcal{I})$ is
a winning pair for $c$.

Then, there is a $Y\subseteq X$ which is homeomorphic to $\mathbb{Q}$ and such
that $|c``[Y]^{2}|\leq 2$.

###### Proof.

Suppose that $(i,j)\in K^{2}$ is such that $(A,\mathcal{I})$ is an
$(i,j)$-winning pair.

Let $\vec{\sigma}=\langle\sigma_{n}\mid n<\omega\rangle$ be an injective
enumeration of ${}^{<\omega}\omega$ satisfying that if
$\rho_{0}\sqsubset\rho_{1}$ are elements of ${}^{<\omega}\omega$, then
$\rho_{0}$ is enumerated before $\rho_{1}$. We will recursively select an
injective sequence $\langle y_{\sigma_{n}}\mid n<\omega\rangle$ of elements of
$X$ and the set of these elements will constitute the desired set $Y$.

We will ensure that for $\rho\in{}^{<\omega}\omega$, the sequence $\langle
y_{\rho{}^{\smallfrown}\langle k\rangle}\mid k<\omega\rangle$ is dense around
$y_{\rho}$: for $k<\omega$, we will enusure that
$y_{\rho{}^{\smallfrown}\langle k\rangle}\in U_{y_{\rho},k}$. Once this is
achieved, $Y$ would be a countable subset of a space with a point-countable
basis and hence also second countable. As $X$ is regular Hausdorff, it would
follow that $Y$ is metrisable by the Urysohn Metrisation Theorem, and as the
construction would ensure that $Y$ is dense-in-itself, $Y$ would be
homeomorphic to $\mathbb{Q}$ by the theorem of Sierpiński we have mentioned
already.

We will also secure that $c``[Y]^{2}=\\{i,j\\}$ (recall that $i$ and $j$ need
not be distinct), and more informatively, we will have that for $m<n<\omega$,

  * •

if $\sigma_{m}\sqsubset\sigma_{n}$, then $c(y_{\sigma_{m}},y_{\sigma_{n}})=i$,
and

  * •

otherwise, if $\sigma_{m}<_{\mathrm{lex}}\sigma_{n}$ then
$c(y_{\sigma_{m}},y_{\sigma_{n}})=j$, and

  * •

otherwise, if $\sigma_{n}<_{\mathrm{lex}}\sigma_{m}$, then
$c(y_{\sigma_{m}},y_{\sigma_{n}})=i$.

From hereon, to simplify the notation, for $n<\omega$ we will also refer to
$y_{\sigma_{n}}$ as $y_{n}$.

We now give details about the recursive procedure. First we talk about the
inductive assumption, then how to perpetuate it. For $n<\omega$, we will have
at the end of stage $n$ the sequence $\langle y_{m}\mid m\leq n\rangle$, the
portion of $Y$ constructed so far, and a subtree of ${}^{<\omega}\omega$ which
we partition into two types of nodes, $\mathcal{B}_{n}$, the _branching nodes_
, and $\mathcal{L}_{n}$, the _leaves_. The nodes of $\mathcal{B}_{n}$ will be
the finite set of $\rho\in{}^{<\omega}\omega$ for which we have already
decided $y_{\rho}$. The nodes of $\mathcal{L}_{n}$ will be the
$\rho\in{}^{<\omega}\omega$ for which we have made some commitment to what
$y_{\rho}$ is, but not decided it totally. So, we have also a labelling
$h_{n}:\mathcal{L}_{n}\rightarrow\mathcal{I}^{+}$ which will give information
about this commitment. Eventually any $\rho\in\mathcal{L}_{n}$ will be added
to $\mathcal{B}_{n^{\prime}}$ for some $n^{\prime}>n$ (we can be precise: the
least such $n^{\prime}<\omega$ is such that $\rho=\sigma_{n^{\prime}}$) and
then we will ensure that for some $M\in h_{n}(\rho)$, $F(M)=y_{\rho}$.
Naturally, as the recursion progresses our approximations to $y_{\rho}$ will
get better and better.

For details, at the end of stage $n$ the objects $\langle y_{m}\mid m\leq
n\rangle,\mathcal{B}_{n},\mathcal{L}_{n},h_{n}$ will satisfy the following
properties.

  1. (i)

(Injectivity requirement) The sequence $\langle y_{m}\mid m\leq n\rangle$ is
an injective sequence of elements of $X$.

  2. (ii)

(Tree requirement) The set $\mathcal{L}_{n}\cup\mathcal{B}_{n}$ satisfies the
following:

     * •

the set $\mathcal{L}_{n}\cup\mathcal{B}_{n}$ is a subtree of
${}^{<\omega}\omega$, that is, closed under initial segments;

     * •

the branching nodes $\mathcal{B}_{n}$ will be the set $\\{\sigma_{m}\mid m\leq
n\\}$;

     * •

for every $\rho\in\mathcal{B}_{n}$, $\\{\rho{}^{\smallfrown}\langle
k\rangle\mid k<\omega\\}\subseteq\mathcal{L}_{n}\cup\mathcal{B}_{n}$;

     * •

the nodes of $\mathcal{L}_{n}$ are terminal nodes of the tree
$\mathcal{L}_{n}\cup\mathcal{B}_{n}$: for $\rho\in\mathcal{L}_{n}$ and
$\rho^{\prime}\in\mathcal{L}_{n}\cup\mathcal{B}_{n}$,
$\rho\not\sqsubset\rho^{\prime}$.

  3. (iii)

(Convergence requirement) For $k<\omega$ such that
$\rho,\rho{}^{\smallfrown}\langle k\rangle$ are both in $\mathcal{B}_{n}$,
$y_{\rho{}^{\smallfrown}\langle k\rangle}\in U_{\rho,k}$.

  4. (iv)

(Ramsey requirement) The sequence $\langle y_{m}\mid m\leq n\rangle$ satisfies
that for $m<m^{\prime}\leq n$,

     * •

if $\sigma_{m}\sqsubset\sigma_{m^{\prime}}$, then
$c(y_{\sigma_{m}},y_{\sigma_{m^{\prime}}})=i$, and

     * •

otherwise, if $\sigma_{m}<_{\mathrm{lex}}\sigma_{m^{\prime}}$, then
$c(y_{\sigma_{m}},y_{\sigma_{m^{\prime}}})=j$, and

     * •

otherwise, $\sigma_{m^{\prime}}<_{\mathrm{lex}}\sigma_{m}$, and then
$c(y_{\sigma_{m}},y_{\sigma_{m^{\prime}}})=i$.

  5. (v)

The labelling $h_{n}:\mathcal{L}_{n}\rightarrow\mathcal{I}^{+}$ satisfies the
following requirements:

     * •

for every $\rho\in\mathcal{L}_{n}$, $h_{n}(\rho)\subseteq A$ and
$h_{n}(\rho)\in\mathcal{I}^{+}$;

     * •

(injectivity promise) for every $\rho\in\mathcal{L}_{n}$, for every $N\in
h_{n}(\rho)$, $F(N)\notin\\{y_{m}\mid m\leq n\\}=\emptyset$;

     * •

(convergence promise) for every $\rho\in\mathcal{B}_{n}$, and every $k<\omega$
such that $\rho{}^{\smallfrown}\langle k\rangle\in\mathcal{L}_{n}$, and every
$N\in h_{n}(\rho{}^{\smallfrown}\langle k\rangle)$, $F(N)\in U_{y_{\rho},k}$;

     * •

(first Ramsey inductive hypothesis) if $\rho_{0}<_{\mathrm{lex}}\rho_{1}$ are
elements of $\mathcal{L}_{n}$, then $(h_{n}(\rho_{1}),h_{n}(\rho_{0}))$ is
$(i,j)$-saturated over $\mathcal{I}$;

     * •

(second Ramsey inductive hypothesis) if $\rho\in\mathcal{B}_{n}$ and
$k<\omega$ and $\rho{}^{\smallfrown}\langle k\rangle\in\mathcal{L}_{n}$, then
for every $N\in h_{n}(\rho{}^{\smallfrown}\langle k\rangle)$,
$c(F(N),y_{\rho})=i$;

     * •

(third Ramsey inductive hypothesis) if $\rho_{0}\in\mathcal{B}_{n}$ and
$\rho_{1}\in\mathcal{L}_{n}$ and $\rho_{0}\not\sqsubset\rho_{1}$ and
$\rho_{0}<_{\mathrm{lex}}\rho_{1}$, then for every $N\in h_{n}(\rho_{1})$,
$c(F(N),y_{\rho_{0}})=j$;

     * •

(fourth Ramsey inductive hypothesis) if $\rho_{0}\in\mathcal{B}_{n}$ and
$\rho_{1}\in\mathcal{L}_{n}$ and $\rho_{0}\not\sqsubset\rho_{1}$ and
$\rho_{1}<_{\mathrm{lex}}\rho_{0}$, then for every $N\in h_{n}(\rho_{1})$,
$c(F(N),y_{\rho_{0}})=i$.

$\blacktriangleright$ The base step. To select $y_{0}$, let $x$ be an
$(i,j)$-winner below $A$ over $\mathcal{I}$, which is witnessed by a sequence
$\langle T_{k}\mid k<\omega\rangle$ of elements of $\mathcal{I}^{+}$ which are
subsets of $A$. We let $y_{0}:=x$. We also take $\mathcal{L}_{0}:=\\{\langle
k\rangle\mid k<\omega\\}$ and let
$h_{0}:\mathcal{L}_{0}\rightarrow\mathcal{I}^{+}$ be the function $\langle
k\rangle\mapsto T_{k}$, and let $\mathcal{B}_{0}:=\\{\langle\rangle\\}$.

$\blacktriangleright$ The recursive step. Suppose that $\langle y_{m}\mid
m\leq n\rangle,\mathcal{B}_{n},\mathcal{L}_{n},h_{n}$ have been constructed
satisfying the above properties. We describe how to select $\langle y_{m}\mid
m\leq n+1\rangle,\mathcal{B}_{n+1},\mathcal{L}_{n+1},h_{n+1}$. Clearly we must
have that

$\mathcal{B}_{n+1}:=\\{\sigma_{m}\mid m\leq n+1\\}.$

Now note that by the nature of the enumeration $\vec{\sigma}$, for every
$\rho\in{}^{<\omega}\omega$ such that $\rho\sqsubset\sigma_{n+1}$, we have
that $\rho\in\mathcal{B}_{n}$. So it follows that
$\sigma_{n+1}\in\mathcal{L}_{n}$. Then by the nature of $\mathcal{L}_{n}$, for
every $\rho\in\mathcal{L}_{n}$, $\rho\not\sqsubset\sigma_{n+1}$ and
$\sigma_{n+1}\not\sqsubset\rho$. So, taking

$\mathcal{L}_{n+1}:=(\mathcal{L}_{n}\cup\\{\sigma_{n+1}{}^{\smallfrown}\langle
k\rangle\mid k<\omega\\})\setminus\\{\sigma_{n+1}\\},$

we see that the tree requirement at stage $n+1$ is taken care of.

In order to ensure continuity between the stages of the recursion, we will
ensure that $y_{\sigma_{n+1}}=F(M)$ for some $M\in h_{n}(\sigma_{n+1})$ and
also, for $\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$,
$h_{n+1}(\rho)\subseteq h_{n}(\rho)$. We will refer to this as the _continuity
promise_. This combined with the convergence promise at stage $n$ will
immediately ensure that the convergence requirement at stage $n+1$ is met.

Let $B:=h_{n}(\sigma_{n+1})$, and for
$\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$, let
$D_{\rho}:=h_{n}(\rho)$. Suppose that
$\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$ is such that
$\sigma_{n+1}<_{\mathrm{lex}}\rho$. The first Ramsey inductive hypothesis at
stage $n$ ensures that $(D_{\rho},B)$ is $(i,j)$-saturated over $\mathcal{I}$.
By Lemma 5.3(i), we have that

$\forall^{\mathcal{I}}M\in B\exists^{\mathcal{I}}N\in
D_{\rho}\,[c(F(M),F(N))=j].$

Similarly, for $\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$ such that
$\rho<_{\mathrm{lex}}\sigma_{n+1}$, the first Ramsey inductive hypothesis at
stage $n$ ensures that $(B,D_{\rho})$ is $(i,j)$-saturated over $\mathcal{I}$.
By Lemma 5.3(i), we have that

$\forall^{\mathcal{I}}M\in B\exists^{\mathcal{I}}N\in
D_{\rho}\,[c(F(M),F(N))=i].$

So, using the fact that $\mathcal{I}$ is $\aleph_{1}$-complete, let
$C\subseteq B$ be an element of $\mathcal{I}^{+}$ such that for every $M\in
C$,

  * •

if $\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$ is such that
$\sigma_{n+1}<_{\mathrm{lex}}\rho$, then

$\exists^{\mathcal{I}}N\in D_{\rho}\,[c(F(M),F(N))=j],$

  * •

and if $\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$ is such that
$\rho<_{\mathrm{lex}}\sigma_{n+1}$, then

$\exists^{\mathcal{I}}N\in D_{\rho}\,[c(F(M),F(N))=i].$

Now, $C\subseteq B\subseteq A$ and $C\in\mathcal{I}^{+}$, so by the hypotheses
we have that there is an $(i,j)$-winner below $C$ over $\mathcal{I}$. So, fix
$M\in C$ such that $F(M)$ is such an $(i,j)$-winner below $C$ over
$\mathcal{I}$ and then let $y_{\sigma_{n+1}}:=F(M)$, and so by our notation,
also $y_{n+1}:=F(M)$.

It should be clear from the injectivity promise at stage $n$ that the sequence
$\langle y_{m}\mid m\leq n+1\rangle$ is injective, so the injectivity
requirement at stage $n+1$ is taken care of. The second, third, and fourth
Ramsey inductive hypotheses at stage $n$ ensure that the Ramsey requirement is
met at stage $n+1$. So we have taken care of the requirements (i)-(iv).

Since $y_{n+1}$ is an $(i,j)$-winner below $C$ over $\mathcal{I}$, fix a
sequence $\langle T_{k}\mid k<\omega\rangle$ such that

  * •

for every $k<\omega$, $T_{k}\subseteq C$ and $T_{k}\in\mathcal{I}^{+}$;

  * •

for every $N\in\bigcup_{k<\omega}T_{k}$, $F(N)\neq y_{n+1}$ and
$c(y_{n+1},F(N))=i$;

  * •

for every $k<l<\omega$, $(T_{l},T_{k})$ is $(i,j)$-saturated over
$\mathcal{I}$;

  * •

for every $k<\omega$, for every $N\in T_{k}$, $F(N)\in U_{y_{n+1},k}$.

Now to define $h_{n+1}:\mathcal{L}_{n+1}\rightarrow J^{+}$. First,
$h_{n+1}\restriction\\{\sigma_{n+1}{}^{\smallfrown}\langle k\rangle\mid
k<\omega\\}$ is just the function $\sigma_{n+1}{}^{\smallfrown}\langle
k\rangle\mapsto T_{k}$. At this point note that for every $k<\omega$,

$h_{n+1}(\sigma_{n+1}{}^{\smallfrown}\langle k\rangle)=T_{k}\subseteq
C\subseteq B=h_{n}(\sigma_{n+1}).$

This, the continuity promise, and the choice of $\langle T_{k}\mid
k<\omega\rangle$ and the first Ramsey inductive hypothesis up to stage $n$
ensures that the first Ramsey inductive hypothesis continues at stage $n+1$.

It should also be clear that the convergence promise has been taken care of by
the continuity promise and the fact that $y_{n+1}$ is an $(i,j)$-winner below
$C$ over $\mathcal{I}$ as witnessed by $\langle T_{k}\mid k<\omega\rangle$.
The same reason tells us that the second Ramsey inductive hypothesis is taken
care of.

Lastly, to describe
$h_{n+1}\restriction(\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\})$. So, let
$\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$. Recall that
$D_{\rho}=h_{n}(\rho)$. By the choice of $C$, and since $M\in C$ and
$y_{n+1}=F(M)$, we have that

  * •

if $\sigma_{n+1}<_{\mathrm{lex}}\rho$, then for $\widetilde{D}_{\rho}:=\\{N\in
D_{\rho}\mid c(F(N),y_{n+1})=j\\}$, $\widetilde{D}_{\rho}\in\mathcal{I}^{+}$,

  * •

if $\rho<_{\mathrm{lex}}\sigma_{n+1}$, then for $\widetilde{D}_{\rho}:=\\{N\in
D_{\rho}\mid c(F(N),y_{n+1})=i\\}$, $\widetilde{D}_{\rho}\in\mathcal{I}^{+}$.

Finally, for $\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$ we let
$h_{n+1}(\rho):=\widetilde{D}_{\rho}$, and then clearly
$h_{n+1}(\rho)\subseteq h_{n}(\rho)$ and we have met the continuity promise.

Now for the third and fourth Ramsey inductive hypotheses. It should be clear
from the construction that we only need to verify them for
$\sigma_{n+1}\in\mathcal{B}_{n+1}$ and
$\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$. However, the definition
of $\widetilde{D}_{\rho}$ was precisely made to ensure this.

Lastly, we come to the injectivity promise. It should be clear from the
construction that we only need to verify this for
$\sigma_{n+1}\in\mathcal{B}_{n+1}$ and
$\rho\in\mathcal{L}_{n}\setminus\\{\sigma_{n+1}\\}$. But clearly for every
$N\in\widetilde{D}_{\rho}$ we have that $c(F(N),y_{n+1})$ is defined, so
$y_{n+1}\neq F(N)$ for every $N\in\widetilde{D}_{\rho}$.

With that the recursive step is completed, and so is the proof. ∎

##  6\. A decisive pair

The main result of this section is Lemma 6.7 which provides us with the
decisive starting point for the main lemma, Lemma 7.1, in Section 7. The
exhaustion argument in Lemma 6.4 is similar to one in Raghavan and Todorčević
[19] though our partial orders are different as they allow the ideal to vary.
This is necessitated by the presence of two ideals in the statement of Lemma
6.7. Normality of the ideals plays a key role in Lemma 6.5. The notion of
unsaturation (see Definition 6.1) plays a more prominent role here than in
[19] for this reason.

Let us recall our collection of fixed objects:

  * •

$(X,\tau)$ is a regular topological space which is not left-separated and has
a point-countable basis $\mathcal{B}$.

  * •

$K$ is a positive natural number and $c:[X]^{2}\rightarrow K$ is a colouring
of pairs from $X$.

  * •

$\Omega$ is an uncountable regular cardinal such that $X,\tau,\mathcal{B}\in
H(\Omega)$.

  * •

$\langle U_{x,k}\mid x\in X,k<\omega\rangle$ is such that for every $x\in X$,
$\langle U_{x,k}\mid k<\omega\rangle$ enumerates $\mathcal{B}_{\\{x\\}}$.

  * •

$\Lambda_{1}\subseteq\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$
is a fixed stationary subset of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$ such that for every
$M\in\Lambda_{1}$, $M\prec H(\Omega)$ and $X,\tau,\mathcal{B}\in M$, and
$\overline{M\cap X}\setminus M\neq\emptyset$.

  * •

$F:\Lambda_{1}\rightarrow X$ is such that for every $M\in\Lambda_{1}$,
$F(M)\in\overline{M\cap X}\setminus M$.

  * •

$\mathbb{I}$ is the collection of fine normal ideals on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$.

Recall that saturation and weak saturation were defined in Definition 5.2. The
following is related to these.

######  Definition 6.1.

Let $(A,B,\mathcal{I})\in\mathcal{C}_{2}$ and $F\subseteq K^{2}$. We say that
$(A,B)$ is _$F$ -unsaturated over $\mathcal{I}$_ if for every $(i,j)\in F$,
$(A,B)$ is not weakly $(i,j)$-saturated over $\mathcal{I}$.

For the next lemma we again remind the reader of our convention that if
$c(x,y)$ is defined, that is, has a specific value, then we implicitly assume
that also $x\neq y$.

######  Lemma 6.2.

For any $(A,B,\mathcal{I})\in\mathcal{C}_{2}$, for some $(i,j)\in K^{2}$ we
have that $(A,B)$ is weakly $(i,j)$-saturated over $\mathcal{I}$.

###### Proof.

Fix for the moment an $M\in A$, and let $x:=F(M)$. By Lemma 4.5, $\\{N\in
B\mid F(N)=x\\}\in\mathcal{I}$. So $C\in\mathcal{I}^{+}$ where

$C:=\\{N\in B\mid F(N)\neq x\\}.$

So for some $i_{M}<K$ we have that

$\\{N\in C\mid c(x,F(N))=i_{M}\\}\in\mathcal{I}^{+}.$

To summarise, for every $M\in A$, for some $i_{M}<K$ we have that

$\\{N\in B\mid F(M)\neq F(N)\ \&\ c(F(M),F(N))=i_{M}\\}\in\mathcal{I}^{+}.$

So stabilising on $i_{M}$ we obtain an $i<K$ such that

$\\{M\in A\mid i_{M}=i\\}\in\mathcal{I}^{+}.$

So,

$\exists^{\mathcal{I}}M\in A\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i].$

Similarly we can find a $j<K$ such that

$\exists^{\mathcal{I}}N\in B\exists^{\mathcal{I}}M\in A\,[c(F(M),F(N))=j],$

and conclude the proof. ∎

######  Lemma 6.3.

Let $(C,D,\mathcal{J})\leq_{2}(A,B,\mathcal{I})$ be elements of
$\mathcal{C}_{2}$. Let $(i,j)\in K^{2}$.

If $(A,B)$ is $\\{(i,j)\\}$-unsaturated over $\mathcal{I}$, then also $(C,D)$
is $\\{(i,j)\\}$-unsaturated over $\mathcal{J}$.

###### Proof.

As $\mathcal{I}\subseteq\mathcal{J}$, we have that
$\mathcal{I}^{+}\supseteq\mathcal{J}^{+}$. Suppose that $(A,B)$ is not weakly
$(i,j)$-saturated over $\mathcal{I}$. There are two possibilities for why. The
first is that

$\\{M\in A\mid\exists^{\mathcal{I}}N\in B\,[c(F(M),F(N))=i]\\}\in\mathcal{I},$

in which case as $C\subseteq A$ and $D\subseteq B$ and
$\mathcal{J}^{+}\subseteq\mathcal{I}^{+}$, also

$\\{M\in C\mid\exists^{\mathcal{J}}N\in D\,[c(F(M),F(N))=i]\\}\in\mathcal{J},$

so that $(C,D)$ is not weakly $(i,j)$-saturated over $\mathcal{J}$ as well.

The other possibility is that

$\\{N\in B\mid\exists^{\mathcal{I}}M\in A\,[c(F(M),F(N))=j]\\}\in\mathcal{I},$

in which case we similarly have that

$\\{N\in D\mid\exists^{\mathcal{J}}M\in C\,[c(F(M),F(N))=j]\\}\in\mathcal{J},$

so again $(C,D)$ is not weakly $(i,j)$-saturated over $\mathcal{J}$. ∎

######  Lemma 6.4.

There is an $(A,\mathcal{I})\in\mathcal{C}_{1}$ and an $F\subsetneq K^{2}$ and
a pair $(i,j)\in K^{2}\setminus F$ such that the following hold:

  1. (i)

for every $(C,D,\mathcal{J})\leq_{2}(A,A,\mathcal{I})$ from $\mathcal{C}_{2}$,
if $(C,D)$ is $F$-unsaturated over $\mathcal{J}$, then $(C,D)$ is
$(i,j)$-saturated over $\mathcal{J}$;

  2. (ii)

for every $(B,\mathcal{J}_{0})\leq_{1}(A,\mathcal{I})$, there is a
$(C,D,\mathcal{J}_{1})\leq_{2}(B,B,\mathcal{J}_{0})$ such that $(C,D)$ is
$F$-unsaturated over $\mathcal{J}_{1}$.

###### Proof.

Below, we will use the letters $a,b$ for members of $\mathcal{C}_{1}$, and
$p,q$ for members of $\mathcal{C}_{2}$. We define a function
$\Sigma_{0}:\mathcal{C}_{2}\rightarrow\operatorname{\mathcal{P}}(K^{2})$ as
follows: for $(A,B,\mathcal{I})\in\mathcal{C}_{2}$,

$\Sigma_{0}(A,B,\mathcal{I}):=\\{(i,j)\in K^{2}\mid(A,B)\text{ is not weakly
$(i,j)$-saturated over }\mathcal{I}\\}.$

Note that by Lemma 6.2, $K^{2}\notin\operatorname{im}(\Sigma_{0})$. Also, for
$q\leq_{2}p$ elements of $\mathcal{C}_{2}$,
$\Sigma_{0}(q)\supseteq\Sigma_{0}(p)$ by Lemma 6.3.

For $q\in\mathcal{C}_{2}$, we will say that it is _maximal for $\Sigma_{0}$_
if for every $q^{\prime}\leq_{2}q$ from $\mathcal{C}_{2}$, we have that

$\Sigma_{0}(q)=\Sigma_{0}(q^{\prime}).$

######  Claim 6.4.1.

For every $p\in\mathcal{C}_{2}$, there is at least one $q\leq_{2}p$ which is
maximal for $\Sigma_{0}$.

###### Proof.

The map
$\Sigma_{0}:\mathcal{C}_{2}\rightarrow\operatorname{\mathcal{P}}(K^{2})$ has
the property that for $q^{\prime}\leq_{2}q$ from $\mathcal{C}_{2}$, we have
that $\Sigma_{0}(q^{\prime})\supseteq\Sigma_{0}(q)$. Since the set $K^{2}$ is
finite, we cannot get an infinite $\leq_{2}$-decreasing sequence of elements
of $\mathcal{C}_{2}$ whose $\Sigma_{0}$-values are strictly
$\subseteq$-increasing. The claim follows. ∎

Now, define a function
$\Sigma_{1}:\mathcal{C}_{2}\rightarrow\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(K^{2}))$
as follows: for $p\in\mathcal{C}_{2}$ and $F\subseteq K^{2}$,
$F\in\Sigma_{1}(p)$ if there is a $q\in\mathcal{C}_{2}$ such that

  * •

$q\leq_{2}p$,

  * •

$\Sigma_{0}(q)=F$,

  * •

$q$ is maximal for $\Sigma_{0}$.

Note that for $p\in\mathcal{C}_{2}$, $K^{2}\notin\Sigma_{1}(p)$. Now we define
a function
$\Gamma:\mathcal{C}_{1}\rightarrow\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(K^{2}))$
as follows: for $(A,\mathcal{I})\in\mathcal{C}_{1}$,

$\Gamma(A,\mathcal{I}):=\Sigma_{1}(A,A,\mathcal{I}).$

Note that by Claim 6.4.1, for every $a\in\mathcal{C}_{1}$,
$\Gamma(a)\neq\emptyset$, and as we have pointed out before,
$K^{2}\notin\Gamma(a)$.

######  Claim 6.4.2.

Let $b\leq_{1}a$ be elements of $\mathcal{C}_{1}$. Then
$\Gamma(b)\subseteq\Gamma(a)$

###### Proof.

Suppose that $b=(B,\mathcal{J})$ and $a=(A,\mathcal{I})$. Then if
$p\leq_{2}(B,B,\mathcal{J})$ and is maximal for $\Sigma_{0}$, then
$p\leq_{2}(A,A,\mathcal{I})$ as well, and naturally $p$ is still maximal for
$\Sigma_{0}$. ∎

######  Claim 6.4.3.

There is an $a\in\mathcal{C}_{1}$ such that for every $b\leq_{1}a$ in
$\mathcal{C}_{1}$, $\Gamma(b)=\Gamma(a)$.

###### Proof.

The map
$\Gamma:\mathcal{C}_{1}\rightarrow\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(K^{2}))$
has the property that for $b\leq_{1}a$ from $\mathcal{C}_{1}$, we have that
$\Gamma(b)\subseteq\Gamma(a)$. Since the set
$\operatorname{\mathcal{P}}(\operatorname{\mathcal{P}}(K^{2}))$ is finite, we
cannot get an infinite $\leq_{1}$-decreasing sequence of elements of
$\mathcal{C}_{1}$ whose $\Gamma$-values are strictly $\subseteq$-decreasing. ∎

From now on we fix such an $a=(A,\mathcal{I})$ in $\mathcal{C}_{1}$. Then
$\Gamma(a)\subseteq\operatorname{\mathcal{P}}(K^{2})$ and
$\Gamma(a)\neq\emptyset$ and $K^{2}\notin\Gamma(a)$.

Fix now $F\in\Gamma(a)$ such that for every $G\in\Gamma(a)$, if $F\subseteq
G$, then $F=G$. As $F\neq K^{2}$, we lastly fix $(i,j)\in K^{2}\setminus F$.
We will now show that $a$ and $F$ and $(i,j)$ are as required by the lemma.

######  Claim 6.4.4.

Let $(C,D,\mathcal{J})\in\mathcal{C}_{2}$ be such that
$(C,D,\mathcal{J})\leq_{2}(A,A,\mathcal{I})$. Then, if $(C,D)$ is
$F$-unsaturated over $\mathcal{J}$, then $(C,D)$ is $(i,j)$-saturated over
$\mathcal{J}$.

###### Proof.

Indeed, suppose that $(C,D)$ is $F$-unsaturated over $\mathcal{J}$ but not
$(i,j)$-saturated over $\mathcal{J}$. In this case, there is
$(C^{\prime},D^{\prime},\mathcal{J})\leq_{2}(C,D,\mathcal{J})$ such that
$(C^{\prime},D^{\prime})$ is not weakly $(i,j)$-saturated over $\mathcal{J}$,
so that $(C^{\prime},D^{\prime})$ is $(F\cup\\{(i,j)\\})$-unsaturated over
$\mathcal{J}$. Now, let $q\leq_{2}(C^{\prime},D^{\prime},\mathcal{J})$ be
maximal for $\Sigma_{0}$. Then $q\leq_{2}(A,A,\mathcal{I})$ and
$\Sigma_{0}(q)\supseteq F\cup\\{(i,j)\\}$. So let $G:=\Sigma_{0}(q)$. Then
$G\in\Gamma(a)$ and $F\subsetneq G$. This contradicts the choice of $F$. ∎

######  Claim 6.4.5.

Let $(B,\mathcal{J}_{0})\leq_{1}(A,\mathcal{I})$. Then there is
$(C,D,\mathcal{J}_{1})\leq_{2}(B,B,\mathcal{J}_{0})$ in $\mathcal{C}_{2}$ such
that $(C,D)$ is $F$-unsaturated over $\mathcal{J}_{1}$.

###### Proof.

By the choice of $(A,\mathcal{I})$, we have that
$\Gamma(B,\mathcal{J}_{0})=\Gamma(A,\mathcal{I})$, so
$F\in\Gamma(B,\mathcal{J}_{0})$, so $F\in\Sigma_{1}(B,B,\mathcal{J}_{0})$. So
there is $(C,D,\mathcal{J}_{1})\leq_{2}(B,B,\mathcal{J}_{0})$ which is maximal
for $\Sigma_{0}$ and such that $\Sigma_{0}(C,D,\mathcal{J}_{1})=F$. So $(C,D)$
is $F$-unsaturated over $\mathcal{J}_{1}$, as desired. ∎

This completes the proof. ∎

######  Lemma 6.5.

Let $\mathcal{I}\subseteq\mathcal{J}$ be elements of $\mathbb{I}$. Let
$(C,D,\mathcal{J})\in\mathcal{C}_{2}$ and $F\subseteq K^{2}$ be such that
$(C,D)$ is $F$-unsaturated over $\mathcal{J}$.

Then, there is $(S,T,\mathcal{J})\leq_{2}(C,D,\mathcal{J})$ such that $(S,T)$
is $F$-unsaturated over $\mathcal{I}$.

###### Proof.

Note that as $\mathcal{I}\subseteq\mathcal{J}$, we have that
$\mathcal{J}^{+}\subseteq\mathcal{I}^{+}$. We will prove the lemma by
induction on the size of $F$. So, suppose first that $F=\\{(i,j)\\}$. So we
have that $(C,D)$ is not weakly $(i,j)$-saturated over $\mathcal{J}$. Since
the other case is symmetric, let us suppose that this is because

$\neg(\exists^{\mathcal{J}}M\in C\exists^{\mathcal{J}}N\in
D\,[c(F(M),F(N))=i]).$

That is, the following set is in $\mathcal{J}$:

$\\{M\in C\mid\\{N\in D\mid c(F(M),F(N))=i\\}\in\mathcal{J}^{+}\\}.$

So, $S\in(\mathcal{J}+C)^{*}$, where

$S:=\\{M\in C\mid\\{N\in D\mid c(F(M),F(N))=i\\}\in\mathcal{J}\\}.$

Note that

$S=\\{M\in C\mid\\{N\in D\mid c(F(M),F(N))\neq i\\}\in(\mathcal{J}+D)^{*}\\}.$

We remind the reader at this point: $c(F(M),F(N))=i$ implies that $F(M)\neq
F(N)$, but $c(F(M),F(N))\neq i$ may also be for the reason that $F(M)=F(N)$.

Let

$H^{\prime}:=\\{x\in H\mid\exists M\in S\,[F(M)=x]\\}.$

For every $x\in H^{\prime}$, let

$T_{x}:=\\{N\in D\mid c(x,F(N))\neq i\\},$

so for every $x\in H^{\prime}$, we have that $T_{x}\in(\mathcal{J}+D)^{*}$.
Now let $T:=\bigtriangleup_{x\in H^{\prime}}T_{x}$, which is in
$(\mathcal{J}+D)^{*}$ as $\mathcal{J}+D$ is normal. Since
$C\in\mathcal{J}^{+}$ and $S\in(\mathcal{J}+C)^{*}$, we have that
$S\in\mathcal{J}^{+}$ and hence also $S\in\mathcal{I}^{+}$. Similarly,
$T\in\mathcal{J}^{+}$ and hence also $T\in\mathcal{I}^{+}$.

Now for every $M\in S$, $T_{F(M)}=\\{N\in D\mid c(F(M),F(N))\neq i\\}$, and
$T_{F(M)}\in(\mathcal{I}+T)^{*}$ as $T\in(\mathcal{I}+T)^{*}$ and by the
definition of the diagonal intersection, see Fact 3.4(vii).

So, for every $M\in S$, $T\cap T_{F(M)}\in(\mathcal{I}+T)^{*}$, so

$\\{N\in T\mid c(F(M),F(N))\neq i\\}\in(\mathcal{I}+T)^{*}.$

So, for every $M\in S$, $\forall^{\mathcal{I}}N\in T\,[c(F(M),F(N))\neq i]$.
So $(S,T,\mathcal{J})\leq_{2}(C,D,\mathcal{J})$ and $(S,T)$ is not weakly
$(i,j)$-saturated over $\mathcal{I}$. This completes the base step.

Now, for the inductive step. Suppose that $|F|>1$ and $(i,j)\in F$. First
appealing to the inductive hypotheses, we obtain
$(S^{\prime},T^{\prime},\mathcal{J})\leq_{2}(C,D,\mathcal{J})$ such that
$(S^{\prime},T^{\prime})$ is $(F\setminus\\{(i,j)\\})$-unsaturated over
$\mathcal{I}$. By Lemma 6.3 we have that $(S^{\prime},T^{\prime})$ is not
weakly $(i,j)$-saturated over $\mathcal{J}$. Then we can appeal to the base
case applied to $(S^{\prime},T^{\prime},\mathcal{J})$ to obtain
$(S,T,\mathcal{J})\leq_{2}(S^{\prime},T^{\prime},\mathcal{J})$ and $(S,T)$ is
not weakly $(i,j)$-saturated over $\mathcal{I}$, and hence $(S,T)$ is
$F$-unsaturated over $\mathcal{I}$ by Lemma 6.3. ∎

######  Remark 6.6.

The above lemma shows that for $\mathcal{I}\subseteq\mathcal{J}$ ideals in
$\mathbb{I}$ and $(A,B,\mathcal{J})\in\mathcal{C}_{2}$ and $(i,j)\in K^{2}$,
if $(A,B)$ is $(i,j)$-saturated over $\mathcal{I}$ then also $(A,B)$ is
$(i,j)$-saturated over $\mathcal{J}$.

######  Lemma 6.7.

There is a pair $(A,\mathcal{I})\in\mathcal{C}_{1}$ and $(i,j)\in K^{2}$ such
that

  * •

for every $\mathcal{J}\in\mathbb{I}$ with $\mathcal{I}\subseteq\mathcal{J}$,

  * •

for every $B\subseteq A$ with $B\in\mathcal{J}^{+}$,

there are $S,T\subseteq B$ with $S,T\in\mathcal{J}^{+}$ such that $(S,T)$ is
$(i,j)$-saturated over $\mathcal{I}$.

###### Proof.

Let $A$, $\mathcal{I}$, $F$, and $(i,j)$ be as in Lemma 6.4. Then for every
$(B,\mathcal{J})\leq_{1}(A,\mathcal{I})$, for some
$(C,D,\mathcal{J}_{1})\leq_{2}(B,B,\mathcal{J})$, we have that $(C,D)$ is
$F$-unsaturated over $\mathcal{J}_{1}$. By Lemma 6.5, we can find
$(S,T,\mathcal{J}_{1})\leq_{2}(C,D,\mathcal{J}_{1})$ such that $(S,T)$ is
$F$-unsaturated over $\mathcal{I}$, and hence, also $(S,T)$ is
$(i,j)$-saturated over $\mathcal{I}$. Since $S,T\in(\mathcal{J}_{1})^{+}$ and
$\mathcal{J}\subseteq\mathcal{J}_{1}$, also $S,T\in\mathcal{J}^{+}$ and so the
proof is completed. ∎

##  7\. The main lemma

We prove now our main lemma, which as we have mentioned earlier uses an idea
we learned from Shelah’s [30].

Let us recall again the objects we have fixed:

  * •

$(X,\tau)$ is a regular topological space which is not left-separated and has
a point-countable basis $\mathcal{B}$.

  * •

$K$ is a positive natural number and $c:[X]^{2}\rightarrow K$ is a colouring
of pairs from $X$.

  * •

$\Omega$ is an uncountable regular cardinal such that $X,\tau,\mathcal{B}\in
H(\Omega)$.

  * •

$\langle U_{x,k}\mid x\in X,k<\omega\rangle$ is such that for every $x\in X$,
$\langle U_{x,k}\mid k<\omega\rangle$ enumerates $\mathcal{B}_{\\{x\\}}$.

  * •

$\Lambda_{1}$ is a fixed stationary subset of
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$ such that for every
$M\in\Lambda_{1}$, $M\prec H(\Omega)$ and $X,\tau,\mathcal{B}\in M$, and
$\overline{M\cap X}\setminus M\neq\emptyset$.

  * •

$F:\Lambda_{1}\rightarrow X$ is such that for every $M\in\Lambda_{1}$,
$F(M)\in\overline{M\cap X}\setminus M$.

  * •

$\mathbb{I}$ is the collection of fine normal ideals on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$.

######  Lemma 7.1.

There is an $\mathcal{I}\in\mathbb{I}$ and an $A\in\mathcal{I}^{+}$ such that
$(A,\mathcal{I})$ is a winning pair for $c$.

###### Proof.

Using Lemma 6.7, let $(A,\mathcal{I})\in\mathcal{C}_{1}$ and $(i,j)\in K^{2}$
be such that

  * •

for every $\mathcal{J}\in\mathbb{I}$ with $\mathcal{I}\subseteq\mathcal{J}$,

  * •

for every $B\subseteq A$ with $B\in\mathcal{J}^{+}$,

there are $S,T\subseteq B$ with $S,T\in\mathcal{J}^{+}$ such that $(S,T)$ is
$(i,j)$-saturated over $\mathcal{I}$.

We will prove that $(A,\mathcal{I})$ is an $(i,j)$-winning pair. Let us recall
from Definition 5.4 what this requires. We need to show that for every
$B\in\mathcal{I}^{+}$ with $B\subseteq A$, there is an $(i,j)$-winner $x$
below $B$ over $\mathcal{I}$. This in turn requires us to show that there is
an $M\in B$ with $F(M)=x$ and there is a sequence $\langle T_{k}\mid
k<\omega\rangle$ such that

  * •

for every $k<\omega$, $T_{k}\subseteq B$ and $T_{k}\in\mathcal{I}^{+}$;

  * •

for every $N\in\bigcup_{k<\omega}T_{k}$, $F(N)\neq x$ and $c(x,F(N))=i$;

  * •

for every $k<l<\omega$, $(T_{l},T_{k})$ is $(i,j)$-saturated over
$\mathcal{I}$;

  * •

for every $k<\omega$, for every $N\in T_{k}$, $F(N)\in U_{x,k}$.

So, fix $B\in\mathcal{I}^{+}$ with $B\subseteq A$. We want to show that there
is an $(i,j)$-winner below $B$ over $\mathcal{I}$, so suppose towards a
contradiction that there is no such element.

Let $\mathcal{T}$ be the set of all finite sequences $\sigma$ such that for
some $l<\omega$,

$\sigma=\langle(S_{k},T_{k})\mid k<l\rangle,$

and such that

  1. (i)

for all $k<l$, $S_{k},T_{k}\in\mathcal{I}^{+}$ with $S_{k}\cup T_{k}\subseteq
B$;

  2. (ii)

for all $k<l$, $(S_{k},T_{k})$ is $(i,j)$-saturated over $\mathcal{I}$;

  3. (iii)

for all $0<k<l$, $S_{k}\cup T_{k}\subseteq S_{k-1}$;

  4. (iv)

for all $k<l$, there is a $U\in\mathcal{B}$ such that for every $M\in
S_{k}\cup T_{k}$, $U=U_{F(M),k}$;

  5. (v)

for all $k<l$ and every $M\in S_{k}$, $\\{N\in T_{k}\mid
c(F(M),F(N))=i\\}\in\mathcal{I}^{+}$.

Note that $\mathcal{T}$ is closed under initial segments, so it is a tree of
finite sequences.

For every $M\in B$, let $\mathcal{T}^{M}$ consist of those
$\sigma\in\mathcal{T}$ such that if $\sigma=\langle(S_{k},T_{k})\mid
k<l\rangle$ for some $l<\omega$, then for every $k<l$, $M\in S_{k}$. Note that
$\mathcal{T}^{M}$ is also closed under initial segments, so it is a subtree of
$\mathcal{T}$. In particular for every $M\in B$, the empty sequence
$\langle\rangle$ is in $\mathcal{T}^{M}$.

######  Claim 7.1.1.

For every $M\in B$, $\mathcal{T}^{M}$ is well-founded.

###### Proof.

Suppose that this is not so, so let $M\in B$ be such that we can find an
infinite sequence

$b=\langle(S_{k},T_{k})\mid k<\omega\rangle$

such that all of the initial segments of $b$ are in $\mathcal{T}^{M}$. It
follows that $M\in\bigcap_{k<\omega}S_{k}$. Let $x:=F(M)$. Now for every
$k<\omega$, let

$\widetilde{T}_{k}:=\\{N\in T_{k}\mid c(x,F(N))=i\\}.$

Then for every $k<\omega$ we have that $\widetilde{T}_{k}\in\mathcal{I}^{+}$
and $\widetilde{T}_{k}\subseteq T_{k}\subseteq B$, and also, for every
$N\in\widetilde{T}_{k}$, $x\neq F(N)$ and $c(x,F(N))=i$.

Also, for $k<l<\omega$, we have that

$\widetilde{T}_{l}\subseteq T_{l}\subseteq S_{l-1}\subseteq S_{k},$

and then as $(S_{k},T_{k})$ is $(i,j)$-saturated over $\mathcal{I}$, also
$(\widetilde{T}_{l},\widetilde{T}_{k})$ is $(i,j)$-saturated over
$\mathcal{I}$.

Lastly, for $k<\omega$, as $\\{M\\}\cup\widetilde{T}_{k}\subseteq S_{k}\cup
T_{k}$, for every $N\in\widetilde{T}_{k}$ we have that
$U_{F(N),k}=U_{F(M),k}=U_{x,k}$, and hence also, $F(N)\in U_{x,k}$.

Altogether, we have that $x$ is an $(i,j)$-winner below $B$ over $\mathcal{I}$
as witnessed by $M$ and the sequence $\langle\widetilde{T}_{k}\mid
k<\omega\rangle$. This is a contradiction.

∎

For every $M\in B$ the tree $\mathcal{T}^{M}$ is well-founded, so we define a
rank $\operatorname{rk}^{M}:\mathcal{T}^{M}\rightarrow\operatorname{Ord}$ via:
for $\sigma\in\mathcal{T}^{M}$,

$\operatorname{rk}^{M}(\sigma):=\sup\\{\operatorname{rk}^{M}(\sigma^{\prime})+1\mid\sigma^{\prime}\in\mathcal{T}^{M},\,\sigma\sqsubset\sigma^{\prime}\\}.$

We now get a chance to use a beautiful idea of Shelah [30]. Let $\mathbf{P}$
be the set of all _creatures_ $\mathfrak{p}$ where a creature is a triple
$\mathfrak{p}=(\sigma,Z,f)$ such that

  * •

$\sigma\in\mathcal{T}$;

  * •

$Z\in\mathcal{I}^{+}$ with $Z\subseteq B$, and more specifically, if
$|\sigma|=0$ then $Z:=B$, and if $\sigma=\langle(S_{k},T_{k})\mid k<l\rangle$
for some $0<l<\omega$, then $Z:=S_{l-1}$;

  * •

$f\in{}^{Z}\operatorname{Ord}$ is the function
$M\mapsto\operatorname{rk}^{M}(\sigma)$, which is legitimate as
$\sigma\in\mathcal{T}^{M}$ for every $M\in Z$.

Note that $\mathbf{P}$ is non-empty, since $(\langle\rangle,B,f)\in\mathbf{P}$
where $f\in{}^{B}\operatorname{Ord}$ is the function
$M\mapsto\operatorname{rk}^{M}(\langle\rangle)$.

Now, given a creature $\mathfrak{p}=(\sigma,Z,f)$ in $\mathbf{P}$, as
$\mathcal{I}+Z$ is an $\aleph_{1}$-complete ideal and
$Z\in(\mathcal{I}+Z)^{*}$ and $f\in{}^{Z}\operatorname{Ord}$, the rank
$\operatorname{rk}_{\mathcal{I}+Z}(f)$ is well-defined, and we will call this
the _hybrid rank_ of $\mathfrak{p}$ to avoid confusion among all the
ranks.161616The terminology is due to Assaf Rinot from [22].

So, among all the creatures in $\mathbf{P}$, let $\mathfrak{p}=(\sigma,Z,f)$
be chosen such that its hybrid rank is minimal. Let $k:=|\sigma|$.

Now consider the ideal $\mathcal{J}:=\mathbf{I}[f,\mathcal{I}+Z]$. By Lemma
3.7,

$\mathcal{I}\subseteq\mathcal{I}+Z\subseteq\mathcal{J},$

and $\mathcal{J}$ is a fine normal ideal on
$\operatorname{{\mathcal{P}_{\\!\omega_{1}}}}(H(\Omega))$ so
$\mathcal{J}\in\mathbb{I}$. As $Z\in\mathcal{J}^{*}$, or more relevant, as
$(Z,\mathcal{J})\in\mathcal{C}_{1}$, by Lemma 4.4 there is $U\in\mathcal{B}$
and $(Y,\mathcal{J})\leq_{1}(Z,\mathcal{J})$ such that for every $M\in Y$,
$U_{F(M),k}=U$. Now, $(Y,\mathcal{J})\leq_{1}(A,\mathcal{I})$, and by the
choice of $(A,\mathcal{I})$ we can find $S^{\prime},T\subseteq Y$ such that
$S^{\prime},T\in\mathcal{J}^{+}$ and $(S^{\prime},T)$ is $(i,j)$-saturated
over $\mathcal{I}$. We have by Lemma 5.3 that

$\forall^{\mathcal{I}}M\in S^{\prime}\exists^{\mathcal{I}}N\in
T[c(F(M),F(N))=i].$

So, let $S\subseteq S^{\prime}$ be such that $S^{\prime}\setminus
S\in\mathcal{I}$ and

$\forall M\in S\exists^{\mathcal{I}}N\in T[c(F(M),F(N))=i].$

Since $\mathcal{I}\subseteq\mathcal{J}$ and $S^{\prime}\in\mathcal{J}^{+}$,
note that $S\in\mathcal{J}^{+}$ too, and of course $(S,T)$ is also
$(i,j)$-saturated over $\mathcal{I}$.

Let $\widetilde{\sigma}:=\sigma{}^{\smallfrown}\langle(S,T)\rangle$, and then
we have that $\widetilde{\sigma}\in\mathcal{T}$. Now, let
$\widetilde{\mathfrak{p}}:=(\widetilde{\sigma},\widetilde{Z},\widetilde{f})$,
where

  * •

$\widetilde{Z}:=S$, and

  * •

$\widetilde{f}\in{}^{S}\operatorname{Ord}$ is the function
$M\mapsto\operatorname{rk}^{M}(\widetilde{\sigma})$.

Clearly $\widetilde{\mathfrak{p}}\in\mathbf{P}$. Now, let us compare
$\operatorname{rk}_{\mathcal{I}+Z}(f)$ and
$\operatorname{rk}_{\mathcal{I}+\widetilde{Z}}(\widetilde{f})$. As
$S\in\mathcal{J}^{+}=(\mathbf{I}[f,\mathcal{I}+Z])^{+}$ and $S\subseteq Z$, we
have that

$\operatorname{rk}_{\mathcal{I}+Z}(f)=\operatorname{rk}_{(\mathcal{I}+Z)+S}(f)=\operatorname{rk}_{\mathcal{I}+S}(f)>\operatorname{rk}_{\mathcal{I}+S}(\widetilde{f})=\operatorname{rk}_{\mathcal{I}+\widetilde{Z}}(\widetilde{f}).$

To see why the inequality is true, note that for every $M\in S$, as
$\sigma,\widetilde{\sigma}\in\mathcal{T}^{M}$ and
$\sigma\sqsubset\widetilde{\sigma}$,

$f(M)=\operatorname{rk}^{M}(\sigma)>\operatorname{rk}^{M}(\widetilde{\sigma})=\widetilde{f}(M),$

so $\widetilde{f}<_{\mathcal{I}+S}f$. So we have that the hybrid rank of
$\widetilde{\mathfrak{p}}$ is strictly less than the hybrid rank of
$\mathfrak{p}$. This contradicts the choice of $\mathfrak{p}$ and we finish. ∎

We are now in a position to prove Theorem D.

######  Corollary 7.2.

Let $(X,\tau)$ be a regular topological space which is not left-separated and
has a point-countable basis. Let $K$ be a positive natural number and
$c:[X]^{2}\rightarrow K$ a colouring of pairs of elements of $X$. Then there
is a $Y\subseteq X$ homeomorphic to $\mathbb{Q}$ such that $|c``[Y]^{2}|\leq
2$.

###### Proof.

By Lemma 7.1, there is an $\mathcal{I}\in\mathbb{I}$ and an
$A\in\mathcal{I}^{+}$ such that $(A,\mathcal{I})$ is a winning pair for $c$.
By Theorem 5.5 this gives the existence of a $Y\subseteq X$ as required. ∎

## Acknowledgements

This work was supported by the Israel Science Foundation (grant agreement
665/20). The result was presented in the Set Theory Seminar of Bar-Ilan
University on 28th March 2024. I would like to thank the participants for this
opportunity and their patience. I would like to thank Assaf Rinot for many
suggestions which have improved the article. I would like to thank Todd
Eisworth for his interest in the ‘very weak’ precipitous game: this will
appear in the final version.

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