# Tiltan and graphs with no infinite paths

Shimon Garti  Institute of Mathematics, The Hebrew University of Jerusalem,
Jerusalem 91904, Israel<EMAIL_ADDRESS>
###### Abstract.

We prove the consistency of tiltan with the positive relation
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$.

###### Key words and phrases:

Tiltan, infinite path, independent sets, generalized Martin’s axiom

###### 2010 Mathematics Subject Classification:

03E02, 03E35, 03E50, 03E75, 05C63

##  0\. Introduction

Let $G=(V,E)$ be a graph. An independent subset of $V$ is a set of vertices
$W\subseteq V$ such that $[W]^{2}\cap E=\varnothing$. An infinite path in $G$
is a sequence of vertices $\langle v_{n}:n\in\omega\rangle$ with no
repetitions such that $\\{v_{n},v_{n+1}\\}\in E$ for every $n\in\omega$.
Intuitively, these concepts are orthogonal. If one wishes to eliminate large
independent sets then one must add edges to many pairs. In such cases it
becomes harder to avoid infinite paths. For making this intuition precise we
need a definition of _large_ independent sets. The most natural suggestion
would be a subset $W$ of $V$ with the same order type.

######  Definition 0.1.

The relation $\tau\rightarrow(\tau,\text{infinite path})^{2}$ means that for
every graph $G=(V,E)$ with ${\rm otp}(V)=\tau$ there exists either an
independent set $W\subseteq V$ so that ${\rm otp}(W)=\tau$ or an infinite
path.

By order type we do not confine ourselves to well-orderings. Rather, we refer
to a variety of structures. We consider ordinals $\alpha$ with their well-
ordering, the backward ordering $\alpha^{*}$ and ordinal products of these
types. All graphs in this paper are undirected.

The notation $\tau\rightarrow(\tau,\text{infinite path})^{2}$ comes from
partition theorems of infinite combinatorics. Given a graph $G$ one may think
of a coloring of its pairs with two colors. The first color is assigned to
every pair of vertices with no edge, and the second color is given to pairs
with an edge. The positive relation states that there exists a full sized
subset with the first color or an infinite sequence with the second color.

We shall focus on the order type $\omega^{*}\cdot\omega_{1}$. For a convenient
and concrete example, if the ambient set is $\omega_{1}\times\omega$ then the
order defined by $(\alpha,m)<^{*}(\beta,n)$ iff
$(\alpha<\beta)\vee(\alpha=\beta$ and $m>n)$ is of type
$\omega^{*}\cdot\omega_{1}$. A convenient way to visualize this type is by
thinking about $\omega_{1}$ many columns, each of which is a copy of
$\omega^{*}$. Regarding the above definition one may wonder whether
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$.

In the parallel abstract situation of infinite combinatorics,
$\lambda\rightarrow(\lambda,\omega)^{2}$ for every infinite cardinal
$\lambda$, and this is known as the Erdős-Dushnik-Miller theorem. However, if
$\alpha$ is an ordinal but not a cardinal then
$\alpha\nrightarrow(\alpha,\omega)^{2}$. These facts motivate the
investigation of more types like $\omega^{*}\cdot\omega_{1}$. We indicate that
an infinite path in a graph is a weaker notion than an infinite monochromatic
set, since the homogeneity is required only at consecutive elements of the
path. There is some evidence that the existence of monochromatic paths is
strictly weaker than the existence of monochromatic sets, see [GMS20] and
[Tod21]. In our context, one may obtain such paths even though the order-type
of the graph is neither a cardinal, nor an ordinal. Put another way, a
mysterious path may show up, as described in [Jan17, page 10]: Moomintroll was
just putting up a swing when Sniff got home. He seemed very interested in the
mysterious path, and directly after lunch they set off to have a look at it.

Back to the context of graph theory, the above relation cannot be decided by
the axioms of set theory. Namely, one can prove the consistency of
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$ in some extension of ZFC on the one hand, and one can show that
$\omega^{*}\cdot\omega_{1}\nrightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$ in another extension on the other hand. The negative direction was
given by Baumgartner and Larson, [BL90], and the positive direction by Larson
in [Lar90]. It is done in these papers through the classical way of
confronting the constructible universe with the universe under Martin’s axiom
with large continuum.

Actually, the full strength of the constructible universe is not required.
Baumgartner and Larson constructed a graph $G=(V,E)$ of type
$\omega^{*}\cdot\omega_{1}$ with no independent subset of this type and no
infinite path merely from the diamond principle at $\aleph_{1}$. Recall that
$\Diamond_{\aleph_{1}}$ says that there exists a sequence of sets $\langle
A_{\alpha}:\alpha\in\omega_{1}\rangle$ such that $A_{\alpha}\subseteq\alpha$
for every $\alpha\in\omega_{1}$ and for every $A\subseteq\omega_{1}$ the set
$S_{A}=\\{\alpha\in\omega_{1}:A\cap\alpha=A_{\alpha}\\}$ is a stationary
subset of $\omega_{1}$. The opposite direction employs Martin’s axiom with
$2^{\omega}>\omega_{1}$, and then
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$. Both directions are elaborated in another paper of Larson,
[Lar06]. In this paper she explains the importance of the type
$\omega^{*}\cdot\omega_{1}$ and poses the problem which stands in the hub of
this paper. We let the following definition into the discussion.

######  Definition 0.2.

Tiltan.
Let $\kappa={\rm cf}(\kappa)>\aleph_{0}$.
The tiltan principle $\clubsuit_{\kappa}$ says that there exists a sequence
$\langle T_{\alpha}:\alpha$ is a limit ordinal of $\kappa\rangle$ such that
each $T_{\alpha}$ is a cofinal subset of $\alpha$ and for every
$A\in[\kappa]^{\kappa}$ the set $S_{A}=\\{\alpha\in\kappa:T_{\alpha}\subseteq
A\\}$ is a stationary subset of $\kappa$.

The common name of this statement is the club principle. It has been
introduced by Ostaszewski, in [Ost76]. We shall call it tiltan111Let us
indicate that in some good old manuscripts the pronunciation is _taltan_ , see
the relevant discussion in [Yei85]. since the word _club_ is extensively used
as an acronym for closed and unbounded sets.

The tiltan follows from the diamond, and it is strictly weaker than the
diamond. In particular, $\Diamond_{\aleph_{1}}\Rightarrow
2^{\omega}=\omega_{1}$ while $\clubsuit_{\aleph_{1}}$ is consistent with
$2^{\omega}>\omega_{1}$. Remark that Martin’s axiom with
$2^{\omega}>\omega_{1}$ implies $\neg\clubsuit_{\aleph_{1}}$. Therefore, the
following question of Larson from [Lar06] is natural:

######  Question 0.3.

Is it consistent that tiltan holds at $\aleph_{1}$ and concomitantly
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$?

We shall give a positive answer to this question. Let us indicate that having
a positive result is a bit surprising. One of the main differences between
tiltan and diamond is that the diamond prediction is based on equality
($A\cap\alpha=A_{\alpha}$) while the tiltan prediction gives only inclusion
($T_{\alpha}\subseteq A$). In the negative arrow relation proved under diamond
in [BL90], only inclusion is needed for the construction of a graph
exemplifying
$\omega^{*}\cdot\omega_{1}\nrightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$. Despite this fact, tiltan is consistent with the positive
relation
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$, as we shall see.

The rest of the paper is arranged in two additional sections. In the first one
we unfold some background material and we try to explicate the idea behind the
proof. In the second, we prove the main theorem. Our notation is mostly
standard. Let us mention the notation $S^{\lambda}_{\kappa}$ which refers to
the set $\\{\delta<\lambda:{\rm cf}(\delta)=\kappa\\}$ where $\kappa$ is a
regular cardinal. We employ the Jerusalem forcing notation, so $p\leq q$ reads
$p$ is weaker than $q$. Consequently we shall speak about a least upper bound
of conditions, a downward closed generic set, and so forth. If $p$ is
compatible with $q$ then we write $p\parallel q$. If $p$ and $q$ are
incompatible then we shall write $p\perp q$. The meaning of the symbol
$\exists^{\infty}$ is that there are infinitely many elements which satisfy
the statement which falls under the scope of this quantifier. We employ this
notation with respect to sets of natural numbers.

I am deeply indebted to the anonymous referee for many mathematical
corrections and a lot of helpful suggestions. The referee pointed out a major
flaw in the original version of the manuscript and enabled me to fix the
problematic issue. I learned several mathematical things from the work of the
referee on my paper, but I learned much more from his/her infinite patience
for paths and infinite path of patience.

##  1\. Background

Larson proved the consistency of
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$ under Martin’s axiom. A central component in our proof is a
similar theorem at the level of $\omega_{2}$. We shall use a generalized form
of Martin’s axiom, and there are several such theorems in the literature. The
most appropriate among which for our proof is Shelah’s version from [She78].

######  Theorem 1.1.

Generalized Martin’s axiom.
One can force $2^{\aleph_{0}}=\aleph_{1},2^{\aleph_{1}}>\aleph_{2}$ and if
$\mathbb{P}$ satisfies:

  1. $(a)$

If $p,q\in\mathbb{P}$ and $p\parallel q$ then there is a least upper bound for
$p,q$ in $\mathbb{P}$.

  2. $(b)$

If $\langle p_{i}:i\in\omega\rangle$ is an increasing sequence of conditions
in $\mathbb{P}$ then it has a least upper bound in $\mathbb{P}$.

  3. $(c)$

If $\\{p_{\alpha}:\alpha\in\omega_{2}\\}\subseteq\mathbb{P}$ then there is a
club $C\subseteq\omega_{2}$ and a regressive function
$f:\omega_{2}\rightarrow\omega_{2}$ such that for every $\alpha,\beta\in C\cap
S^{\aleph_{2}}_{\aleph_{1}}$ if $f(\alpha)=f(\beta)$ then $p_{\alpha}\parallel
p_{\beta}$.

then for every $\kappa<2^{\aleph_{1}}$ such that
$\gamma<\kappa\Rightarrow\gamma^{\aleph_{0}}<\kappa$ and any collection
$\mathcal{D}=\\{D_{\eta}:\eta\in\kappa\\}$ of dense subsets of $\mathbb{P}$
there exists a filter $G\subseteq\mathbb{P}$ so that $G\cap
D_{\eta}\neq\varnothing$ for every $\eta\in\kappa$.

$\qed_{\ref{thmgma}}$

The forcing conditions in Larson’s proof are finite independent sets. In our
proof the conditions are countable. As a first step we shall use the
generalized Martin’s axiom in order to force
$\omega^{*}\cdot\omega_{2}\rightarrow(\omega^{*}\cdot\omega_{2},\text{infinite
path})^{2}$, and requirement $(b)$ above forces us to force with countable
conditions. This is one major difference between Martin’s axiom and the
generalized Martin’s axiom which complicates the density argument.

Another problem is the chain condition. Martin’s axiom applies to any $ccc$
forcing notion, but all the generalizations to higher cardinals require more
than $\kappa$-cc, and it is known that $\kappa$-cc is insufficient. In
Shelah’s version, the strengthening of the chain condition is reflected in
requirement $(c)$. Our proof of $(c)$ in the specific forcing notion of this
paper is based on the ordinary partition relation
$\omega_{2}\rightarrow(\omega_{2}-st,\omega_{1})^{2}$ which says that for
every coloring $d:[\omega_{2}]^{2}\rightarrow\\{0,1\\}$ one can find either a
$1$-monochromatic sequence of type $\omega_{1}$ or a stationary
$0$-monochromatic subset of $\omega_{2}$. The following is a folklore but we
give the proof since we will use both the statement and the argument within
the proof.

######  Lemma 1.2.

Assume $2^{\omega}=\omega_{1}$. Then
$\omega_{2}\rightarrow(\omega_{2}-st,\omega_{1})^{2}$. Moreover,
$T\rightarrow(\omega_{2}-st,\omega_{1})^{2}$ whenever $T\subseteq
S^{\omega_{2}}_{\omega_{1}}$ is stationary.

_Proof_.
Let $d:[\omega_{2}]^{2}\rightarrow\\{0,1\\}$ be a coloring. If there is a
$1$-monochromatic sequence of length $\omega_{1}$ then we are done. Suppose
that there is no such a sequence. For every $\delta\in
S^{\omega_{2}}_{\omega_{1}}$ choose a sequence $c_{\delta}$ of ordinals below
$\delta$ such that $c_{\delta}\cup\\{\delta\\}$ is $1$-monochromatic and
$c_{\delta}$ is maximal with this property.

By our assumption, $c_{\delta}$ is bounded below $\delta$ since ${\rm
cf}(\delta)=\omega_{1}$. Hence the mapping $h(\delta)=\sup(c_{\delta})$ is
regressive on $S^{\omega_{2}}_{\omega_{1}}$. Choose $\eta\in\omega_{1}$ and a
stationary $S\subseteq S^{\omega_{2}}_{\omega_{1}}$ such that $h(\delta)=\eta$
for every $\delta\in{S}$. Notice that $\eta<\min(S)$. Since
$2^{\omega}=\omega_{1}$, there are only $\aleph_{1}$-many sequences of the
form $c_{\delta}$ (recall that $\eta$ is an ordinal less than $\omega_{2}$ and
each $c_{\delta}$ is a subset of $\eta$). Hence by shrinking $S$ if needed we
may assume that there is a fixed sequence $c$ such that $c_{\delta}=c$ for
every $\delta\in{S}$.

We claim that $S$ is $0$-monochromatic under $d$. To see this, suppose that
$\delta,\varepsilon\in{S}$ and $\delta<\varepsilon$. If
$d(\delta,\varepsilon)=1$ then $c\cup\\{\delta\\}$ is $1$-monochromatic with
$\varepsilon$ and then $h(\varepsilon)\geq\delta>\eta$. This is impossible
since $\varepsilon\in{S}$. Hence necessarily $d(\delta,\varepsilon)=0$
whenever $\\{\delta,\varepsilon\\}\subseteq{S}$, so we are done. The
additional part of the lemma is proved in the same way, upon replacing
$S^{\omega_{2}}_{\omega_{1}}$ by $T$.

$\qed_{\ref{lemstat}}$

The next issue is a special kind of tiltan which we shall need for our proof.
Definition 0.2 is phrased with respect to $\kappa$, but one can replace
$\kappa$ by any stationary subset $S\subseteq\kappa$. Clearly, if
$S_{0}\subseteq S_{1}$ are stationary then
$\clubsuit_{S_{0}}\Rightarrow\clubsuit_{S_{1}}$ and hence
$\clubsuit_{S}\Rightarrow\clubsuit_{\kappa}$ whenever $S$ is a stationary
subset of $\kappa$. The following theorem from [She98] served for proving the
consistency of tiltan at $\aleph_{1}$ with $2^{\omega}>\omega_{1}$, and we
shall use it with respect to infinite graphs.

######  Theorem 1.3.

Assume that $\Diamond_{S}$ holds at every stationary subset $S$ of
$\aleph_{1}$ and $\aleph_{2}$. Then one can define a tiltan sequence on
$S^{\aleph_{2}}_{\aleph_{0}}$ which is indestructible upon any further forcing
extension with an $\aleph_{1}$-complete forcing notion.

$\qed_{\ref{thmindest}}$

We indicate that the proof of the generalized Martin’s axiom employs an
$\aleph_{1}$-complete forcing notion, hence preserves instances of
indestructible tiltan. We shall use this fact in the proof of the main
theorem.

We mention three additional classical theorems, to be used within our proof.
Firstly, Ramsey’s theorem which says that
$\omega\rightarrow(\omega)^{2}_{\ell}$ for every $\ell\in\omega$. Namely, any
coloring $c:[\omega]^{2}\rightarrow\ell$ admits a monochromatic infinite set.
Secondly, Hajnal’s free set theorem which says that if
$\kappa<\lambda,|A|=\lambda,f:A\rightarrow\mathcal{P}(A)$ is a set-mapping
(i.e. $a\notin f(a)$ for every $a\in A$) and $|f(a)|<\kappa$ for every $a\in
A$ then there exists an $f$-free subset $B\subseteq A$ of size $\lambda$.
Recall that $B$ is $f$-free iff $B\cap f(b)=\varnothing$ whenever $b\in B$.
For the third theorem recall that if $\kappa$ is an infinite cardinal then
${\rm log}_{\kappa}(\kappa^{+})=\min\\{\theta:\kappa^{\theta}>\kappa\\}$. One
can show that if $\kappa\geq\omega$ then
$\kappa^{+}\rightarrow(\kappa^{+},{\rm log}_{\kappa}(\kappa^{+})+1)^{2}$, see
[EHMR84]. In particular, if $2^{\omega}=\omega_{1}$ then ${\rm
log}_{\omega_{1}}(\omega_{2})=\omega_{1}$ and hence
$\omega_{2}\rightarrow(\omega_{2},\omega_{1})^{2}$. In fact, one has the
stronger relation $\omega_{2}\rightarrow(\omega_{2}-st,\omega_{1})^{2}$ as
proved above.

We shall also need a statement concerning path relations in the following weak
form. Call a coloring $d:\kappa\times\kappa\rightarrow\omega\times\omega$
_anti-symmetric_ iff $d(\alpha,\beta)=(i,j)\Leftrightarrow
d(\beta,\alpha)=(j,i)$ whenever $\alpha,\beta\in\kappa$. Let us say that
$\kappa\rightarrow_{\rm asp}(\omega)^{2}_{\omega\times\omega}$ iff for every
anti-symmetric coloring $d:\kappa\times\kappa\rightarrow\omega\times\omega$
one can find an infinite path $\psi=(\alpha_{m}:m\in\omega)$, the elements of
$\psi$ being ordinals of $\kappa$ and for every $m\in\omega$ if
$d(\alpha_{m},\alpha_{m+1})=(i,j)\wedge d(\alpha_{m+1},\alpha_{m+2})=(k,\ell)$
then $j=k$.

######  Lemma 1.4.

$\omega_{1}\rightarrow_{\rm asp}(\omega)^{2}_{\omega\times\omega}$.

_Proof_.
Let $d:\omega_{1}\times\omega_{1}\rightarrow\omega\times\omega$ be anti-
symmetric. Let $\chi$ be a sufficiently large regular cardinal and choose a
countable elementary submodel $M\prec\mathcal{H}(\chi)$ so that $d\in M$. Let
$\delta=M\cap\omega_{1}$ be the characteristic ordinal of $M$ and notice that
${\rm cf}(\delta)=\omega$.

Fix an ordinal $\alpha_{0}\in\delta$ and assume that
$d(\alpha_{0},\delta)=(i,j)$. Denote the set
$\\{\alpha\in\delta:d(\alpha,\delta)=(i,j)\\}$ by $B$ and notice that $B$ is
unbounded in $\delta$ by elementarity. By definition, $\alpha_{0}\in B$.
Choose $\alpha_{1}>\alpha_{0}$ so that $\alpha_{1}\in B$. This means that
$d(\alpha_{0},\delta)=d(\alpha_{1},\delta)=(i,j)$ so by elementarity one can
find $\beta_{1}>\alpha_{1}$ such that $\beta_{1}<\delta$ and
$d(\alpha_{0},\beta_{1})=d(\alpha_{1},\beta_{1})=(i,j)$. We choose now another
element $\alpha_{2}\in B$ so that $\alpha_{2}>\beta_{1}$. Since
$d(\alpha_{2},\delta)=(i,j)$ one can choose $\beta_{2}<\delta$ such that
$\beta_{2}>\alpha_{2}$ and
$d(\alpha_{1},\beta_{2})=d(\alpha_{2},\beta_{2})=(i,j)$. We render this
process in the same way by induction on $n\in\omega$ and finally define:

$\psi=(\alpha_{0},\alpha_{n},\beta_{n}:0<n<\omega).$

One can verify that $\psi$ forms an infinite path in the sense defined before
the statement of the lemma. We conclude, therefore, that
$\omega_{1}\rightarrow_{\rm asp}(\omega)^{2}_{\omega\times\omega}$ as
required.

$\qed_{\ref{lemwp}}$

##  2\. Graphs with no infinite path

In this section we prove the main result of the paper, namely tiltan is
consistent with
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$. Let us describe the architecture of the proof. The first step is
to fix an indestructible tiltan sequence at $S=S^{\aleph_{2}}_{\aleph_{0}}$.
The second step is to force the generalized Martin’s axiom so that
$2^{\omega}=\omega_{1},2^{\omega_{1}}>\omega_{2}$ and the tiltan from the
first step is preserved. The main theorem at this stage is the positive
relation
$\omega^{*}\cdot\omega_{2}\rightarrow(\omega^{*}\cdot\omega_{2},\text{infinite
path})^{2}$. This relation will follow from the generalized Martin’s axiom.
The final step is to collapse $\aleph_{1}$ by making it a countable ordinal.

It is easy to show that the tiltan is preserved by this collapse, in the sense
that it holds in the generic extension over some stationary subset of
$\aleph_{1}$. Likewise, the above positive relation obtained by the
generalized Martin’s axiom becomes after the collapse
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$. This general plan has been used by Shelah, [She98], in his proof
of the consistency of tiltan with $2^{\omega}>\omega_{1}$. At the end of the
paper we shall try to explain what are the features of a statement that one
should expect to hold (consistently) with tiltan.

We commence with the concept of clean columns, as defined by Larson. The
definitions and claims in our context are adapted to the level of
$\aleph_{2}$. In the definition and lemma below we follow in the footsteps of
Larson.

######  Definition 2.1.

Clean columns.
Let $G=(V,E)$ be a graph where $V\subseteq\omega_{2}\times\omega$.

  1. $(\aleph)$

For every $\beta\in\omega_{2}$, the $\beta$th column of the graph is the set
$V(\beta)=V\cap(\\{\beta\\}\times\omega)$.

  2. $(\beth)$

$G$ has clean columns iff the following three properties hold for every
$\beta\in\omega_{2}$:

    1. $(a)$

Either $V(\beta)=\varnothing$ or $|V(\beta)|=\aleph_{0}$.

    2. $(b)$

$[V(\beta)]^{2}\cap E=\varnothing$.

    3. $(c)$

For every $(\alpha,n)\in V$ there is at most one pair $(\beta,\ell)$ such that
$\\{(\alpha,n),(\beta,\ell)\\}\in E$.

Graphs with clean columns simplify considerably the treatment of independent
subsets and related notions. Of course, a graph $G$ may lack this property.
However, we focus on graphs of type $\omega^{*}\cdot\omega_{2}$ with no
infinite path. In such graphs one can always pass to a subgraph of the same
order type with clean columns. Ahead of proving this assertion, we need a
simple lemma.

######  Lemma 2.2.

Let $G=(V,E)$ be a graph over $\omega_{2}\times\omega$ with no infinite path,
and let $\beta\in\omega_{2}$. Assume that $C\subseteq\\{\beta\\}\times\omega$
and $|C|=\aleph_{0}$.
There exists a finite set $A\subseteq\omega_{2}\times\omega$ and an infinite
set $B\subseteq C$ such that:

  1. $(a)$

Every element of $A$ is connected with every element of $B$.

  2. $(b)$

If $(\alpha,m)\notin A$ then there is at most one element $(\beta,n)\in B$
such that $\\{(\alpha,m),(\beta,n)\\}\in E$.

  3. $(c)$

$[B]^{2}\cap E=\varnothing$.

_Proof_.
We try to define by induction on $i\in\omega$ pairs of the form
$(D_{i},a_{i})$ such that $D_{i}\subseteq C$ is infinite and
$a_{i}\in\omega_{2}\times\omega$. We indicate that this attempt is doomed to
failure after finitely many steps.

At the stage of $i=0$ we choose any infinite independent $D_{0}\subseteq C$.
The existence of such a set follows from Ramsey’s theorem upon defining
$d:[C]^{2}\rightarrow\\{0,1\\}$ by $d(x,y)=0$ iff $\\{x,y\\}\notin E$.
Ramsey’s theorem ensures the existence of an infinite monochromatic set
$D_{0}\subseteq C$. The assumption that $G$ has no infinite path implies that
$D_{0}$ must be $0$-monochromatic, that is an independent set.

Now we ask whether there is an element $a\in\omega_{2}\times\omega$ so that
$a$ is connected with $\aleph_{0}$-many elements from $D_{0}$. If the answer
is positive then let $a_{0}$ be the $<_{\rm lex}$-first such element. If the
answer is negative then the process is terminated.

At the stage of $i+1$ we let $D_{i+1}=D_{i}\cap E(a_{i})$. By the induction
hypothesis at the $i$th stage, $D_{i+1}$ is infinite. Now we ask if there
exists some $a\in\omega_{2}\times\omega$ connected with $\aleph_{0}$-many
elements from $D_{i+1}$ such that $a\neq a_{j}$ for every $j\leq i$. If yes,
let $a_{i+1}$ be the $<_{\rm lex}$-first with this property. If not, the
process is terminated.

Remark that for some $\ell\in\omega$ we will be able to define $D_{\ell}$ but
not $a_{\ell}$. Otherwise, for every $i\in\omega$ choose an element $d_{i}\in
D_{i+2}-\\{d_{j}:j<i\\}$ (here we use the infinitude of each $D_{i}$) and then
$\langle a_{n},d_{n}:n\in\omega\rangle$ forms an infinite path, a
contradiction.

Set $D=D_{\ell},A=\\{a_{i}:i<\ell\\}$. Define a coloring
$c:[D]^{2}\rightarrow\\{0,1\\}$ as follows. Let $c(x,y)=0$ iff there exists
$v\notin A$ such that both $\\{x,v\\},\\{y,v\\}\in E$. By another application
of Ramsey’s theorem there is an infinite $B\subseteq D$ which is monochromatic
under $c$. Observe that $B$ must be $1$-monochromatic, since if $x,y\in
B,c(x,y)=0$ then one can produce an infinite path from the elements of $B$ and
the elements $v\notin A$ which connect them. This argument uses the fact that
every such $v$ is not in $A$ hence connected with only finitely many elements
of $B$.

Now the sets $A,B$ are as required. First, $A$ is finite and $B$ is infinite.
Second, $(a)$ follows from the choice of the elements of $A$, $(b)$ follows
from the fact that $B$ is $c$-monochromatic and $(c)$ from the fact that
$B\subseteq D$ and $D$ is independent.

$\qed_{\ref{lemfinite}}$

Equipped with the above lemma, we can proceed to the following.

######  Claim 2.3.

Assume that:

  1. $(a)$

$V\subseteq\omega_{2}\times\omega$ and ${\rm
otp}(V)=\omega^{*}\cdot\omega_{2}$.

  2. $(a)$

$G=(V,E)$ is a graph with no infinite path.

Then there exists $W\subseteq V,{\rm otp}(W)=\omega^{*}\cdot\omega_{2}$ such
that the graph $H=(W,E\cap[W]^{2})$ has clean columns.

_Proof_.
We may assume that all the columns of $V$ are infinite, since ${\rm
otp}(V)=\omega^{*}\cdot\omega_{2}$ and hence it will remain with the same
order type after removing all the finite columns. We apply Lemma 2.2 to every
column of $V$, and we get some $U\subseteq V,{\rm
otp}(U)=\omega^{*}\cdot\omega_{2}$, every nonempty column of $U$ is infinite
and edge-free and for each $U(\beta)$ there is a finite set $A(\beta)$ as in
the lemma.

Denote the left component $\\{\beta:|U(\beta)|=\aleph_{0}\\}$ by $I$, and
define $f:I\rightarrow[I]^{<\omega}$ by $f(\beta)=\\{\alpha\in I:\exists
m\in\omega,(\alpha,m)\in A(\beta)\\}$. Notice that $\beta\notin f(\beta)$ for
every $\beta\in I$, since $U(\beta)$ is edge-free and hence no pair of the
form $(\beta,m)$ can be an element of $A(\beta)$. This means that $f$ is a
set-mapping. Further, for every $\beta\in I$ one can see that $f(\beta)$ is a
finite set. This is simply because $A(\beta)$ is finite, due to Lemma 2.2. By
Hajnal’s free set theorem there exists $J\subseteq I,|J|=\aleph_{2}$ such that
$J$ is $f$-free, that is $\alpha\notin f(\beta)$ whenever $\alpha,\beta\in J$.

Define $W=\bigcup\\{U(\beta):\beta\in J\\}$ and observe that ${\rm
otp}(W)=\omega^{*}\cdot\omega_{2}$. The fact that $H=(W,[W]^{2}\cap E)$ has
clean columns comes from the properties of each $U(\beta)$ as guranteed by the
lemma, so we are done.

$\qed_{\ref{clmclean}}$

The ability to clean the columns is helpful in the proof of the main theorem.
The proof depends on two additional lemmata. The second lemma will be
postponed after the proof of the main theorem. For the first lemma let us
define the concept of _a replete ordinal_. Let $G=(V,E)$ be a graph with
$V\subseteq\omega_{2}\times\omega$, and assume that
$\psi=\\{(\alpha_{i},m_{i}):i\in\omega\\}\subseteq V$. An ordinal
$\beta\in\omega_{2}$ will be called $\psi$-replete iff there exists
$n(\beta)\in\omega$ such that for every $k\in[n(\beta),\omega)$ there is
$i_{k}\in\omega$ for which $\\{(\alpha_{i_{k}},m_{i_{k}}),(\beta,k)\\}\in E$.

######  Lemma 2.4.

Suppose that:

  1. $(a)$

$2^{\omega}=\omega_{1}$.

  2. $(b)$

$V\subseteq\omega_{2}\times\omega$ and ${\rm
otp}(V)=\omega^{*}\cdot\omega_{2}$.

  3. $(c)$

$H=(V,E)$ is a graph with clean columns.

  4. $(d)$

There is no independent subset of $V$ of type $\omega^{*}\cdot\omega_{2}$.

  5. $(e)$

$\psi=\\{(\alpha_{i},m_{i}):i\in\omega\\}\subseteq V$.

  6. $(f)$

$R$ is an unbounded subset of $\omega_{2}$ such that every $\beta\in{R}$ is
$\psi$-replete.

Then there exists an infinite path in $H$.

_Proof_.
For every $\beta\in R$ let $n(\beta)\in\omega$ be as in the definition of
repleteness and let $A_{\beta}\in[\omega]^{\omega}$ be such that if $i\in
A_{\beta}$ then there is $k\in\omega$ so that $i=i_{k}$, that is
$\\{(\alpha_{i},m_{i}),(\beta,k)\\}\in E$. Since $2^{\omega}=\omega_{1}$ and
$|R|=\aleph_{2}$ we may assume that $A_{\beta}=A$ for every $\beta\in R$,
where $A$ is some fixed element of $[\omega]^{\omega}$. Similarly, we can
assume that $n(\beta)\in\omega$ is the same natural number for every $\beta\in
R$, and without loss of generality $n(\beta)=0$ for every $\beta\in R$.

Define $c:[R]^{2}\rightarrow 2$ by $c(\beta,\gamma)=0$ iff there is no edge
from $(\beta,k)$ to $(\gamma,\ell)$ whenever $k,\ell\in\omega$. Put another
way, $c(\beta,\gamma)=1$ iff there are $k,\ell\in\omega$ for which
$\\{(\beta,k),(\gamma,\ell)\\}\in E$. By the Erdős-Dushnik-Miller theorem
either some $S\in[R]^{\omega_{2}}$ is $0$-monochromatic or some
$\\{\beta_{m}:m\in\omega\\}\subseteq R$ is $1$-monochromatic. In the first
case $S\times A$ forms an independent subset of $V$222We may assume, without
loss of generality, that $S\times{A}\subseteq{V}$. of type
$\omega^{*}\cdot\omega_{2}$, contradicting $(d)$. We conclude, therefore, that
$\\{\beta_{m}:m\in\omega\\}\subseteq R$ is $1$-monochromatic for some infinite
subset of $R$.

By induction on $m\in\omega$ we try to choose an element $t_{m}\in\psi$ such
that $\\{t_{m},(\beta_{m},k_{m})\\}\in E$ and $m<n<\omega\Rightarrow t_{m}\neq
t_{n}$ and for some $\ell$ we have
$\\{(\beta_{m},k_{m}),(\beta_{m+1},\ell)\\}\in E$. This is possible since
$c(\beta_{m},\beta_{m+1})=1$ so we fix $k_{m}$ and $\ell$ for which
$\\{(\beta_{m},k_{m}),(\beta_{m+1},\ell)\\}\in E$ and then we can choose
$t_{m}$ and $t_{m+1}$ from $\psi$ using the fact that both $\beta_{m}$ and
$\beta_{m+1}$ are $\psi$-replete.333By a careful choice of $A$ we may assume
that $t_{m}\neq t_{m+1}$. Now the sequence
$\langle(\beta_{m},k_{m}),(\beta_{m+1},k_{m+1}),t_{m+1}:m\in\omega\rangle$
forms an infinite path in $H$ so the proof is accomplished.

$\qed_{\ref{lemsecond}}$

We can prove now the substantial result which reads as follows:

######  Theorem 2.5.

Assume $2^{\omega}=\omega_{1},2^{\omega_{1}}>\omega_{2}$ and the generalized
Martin’s axiom holds. Then
$\omega^{*}\cdot\omega_{2}\rightarrow(\omega^{*}\cdot\omega_{2},\text{infinite
path})^{2}$.

_Proof_.
Let $H=(V,E)$ be a graph with no infinite path such that ${\rm
otp}(V)=\omega^{*}\cdot\omega_{2}$. We are assuming toward contradiction that
there is no independent subset of $V$ of type $\omega^{*}\cdot\omega_{2}$. By
Claim 2.3 we may assume that $H$ has clean columns. As annotated above, let
$I=\\{\beta\in\omega_{2}:|V(\beta)|=\aleph_{0}\\}$.

We define a forcing notion $\mathbb{P}$. A condition $p\in\mathbb{P}$ is a
countable independent subset of $V$. If $p,q\in\mathbb{P}$ then
$p\leq_{\mathbb{P}}q$ iff $p\subseteq q$. By Lemma 2.6 below, $\mathbb{P}$
satisfies $(c)$ of Theorem 1.1. If $p,q\in\mathbb{P}$ and $p\parallel q$ then
$p\cup q\in\mathbb{P}$ and it is a least upper bound by the definition of
$\leq_{\mathbb{P}}$. Similarly, if $(p_{j}:j\in\omega)$ is
$\leq_{\mathbb{P}}$-increasing then $\bigcup_{j\in\omega}p_{j}\in\mathbb{P}$,
being countable and independent, and it is a least upper bound. Hence
$\mathbb{P}$ satisfies the requirements of Theorem 1.1.

For every $\beta\in I$ let
$D_{\beta}=\\{p\in\mathbb{P}:\exists\gamma>\beta,\exists^{\infty}\ell,(\gamma,\ell)\in
p\\}$. We claim that each $D_{\beta}$ is a dense open subset of $\mathbb{P}$.
The fact that $D_{\beta}$ is open follows from the definition, so let us prove
density. Suppose that $p=\\{(\alpha_{i},m_{i}):i\in\omega\\}\notin D_{\beta}$,
but $p\in\mathbb{P}$. We know, therefore, that $p$ is independent and we
observe that assumptions $(a)-(e)$ of Lemma 2.4 hold with $p$ here stands for
$\psi$ there (note that $(d)$ is our assumption toward contradiction). Since
the conclusion of the lemma fails we see that necessarily assumption $(f)$ of
the lemma fails. That is, the set $R$ of $p$-replete ordinals is bounded in
$\omega_{2}$. Choose $\gamma\in\omega_{2}$ such that $\gamma>\beta$ and
$\gamma>\sup(R)$. Namely, $\gamma$ is not $p$-replete and hence
$(\gamma,\ell)$ is not connected with an element of $p$ for an infinite set
$a$ of $\ell$s. Define $q=p\cup\\{(\gamma,\ell):\ell\in a\\}$. Since $p\leq
q\in D_{\beta}$ we see that $D_{\beta}$ is dense.

The collection $\mathcal{D}=\\{D_{\beta}:\beta\in I\\}$ is of size
$\aleph_{2}$ and $2^{\omega_{1}}>\omega_{2}$. Further,
$\alpha<\aleph_{2}\Rightarrow\alpha^{\aleph_{0}}<\aleph_{2}$ since
$2^{\omega}=\omega_{1}$. Hence there exists a generic filter
$G\subseteq\mathbb{P}$ such that $G\cap D_{\beta}\neq\varnothing$ for every
$\beta\in I,n\in\omega$. Define $W=\bigcup\\{p:p\in G\\}$, and notice that
${\rm otp}(W)=\omega^{*}\cdot\omega_{2}$. Since every $p\in G$ is independent
and $G$ is a directed set, $W$ is independent as well, so we arrived at a
contradiction and hence we are done.

$\qed_{\ref{thmmtgma}}$

We accomplish the above proof by the following lemma.

######  Lemma 2.6.

Let $H=(V,E)$ be a graph with no infinite path, and let $\mathbb{P}$ be the
associated forcing whose conditions are countable independent sets. Assume
that $2^{\omega}=\omega_{1}$. Then $\mathbb{P}$ satisfies requirement $(c)$ of
Theorem 1.1.

_Proof_.
Let $\\{p_{\alpha}:\alpha\in\omega_{2}\\}$ be a set of conditions in
$\mathbb{P}$. By induction on $\beta\in\omega_{2}$ we choose a stationary
subset $A_{\beta}$ of $S^{\omega_{2}}_{\omega_{1}}$ such that:

  1. $(a)$

If $i,j\in A_{\beta}$ then $p_{i}\parallel p_{j}$.

  2. $(b)$

The set $W=S^{\omega_{2}}_{\omega_{1}}-\bigcup A_{\beta}$ is not stationary in
$\omega_{2}$.

We can choose $A_{\beta}$ by applying Lemma 1.2 inductively over the set
$S^{\omega_{2}}_{\omega_{1}}-\bigcup_{\gamma<\beta}A_{\gamma}$. Let us
describe the construction explicitly. For $\beta=0$ define
$d:[S^{\omega_{2}}_{\omega_{1}}]^{2}\rightarrow{2}$ as follows. If $i,j\in
S^{\omega_{2}}_{\omega_{1}}$ then let $d(i,j)=0$ iff $p_{i}\parallel{p_{j}}$.
By Lemma 1.2, either there is a stationary set $T\subseteq
S^{\omega_{2}}_{\omega_{1}}$ for which $d^{\prime\prime}[T]^{2}=\\{0\\}$ or an
$\omega_{1}$-sequence of elements of $S^{\omega_{2}}_{\omega_{1}}$ for which
the range of $d$ on its pairs is constantly one. The second option is
impossible since by an application of Lemma 1.4 it means that there is an
infinite path in $H$, contrary to our assumptions. Thus, the first option
holds. Set $A_{0}=T$.

For $\beta>0$, without loss of generality
$S^{\omega_{2}}_{\omega_{1}}-\bigcup_{\gamma<\beta}A_{\gamma}$ is stationary.
Apply Lemma 1.2 in the same way to obtain $A_{\beta}$. We emphasize that the
choice of the $A_{\beta}$s depend on $\\{p_{\alpha}:\alpha\in\omega_{2}\\}$
and, moreover, on the enumeration of its elements. Since we wish to use the
argument given in that lemma, for every $\beta\in\omega_{2}$ there will be an
ordinal $\eta_{\beta}\in\omega_{2}$ and a fixed sequence
$c_{\beta}\subseteq\eta_{\beta}$ as described in the proof of Lemma 1.2.
Recall that $\eta_{\beta}<\min(A_{\beta})$. Notice that if
$c_{\beta}=c_{\gamma}$ then $\eta_{\beta}=\eta_{\gamma}$ and $A_{\beta}\cup
A_{\gamma}$ is linked.444The adjective _linked_ means that every two elements
from $A_{\beta}\cup A_{\gamma}$ are compatible. Hence by taking unions of
$A_{\beta}$s whenever possible we may assume that $\beta\neq\gamma\Rightarrow
c_{\beta}\neq c_{\gamma}$.

Let $T=\\{\min(A_{\beta}):\beta\in\omega_{2}\\}$. We claim that $T$ is not
stationary. Suppose not, and define $g:T\rightarrow\omega_{2}$ by
$g(\min(A_{\beta}))=\eta_{\beta}$. This is a regressive function so there is a
stationary set $T^{\prime}\subseteq{T}$ and a fixed $\eta\in\omega_{2}$ such
that $\beta\in T^{\prime}$ implies $\eta_{\beta}=\eta$. Moreover, since each
$c_{\beta}$ is countable we may shrink $T^{\prime}$ to a stationary set for
which all the $c_{\beta}$s are the same fixed sequence, say $c$. However, this
is impossible since by the choice of the $A_{\beta}$s if $\beta\neq\gamma$
then $c_{\beta}\neq c_{\gamma}$. We conclude, therefore, that $T$ is not
stationary.

Re-enumerate the family of sets $(A_{\beta}:\beta\in\omega_{2})$ by
$(S_{i}:i\in\omega_{2})$ in such a way that
$i<j\Rightarrow\min(S_{i})<\min(S_{j})$. Notice that this implies
$i\leq\min(S_{i})$ for every $i\in\omega_{2}$. Define
$f:\omega_{2}\rightarrow\omega_{2}$ by letting $f(\alpha)=i$ iff $\alpha\in
S_{i}$. Let $C$ be a club of $\omega_{2}$ disjoint from $T\cup{W}$. If
$\alpha\in C\cap S^{\omega_{2}}_{\omega_{1}}$ then $f(\alpha)<\alpha$ since
$f(\alpha)\leq\alpha$ by the fact that $i\leq\min(S_{i})$ and
$f(\alpha)=\alpha$ is possible only if $\alpha\in{T}$. Therefore, $f$ is
regressive on $C\cap S^{\omega_{2}}_{\omega_{1}}$. If
$\alpha_{0},\alpha_{1}\in C\cap S^{\omega_{2}}_{\omega_{1}}$ and
$f(\alpha_{0})=f(\alpha_{1})$ then $\alpha_{0}$ and $\alpha_{1}$ belong to the
same $S_{i}$, so $p_{\alpha_{0}}\parallel p_{\alpha_{1}}$. This concludes the
proof of the lemma.

$\qed_{\ref{lemmetukan}}$

We can prove, finally, the main result of this paper:

######  Theorem 2.7.

Tiltan at $\aleph_{1}$ is consistent with the positive relation
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$.

_Proof_.
Begin with a universe in which there are enough diamonds so that one can
create a tiltan sequence at some stationary set $S\subseteq
S^{\omega_{2}}_{\omega}$, and this instance of tiltan is indestructible upon
any $\aleph_{1}$-complete forcing notion. The description of this construction
and the indestructibility proof appear in [She98]. We may assume that
$2^{\omega}=\omega_{1}$ along with this construction (actually, this is the
natural situation).

We force now the generalized Martin’s axiom of Theorem 1.1 with
$2^{\omega_{1}}>\omega_{2}$. This is done by an $\aleph_{1}$-complete forcing
notion, hence the tiltan over $S$ is preserved. Observe that this means also
tiltan at $\aleph_{2}$, since $S\subseteq\omega_{2}$. By Theorem 2.5 we see
that the relation
$\omega^{*}\cdot\omega_{2}\rightarrow(\omega^{*}\cdot\omega_{2},\text{infinite
path})^{2}$ holds at this stage.

Now we force with $L\acute{e}vy(\aleph_{0},\aleph_{1})$. We claim that the
statement of our theorem holds in the generic extension by this collapse. It
is easy to see that the tiltan holds at $\aleph_{1}$. Basically, the reason is
that unbounded subsets of $\aleph_{1}$ after the collapse contain unbounded
sets of size $\aleph_{2}$ from the previous stage. The formal proof is
elaborated in [She98]. By a similar argument, the relation
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$ holds after the collpase.

For this, assume that $G=(V,E)$ is a graph over the set
$\omega_{1}\times\omega$ of order type $\omega^{*}\cdot\omega_{1}$. In the
stage just before the collapse there is a graph $H=(W,F)$ such that ${\rm
otp}(W)=\omega^{*}\cdot\omega_{2}$ and such that $W$ is forced to be a subset
of $V$ by the collapse. We indicate that the copy of $\omega^{*}$ at each
column in the generic extension may issue from a copy of type $\omega_{1}^{*}$
before the collapse, in which case we can take a subset of it of order type
$\omega^{*}$. Now if there is an infinite path in $H$ then it will remain
infinite after the collapse as an infinite sequence cannot become finite in
the generic extension. If there is no infinite path in $H$ then there is an
independent set of type $\omega^{*}\cdot\omega_{2}$, and after the collapse it
will be of type $\omega^{*}\cdot\omega_{1}$, so we are done.

$\qed_{\ref{thmmt}}$

We conclude with a modest attempt to understand the difference between tiltan
and diamond as reflected in our result. Apparently, tiltan should work in the
diamond construction of Baumgartner and Larson from [BL90] since one has to
add only one edge to each independent set. Nonetheless, tiltan consistently
fails and the reason is that rather than guessing uncountable subsets of
$\aleph_{1}$ we try to guess subsets of $\omega_{1}\times\omega$. Moreover, we
must make sure that every strongly cofinal subset of $\omega_{1}\times\omega$
is predicted by strongly cofinal subsets of $\beta\times\omega$ for
unboundedly many $\beta\in\omega_{1}$.

Practically, we translate subsets of $\omega_{1}\times\omega$ to subsets of
$\omega_{1}$, apply our tiltan sequence, and then send it back to
$\omega_{1}\times\omega$. But we loose information in this process, and in
particular it may happen that a strongly cofinal set will be predicted by
narrow sets which will not be strongly cofinal after the translation to
$\omega_{1}\times\omega$. Larson considered in [Lar06] a stronger version of
tiltan in which strongly cofinal sets are predicted by strongly cofinal tiltan
elements. Unfortunately, such a prediction principle implies the continuum
hypothesis (hence diamond)555Recall that tiltan with the continuum hypothesis
imply diamond. and Larson concluded that this direction will not settle the
problem of tiltan and
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$. On the other hand, this fact motivated our attempt to force the
consistency of tiltan with
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$.

It would be interesting to ask a similar question with respect to _superclub_
, a prediction principle defined by Primavesi in [Pri11]. A superclub sequence
$\langle S_{\alpha}:\alpha\in\lim(\omega_{1})\rangle$ satisfies
$S_{\alpha}\subseteq\alpha$ for every $\alpha$ and for every
$A\in[\omega_{1}]^{\omega_{1}}$ one can find $B\subseteq A,|B|=\aleph_{1}$
such that $\\{\delta\in\omega_{1}:B\cap\delta=S_{\delta}\\}$ is a stationary
subset of $\omega_{1}$. Superclub holds iff there exists a superclub sequence.
One can see that diamond implies superclub and superclub implies tiltan. Both
implications are irreversible. In particular, superclub is strictly stronger
than tiltan, see [Gar18] and [GS23].

######  Question 2.8.

Is it consistent that superclub holds at $\aleph_{1}$ and
$\omega^{*}\cdot\omega_{1}\rightarrow(\omega^{*}\cdot\omega_{1},\text{infinite
path})^{2}$?

We indicate that if one begins with superclub (or even diamond) at
$\omega_{2}$ and collapses $\aleph_{1}$ then superclub fails at $\aleph_{1}$
in the generic extension, as follows from [Gar18]. In other words, the method
of the current paper does not resolve the above problem.

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