# Endpoint-homogeneous fans

Will Brian  W. R. Brian
Department of Mathematics and Statistics
University of North Carolina at Charlotte
Charlotte, NC, USA<EMAIL_ADDRESS>wrbrian.wordpress.com and  Rene Gril
Rogina  R. Gril Rogina
Faculty of Sciences and Mathematics
University of Maribor
Maribor, Slovenia<EMAIL_ADDRESS>
###### Abstract.

A fan $F$ is _endpoint-homogeneous_ if for any two endpoints $e,e^{\prime}$ of
$F$, there is a homeomorphism $h:F\to F$ such that $h(e)=e^{\prime}$.

We prove there are uncountably many distinct homeomorphism types of endpoint-
homogeneous smooth fans. To do this, we associate to each such fan $F$ a
topological invariant, in the form of a characteristic subset
$EPG(F)\subseteq[0,1]$ describing how the endpoints of $F$ limit onto any
given blade of $F$. We describe precisely all the uncountably many different
$X\subseteq[0,1]$ that can arise as $EPG(F)$ for some endpoint-homogeneous
smooth fan $F$. We also prove the existence of $\frac{1}{n}$-homogeneous
smooth fans for all $n\geq 5$.

###### Key words and phrases:

smooth fans, Cantor fan, Lelek fan, homogeneity

###### 2020 Mathematics Subject Classification:

54F15, 54F65

The first author is supported in part by NSF grant DMS-2154229. The second
author is supported in part by the Slovenian Research Agency (research program
J1-4632).

##  1\. Introduction

A topological space $X$ is _homogeneous_ if for any $x,y\in X$, there is a
homeomorphism $h:X\to X$ such that $h(x)=y$. This property expresses that $X$
has an especially rich group of self-homeomorphisms.

Many continua fail to be homogeneous, but still have very rich self-
homeomorphism groups. The class of _fans_ (which is defined precisely in the
next section) contains many examples of this kind. A fan cannot be
homogeneous. It has one “top” point, many endpoints, and many points that are
neither. These three classes of points are topologically defined, and
consequently are invariant under homeomorphisms. Nevertheless, some fans, such
as the Cantor fan and the Lelek fan, have extremely rich homeomorphism groups
(see [2], [5]).

A fan $F$ is called _endpoint-homogeneous_ if for any two endpoints
$e,e^{\prime}$ of $F$, there is a homeomorphism $h:F\to F$ such that
$h(e)=e^{\prime}$. Equivalently, $F$ is endpoint-homogeneous if for any two
“blades” of the fan (arcs connecting the top of the fan to some endpoint),
there is a self-homeomorphism of $F$ that restricts to a homeomorphism between
these two blades.

Let $F$ be an endpoint-homogeneous fan and $E$ its set of endpoints. Because
the blades of $F$ are all topologically indistinguishable, there is a single
set $EPG(F)\subseteq[0,1]$ such that for every blade $B$ of $F$, there is a
homeomorphism $B\to[0,1]$ mapping $B\cap\overline{E\setminus B}$ onto
$EPG(F)$. In other words, $EPG(F)$ is a topological invariant describing how
the endpoints of $F$ limit onto any given blade of $F$.

In this paper we work towards a classification of endpoint-homogeneous smooth
fans. Our main results are:

  1. (1)

A set $X\subseteq[0,1]$ is equal to the invariant $EPG(F)$ of some smooth
endpoint-homogeneous fan if and only if:

     * $\circ$

$X=\emptyset$, $\\{1\\}$, or $[0,1]$, or

     * $\circ$

$X$ is a closed subset of $[0,1]$ with $0\in X$ and $1\notin X$, or

     * $\circ$

$X$ is a closed subset of $[0,1]$ with $0,1\in X$, and $1$ is an isolated
point of $X$.

  2. (2)

In particular, it follows that there are uncountably many homeomorphism types
of endpoint-homogeneous smooth fans.

We conjecture that if $F$ and $G$ are smooth fans and
$EPG(F)=EPG(F)\neq\emptyset$, then $F$ and $G$ are homeomorphic. The case
$EPG(F)=\emptyset$ is something of a degenerate case; there are countably many
homeomorphism types of smooth fans for which the endpoints do not accumulate
onto any blades (namely those fans with finitely many blades, also known as
the $n$-ods, with $n\geq 3$). If true, this conjecture, combined with
statement $(1)$ above, would give an exact classification of all endpoint-
homogeneous smooth fans.

We do not say much about non-smooth fans, but we do provide examples of non-
smooth endpoint-homogeneous fans, and show

  1. $(3)$

There are uncountably many distinct homeomorphism types of non-smooth
endpoint-homogeneous fans, including some that violate statement $(1)$ above.

Another natural weakening of homogeneity that can apply to fans, or dendroids
more generally, is $\frac{1}{3}$-homogeneity. A space $X$ is _$\frac{1}{n}$
-homogeneous_ if there is a partition of $X$ into $n$ sets (but no fewer) such
that if $x,y\in X$ lie in the same partition piece, then there is a
homeomorphism $h:X\to X$ with $h(x)=y$. Equivalently, $X$ is
$\frac{1}{n}$-homogeneous if the homeomorphism group of $X$ has $n$ distinct
orbits with respect to its natural action on $X$.

As mentioned above, fans have $3$ distinct classes of points that are
invariant under homeomorphisms: the top, the endpoints, and everything else
(usually called the _ordinary_ points). Thus $\frac{1}{3}$-homogeneity is
essentially the strongest weakening of homogeneity that can apply to fans. But
it is perhaps too strong. In [2], Acosta, Hoehn, and Juárez completely
classify all $\frac{1}{3}$-homogeneous smooth fans. Other than the simple
$n$-ods (again something of a degenerate case), the $\frac{1}{3}$-homogeneous
smooth fans come in precisely $4$ homeomorphism types: the Cantor fan, the
Lelek fan, and two others. They also prove that no smooth fan is
$\frac{1}{4}$-homogeneous. In addition to the results listed above, we show:

  1. $(4)$

For every $n\geq 5$, there is an endpoint-homogeneous,
$\frac{1}{n}$-homogeneous smooth fan.

Note that endpoint-homogeneity is a strict weakening of
$\frac{1}{3}$-homogeneity: we require only that the endpoints (but not the
ordinary points) be an orbit of the homeomorphism group. This way of weakening
$\frac{1}{3}$-homogeneity brings us from having only $4$ nontrivial examples
to having uncountably many.

It is worth mentioning that $\frac{1}{n}$-homogeneity is a broadly studied
topic in continuum theory, not just with $n=3$ and not just with fans. One
notable example is [3], which provides a complete classification of all
$\frac{1}{3}$-homogeneous dendrites. The introduction of [3] contains an
extensive list of papers on the topic of $\frac{1}{n}$-homogeneity.

It is also worth mentioning that the opposite of endpoint-homogeneous fans was
studied by Hernández-Gutiérrez and Hoehn in [13]. They define a fan $F$ to be
_endpoint-rigid_ if every self-homeomorphism of $F$ restricts to the identity
map on the set of endpoints of $F$. They construct many examples of such fans,
including several where the set of endpoints of the fan (with the rest of the
fan removed) is a homogeneous space.

The characteristic set $EPG(F)$, as described above, is not well-defined for
all smooth fans, but only those for which no two blades can be distinguished
by looking at limits of endpoints. (We do not know whether this latter class
is actually larger than the class of smooth endpoint-homogeneous fans.) The
fans with this property were first defined in [12], where the fans $F$ such
that $EPG(F)=X\subseteq[0,1]$ are called the $X$_-endpoint generated_ fans (or
$X$-EPG fans). These arise in a classification of the Mahavier products
obtained from a pair of lines meeting at the origin. (In all but a few cases,
it is an $X$-EPG fan.) This classification is the main result of [12] and [6].
The work in the present paper arose from the desire to classify those $X$ for
which smooth $X$-EPG fans exist, followed by the realization that all known
constructions of these fans are endpoint-homogeneous.

##  2\. Definitions and notation

In this section we give definitions, notation, and preliminary results that
are used throughout the paper.

######  Definition 2.1.

A _continuum_ is a non-empty compact connected metric space. A _subcontinuum_
of some continuum $X$ is a subspace of $X$ that is itself a continuum.

In what follows, we sometimes write $I$ to denote the unit interval $[0,1]$.
Any subset of a continuum homeomorphic to $I$ is an _arc_. If $X$ is a
continuum and $A\subseteq X$ is an arc, then the _endpoints_ of $A$ are the
two points $a,b\in A$ such that $A\setminus\\{a\\}$ and $A\setminus\\{b\\}$
are connected. A _simple triod_ is a continuum homeomorphic to
$([-1,1]\times\\{0\\})\cup(\\{0\\}\times I)$, i.e., three copies of $I$ glued
together at an endpoint. Similarly, a _simple $n$-od_ is $n$ copies of $I$
glued together at an endpoint. If $T$ is a simple triod (or a simple $n$-od),
then the _top point_ of $T$ is the unique point $t\in T$ such that
$T\setminus\\{t\\}$ has three connected components (or $n$ components for a
simple $n$-od).

######  Definition 2.2.

  1. (1)

A continuum $X$ is _unicoherent_ , if whenever $A$ and $B$ are subcontinua of
$X$ such that $X=A\cup B$, then $A\cap B$ is connected.

  2. (2)

A continuum $X$ is _hereditarily unicoherent_ if every subcontinuum of $X$ is
unicoherent.

  3. (3)

A continuum $X$ is a _dendroid_ , if it is an arcwise connected, hereditarily
unicoherent continuum.

  4. (4)

Let $X$ be a dendroid. A point $x\in X$ is an _endpoint of $X$_, if it is an
endpoint of every arc in $X$ that contains it. The set of all endpoints of $X$
is denoted $E(X)$.

  5. (5)

Let $X$ be a dendroid. A point $x\in X$ is a _ramification point_ of $X$ if
there is a simple triod $T\subseteq X$ with top $x$. The set of all
ramification points of $X$ is denoted $R(X)$. A point in $X\setminus(E(X)\cup
R(X))$ is an _ordinary point_ ; the set of these points is denoted $O(X)$.

  6. (6)

A _fan_ is a dendroid with exactly one ramification point $t$, which is called
the _top_ of the fan.

  7. (7)

A fan $F$ is _endpoint-homogeneous_ if for any $e,e^{\prime}\in F$, there is a
homeomorphism $h:F\to F$ such that $h(e)=e^{\prime}$.

######  Observation 2.3.

Let $X$ and $Y$ be dendroids and $h:X\to Y$ a homeomorphism. Then $h$ maps
$E(X)$ onto $E(Y)$, $R(X)$ onto $R(Y)$, and $O(X)$ onto $O(Y)$. In particular,
no fan is homogeneous.

Two important examples of fans are the Cantor fan and the Lelek fan. The
_Cantor fan_ is any continuum homeomorphic to $\bigcup_{x\in K}A_{x}$, where
$K\subseteq I$ is the Cantor set, and for each $x\in K$, $A_{x}$ is the
straight line segment in the plane from $(\nicefrac{{1}}{{2}},1)$ to $(x,0)$.
More succinctly, the Cantor fan is the one-point compactification of
$K\times[0,1)$, also known as the cone over $K$.

A _Lelek fan_ is any subcontinuum $F$ of a Cantor fan such that $F$ is a fan
and $E(F)$ is dense in $F$. Charatonik proved in [10] that this description
uniquely specifies a continuum up to homeomorphism, i.e., any two Lelek fans
are homeomorphic (see [4], [7], and [13] for further characterizations).

Lelek fanCantor fan

Observe that dendroids generally, and fans in particular, are _uniquely
arcwise connected_ , which means that if $X$ is a dendroid and $x,y\in X$ with
$x\neq y$, then there is exactly one arc in $X$ whose endpoints are $x$ and
$y$. (Some such arc must exist, because $X$ is arcwise connected, but any such
arc is unique because $X$ is hereditarily unicoherent.)

######  Definition 2.4.

Let $X$ be a dendroid. Given $x,y\in X$ with $x\neq y$, let $[x,y]$ denote the
unique arc in $X$ with endpoints $x$ and $y$. If $x=y$, let $[x,y]=\\{x\\}$.

If $t$ is the top of a fan $F$ and $e\in E(F)$, then $[t,e]$ is a _blade_ of
$F$. The set of all blades of $F$ is denoted $\mathcal{B}(F)$.

######  Lemma 2.5.

A fan $F$ is endpoint-homogeneous if and only if for any blades
$B,B^{\prime}\in\mathcal{B}(F)$, there is a homeomorphism $F\to F$ mapping $B$
onto $B^{\prime}$.

###### Proof.

For the “only if” direction, suppose $F$ is an endpoint-homogeneous fan. Given
two blades $B$ and $B^{\prime}$ of $F$, there are (by the definition of a
blade) $e,e^{\prime}\in E(F)$ such that $B=[t,e]$ and
$B^{\prime}=[t,e^{\prime}]$, where $t$ denotes the top of $F$. But then there
is a homeomorphism $h:F\to F$ such that $h(e)=e^{\prime}$. Because $h(t)=t$,
this means $h$ maps $B=[t,e]$ onto $[h(t),h(e)]=[t,e^{\prime}]=B^{\prime}$.

For the “if” direction, suppose that for any two blades $B,B^{\prime}$ of $F$,
there is a homeomorphism $F\to F$ mapping $B$ onto $B^{\prime}$. Let
$e,e^{\prime}\in E(F)$, and fix a homeomorphism $h:F\to F$ mapping $[t,e]$
onto $[t,e^{\prime}]$, where $t$ denotes the top of $F$. Because $h(t)=t$, we
must have $h(e)=e^{\prime}$. ∎

######  Definition 2.6.

A fan $F$ with top $t$ is _smooth_ if for any convergent sequence
$\left\langle x_{n}\colon n\in\mathbb{N}\right\rangle$ of points in $F$,

$\textstyle\lim_{n\to\infty}[t,x_{n}]\,=\,[t,\lim_{n\to\infty}x_{n}]$

where the limit on the left is taken with respect to the Hausdorff metric.

...... Two non-smooth fans

We make repeated use of the following fact, a proof of which can be found in
[9, Theorem 9], [8, Proposition 4], or [11, Corollary 4].

######  Theorem 2.7.

A fan is smooth if and only if it is (homeomorphic to) a subcontinuum of the
Cantor fan.

The following definition first appeared in [12].

######  Definition 2.8.

Let $F$ be a fan with top $t$. Given $X\subseteq[0,1]$, we say that $F$ is an
_$X$ -endpoint generated_ fan (or an $X$_-EPG_ fan) if for every
$B\in\mathcal{B}(F)$, there is a homeomorphism $\phi:I\to B$ such that
$\phi(0)=t$ and $\phi[X]=B\cap\overline{E(F)\setminus B}$.

For example, $n$-ods are $\emptyset$-EPG fans, the Cantor fan is a
$\\{1\\}$-EPG fan, and the Lelek fan is a $[0,1]$-EPG fan. In fact,
Charatonik’s characterization of the Lelek fan shows it is the only
$[0,1]$-EPG smooth fan.

######  Theorem 2.9.

If $F$ is an endpoint-homogeneous fan, then there is some $X\subseteq[0,1]$
such that $F$ is an $X$-EPG fan.

###### Proof.

Let $F$ be a fan with top $t$, fix some $B_{0}\in\mathcal{B}(F)$, and let
$\phi:I\to B$ be a homeomorphism with $\phi(0)=t$. Let
$X=\phi^{-1}\big{(}B_{0}\cap\overline{E(F)\setminus B_{0}}\big{)}$. Given any
$B\in\mathcal{B}(F)$, there is a homeomorphism $h:F\to F$ that maps $B_{0}$
onto $B$. But then $h\circ\phi$ is a homeomorphism $I\to B$ with
$h\circ\phi(t)=t$, and, because $h$ fixes $E(F)$,
$h\circ\phi[X]=h[B_{0}\cap\overline{E(F)\setminus
B_{0}}]=B\cap\overline{E(F)\setminus B}$. Hence $F$ is $X$-EPG. ∎

This theorem shows that to understand endpoint-homogeneous fans, it is
necessary first to understand $X$-EPG fans. This raises several questions:

  * $\circ$

For which $X\subseteq[0,1]$ is there an $X$-EPG fan?

  * $\circ$

For which $X\subseteq[0,1]$ can such a fan be endpoint-homogeneous?

  * $\circ$

If there is an $X$-EPG fan for some nonempty $X\subseteq[0,1]$, is it unique
up to homeomorphism?

As mentioned in the introduction, a significant part of this paper is devoted
to fully answering the first two of these questions. We record for now two
basic observations in this direction.

######  Observation 2.10.

If there is an $X$-EPG fan, then $X$ is closed.

######  Observation 2.11.

Let $X$ and $Y$ be closed subsets of $I$. The following statements are
equivalent:

  1. (1)

There is an order-preserving homeomorphism $h:I\to I$ such that $h[X]=Y$.

  2. (2)

Some $X$-EPG fan is a $Y$-EPG fan.

  3. (3)

Every $X$-EPG fan is a $Y$-EPG fan.

In light of this observation, we define the following topological invariant.

######  Definition 2.12.

Two subsets of $[0,1]$ are _equivalently embedded_ if there is an order-
preserving homeomorphism $I\to I$ mapping one onto the other. This is an
equivalence relation on subsets of $I$.

If $X\subseteq I$ and $F$ is an $X$-EPG fan, we define $EPG(F)$ to be the
equivalence class of $X$ under the “equivalently embedded” relation.

Abusing notation, we will write $EPG(F)=X$ when $F$ is an $X$-EPG fan. But it
is important to bear in mind that the property of being $X$-EPG does not
depend specifically on $X$, but rather on how $X$ is embedded in $[0,1]$.

Note that homeomorphic subsets of $I$ are not necessarily equivalently
embedded. For example, $X=\\{0\\}$, $Y=\\{\nicefrac{{1}}{{2}}\\}$, and
$Z=\\{1\\}$ are all homeomorphic to each other, but not equivalently embedded
in $I$. As we shall see, there are $X$-EPG and $Z$-EPG smooth fans, but they
look very different (the former have countably many blades and the latter
uncountably many, for example); and there are no $Y$-EPG smooth fans at all.

We make use in several places of Brouwer’s characterization of the Cantor
space: up to homeomorphism, the Cantor space is the unique nonempty zero-
dimensional compact metrizable space containing no isolated points (see, e.g.,
[14, Theorem 7.14, page 109]).

##  3\. Smooth fans from combs

This section presents a general way of describing smooth fans, which we later
use to build endpoint-homogeneous, $X$-EPG smooth fans for every $X$ such that
an $X$-EPG smooth fan exists. This technique has been used elsewhere, e.g. by
Aarts and Oversteegen to study the Lelek fan in [1]. We present it here
nonetheless, in order to keep our paper reasonably self-contained. Several of
the proofs in this section have been omitted on the grounds that they are
routine exercises in general topology.

######  Definition 3.1.

Let $\pi_{1}$ and $\pi_{2}$ denote the natural coordinate projection maps
$I\times I\to I$; that is, $\pi_{1}(x,y)=x$ and $\pi_{2}(x,y)=y$ whenever
$(x,y)\in I^{2}$.

Furthermore, if $C\subseteq I^{2}$ and $y\in I$, then
$C_{y}=\left\\{x\colon(x,y)\in C\right\\}=\pi_{2}^{-1}(y)\cap C$.

######  Definition 3.2.

Let $K\subseteq[0,1]$ denote the Cantor set. A _comb over $K$_ (or, in
context, simply a _comb_) is a closed subset $C\subseteq I^{2}$ such that

  1. (1)

$C_{0}=I$,

  2. (2)

$C_{y}$ is a subset of the Cantor set for all $y\in(0,1]$,

  3. (3)

if $y_{1}<y_{2}$ then $C_{y_{1}}\supseteq C_{y_{2}}$, and

  4. (4)

$|C_{y}|>2$ for some $y\in(0,1]$.

Given a comb $C$, let $F_{C}$ denote the space obtained by collapsing
$[0,1]\times\\{0\\}$ to a point. This space is called the _fan of the comb
$C$_.

Formally, we take $F_{C}=\\{t\\}\cup C\setminus[0,1]\times\\{0\\}$. In other
words, the points of $F_{C}$ are literally equal to the points of $C$ with the
$X$-axis removed, plus one extra point $t$, the top of the fan. The natural
quotient map is

$q_{C}(z)\,=\,\begin{cases}z&\text{ if }\pi_{2}(z)>0,\\\ t&\text{ if
}\pi_{2}(z)=0.\end{cases}$

$q_{C}$

######  Lemma 3.3.

There is a largest comb, namely $C_{K}=[0,1]\times\\{0\\}\cup K\times(0,1]$,
and the fan of this comb is (homeomorphic to) the Cantor fan.

###### Proof.

It is clear from the definition above that $C_{K}$ is a comb, and that every
other comb is a subset of $C_{K}$. Furthermore, the fan of $C_{K}$ is the cone
over $K$, i.e., the Cantor fan. ∎

######  Theorem 3.4.

If $C$ is a comb, then $F_{C}$ is a smooth fan. Conversely, if $F$ is a smooth
fan then there is a comb $C\subseteq I^{2}$ such that $F_{C}\approx F$.

###### Proof.

To prove the first assertion of the theorem, let $C$ be a comb. Then $C$ is a
closed subset of $C_{K}$ (the maximal comb described in the previous lemma),
and this implies that $F_{C}$ is a closed subset of the Cantor fan. Thus if
$F_{C}$ is a fan, then it must be a smooth fan.

To check that $F_{C}$ is a fan, first let us check that it is a continuum. As
a closed subspace of the Cantor fan, it is clearly compact and metrizable. We
claim $F_{C}$ is also path-connected. Let $z\in F_{C}$ with $z\neq t$. Note
that $z=(x,y)\in C$ for some $y\in(0,1]$. By condition $(3)$ in the definition
of a comb, $\\{x\\}\times[0,y]\subseteq C$. In $F_{C}$, this set is an arc
with endpoints $t$ and $z$. This means $F_{C}$ is path-connected, because if
$z,z^{\prime}\in F_{C}$, then there is an arc connecting $z$ to $t$ and an arc
connecting $t$ to $z^{\prime}$.

Observe that $F_{C}$ is hereditarily unicoherent, because it is a closed
subset of the Cantor fan, which is hereditarily unicoherent. Together with the
previous paragraph, this means $F_{C}$ is a dendroid.

By requirement $(4)$ in the definition of a comb, there is some $y>0$ with
$|C_{y}|>2$. If $z_{1}$, $z_{2}$, and $z_{3}$ are distinct points in $C_{y}$,
then

$[0,1]\times\\{0\\}\cup\\{z_{1}\\}\times(0,1]\cup\\{z_{2}\\}\times(0,1]\cup\\{z_{3}\\}\times(0,1]$

collapses to a triod in $F_{C}$ with top $t$. Hence $t$ is a ramification
point of $F_{C}$. There are no other ramification points of $F_{C}$, because
if $t^{\prime}$ were a ramification point of $F_{C}$ (i.e., it is the top of a
triod $T\subseteq F_{C}$), then it would be a ramification point of the Cantor
fan containing $F_{C}$; but the Cantor fan has only one ramification point.

Thus $F_{C}$ is a fan, and as we noted earlier, this means it is a smooth fan.

For the second assertion of the theorem, suppose $F$ is a smooth fan. By
Theorem 2.7, $F$ is a subcontinuum of the Cantor fan, which we may (and do)
view as the fan of the comb $C_{K}$ described in the previous lemma. Let
$C=(F\cap C_{K}\setminus\\{t\\})\cup[0,1]\times\\{0\\}$ (where $t$ denotes the
top of $C_{K}$). It is not difficult to check that $C$ is a comb over $K$, and
the fan of this comb is $F$. ∎

Observe that every comb is in fact a dendroid. In particular, we may (and do)
use the notation $[z,z^{\prime}]$ to denote the unique arc in a comb $C$
containing $z$ and $z^{\prime}$ as endpoints.

######  Definition 3.5.

Let $C$ be a comb, and suppose that $(x,y)\in C$ for some $y>0$, but
$(x,y^{\prime})\notin C$ for any $y^{\prime}>y$. Then $[(x,0),(x,y)]$ is
called a _blade_ of $C$, specifically the _blade of $C$ at $x$_, and $(x,y)$
is called the _tip_ of the blade. If $C$ has a blade at some $x\in K$, then
the blade at $x$ is denoted $B_{x}$.

The set of all blades of $C$ is denoted $\mathcal{B}(C)$, and the set of all
points that are the tip of some blade is denoted $\mathcal{E}(C)$.

Observe that $|\mathcal{B}(C)|\geq 3$, for any comb $C$, by part $(4)$ of the
definition of a comb. Of course combs have teeth, not blades, but the
following lemma shows why we have chosen this terminology.

######  Lemma 3.6.

Let $C\subseteq I^{2}$ be a comb. Then

  1. (1)

If $B\in\mathcal{B}(C)$ then $q_{C}[B]\in\mathcal{B}(F_{C})$, and if
$B^{\prime}\in\mathcal{B}(F_{C})$ then there is some $B\in\mathcal{B}(C)$ with
$q_{C}[B]=B^{\prime}$.

  2. (2)

$\mathcal{E}(C)=E(F_{C})$.

In other words, blades of $C$ correspond naturally to blades of $F_{C}$, and
the tips of those blades correspond naturally (in fact, are equal to) the
endpoints of $F_{C}$. Note that $\mathcal{E}(C)$ and $E(C)$ may not exactly be
equal, but they are nearly so: $\mathcal{E}(C)\subseteq
E(C)\subseteq\mathcal{E}(C)\cup\\{(0,0),(1,0)\\}$. In other words, it is
possible that $(0,0)$ and $(1,0)$ are endpoints of $C$, but these two points
are the only possible difference between $E(C)$ and $\mathcal{E}(C)$. Whether
these points are in $E(C)$ depends on whether $C$ has blades at $0$ or $1$.

######  Observation 3.7.

$\pi_{1}[\mathcal{E}(C)]\,=\,\left\\{x\in I\colon C\text{ has a blade at
}x\right\\}$.

We close this section with a lemma roughly stating that in order for $F_{C}$
to be an $X$-EPG fan (for some given $X\subseteq[0,1]$), the comb $C$ must
have an analogous property.

######  Lemma 3.8.

Let $C\subseteq I^{2}$ be a comb, and let $X\subseteq I$.

  1. (1)

If $0\notin X$, then $F_{C}$ is an $X$-EPG fan if and only if
$\overline{E(C)}\cap I\times\\{0\\}=\emptyset$, and for each
$B\in\mathcal{B}(C)$ there is a homeomorphism $\phi_{B}:I\to B$ such that
$\phi$ maps $1$ to the tip of $B$, and $\phi[X]=B\cap\overline{E(C)\setminus
B}$.

  2. (2)

If $0\in X$, then $F_{C}$ is an $X$-EPG fan if and only if
$\overline{E(C)}\cap I\times\\{0\\}\neq\emptyset$, and for each
$B\in\mathcal{B}(C)$ there is a homeomorphism $\phi_{B}:I\to B$ such that
$\phi$ maps $1$ to the tip of $B$, and
$\phi[X\setminus\\{0\\}]=B\cap\overline{E(C)\setminus
B}\setminus(I\times\\{0\\})$.

##  4\. Sets that do not work

In this section and the next, we exactly characterize the sets
$X\subseteq[0,1]$ such that there is an $X$-EPG smooth fan. As mentioned in
the introduction, the complete list is:

  * $\circ$

$X=\emptyset$, $\\{1\\}$, or $[0,1]$, or

  * $\circ$

$X$ is a closed subset of $[0,1]$ with $0\in X$ and $1\notin X$, or

  * $\circ$

$X$ is a closed subset of $[0,1]$ with $0,1\in X$, and $1$ is an isolated
point of $X$ (i.e., a set of the second type plus the point $1$).

In this section we prove that there is no $X$-EPG fan for any
$X\subseteq[0,1]$ not appearing on this list. In the next section we construct
examples of $X$-EPG fans for every $X$ that does appear on this list.

The following two lemmas rely on two well-known facts from general topology:
$(1)$ any relatively discrete subset of a compact metric space is countable,
and $(2)$ any nonempty, closed subset of the Cantor set $K$ with no isolated
points is homeomorphic to $K$.

######  Theorem 4.1.

Let $X\subseteq[0,1]$, and suppose that $F$ is an $X$-EPG fan. If $1\notin X$,
then $E(F)$ is countable.

###### Proof.

Note that $F$ is a compact metric space. Because $F$ is $X$-EPG and $1\notin
X$, no member of $E(F)$ is a limit point of $E(F)$; i.e., $E(F)$ is a
relatively discrete subset of $F$. Hence $E(F)$ is countable. ∎

######  Lemma 4.2.

Let $C$ be a comb, and suppose $F_{C}$ is an $X$-EPG smooth fan for some
$X\subseteq[0,1]$. If $0\notin X$ and $X\neq\emptyset$, then
$\pi_{1}[\mathcal{E}(C)]$ is homeomorphic to the Cantor space.

###### Proof.

Observe that every blade of $C$ contains exactly one member of
$\mathcal{E}(C)$. Consequently, $\pi_{1}$ restricts to an injection on
$\mathcal{E}(C)$. We claim that $\pi_{1}[\mathcal{E}(C)]$ is closed and has no
isolated points.

To see that it is closed, fix a sequence $\left\langle x_{n}\colon
n\in\mathbb{N}\right\rangle$ of points in $\pi_{1}[\mathcal{E}(C)]$. For each
$n$, there is some $y_{n}$ such that $(x_{n},y_{n})\in\mathcal{E}(C)$. Because
$C$ is compact, $\left\langle(x_{n},y_{n})\colon n\in\mathbb{N}\right\rangle$
has a subsequence converging to some $(x,y)\in C$. Because $0\notin X$, Lemma
3.8 implies $y>0$. But then $C$ has a blade at $x$, and the tip of that blade
is in $\mathcal{E}(C)$, hence $x\in\pi_{1}[\mathcal{E}(C)]$. Thus we have a
subsequence of $\left\langle x_{n}\colon n\in\mathbb{N}\right\rangle$
converging to $x\in\pi_{1}[\mathcal{E}(C)]$. This shows that every sequence in
$\pi_{1}[\mathcal{E}(C)]$ has a convergent subsequence, which means
$\pi_{1}[\mathcal{E}(C)]$ is closed.

To see that $\pi_{1}[\mathcal{E}(C)]$ has no isolated points, fix some
$x\in\pi_{1}[\mathcal{E}(C)]$, and fix $y>0$ with $(x,y)\in\mathcal{E}(C)$.
Because $X\neq\emptyset$, there is (by Lemma 3.8) some sequence of distinct
points $\left\langle(x_{n},y_{n})\colon n\in\mathbb{N}\right\rangle$ in
$\mathcal{E}(C)$ converging to $(x,y^{\prime})$ for some $y^{\prime}\leq y$.
Because $\pi_{1}$ is continuous and (as noted above) injective on
$\mathcal{E}(C)$, this means $\left\langle x_{n}\colon
n\in\mathbb{N}\right\rangle$ is a sequence of distinct points in
$\pi_{1}[\mathcal{E}(C)]$ converging to $\pi_{1}((x,y^{\prime}))=x$. Hence $x$
is not isolated in $\pi_{1}[\mathcal{E}(C)]$.

Thus $\pi_{1}[\mathcal{E}(C)]$ is a closed, nonempty subset of $K$ (the Cantor
set) with no isolated points. By Brouwer’s characterization of the Cantor
space, this implies $\pi_{1}[\mathcal{E}(C)]$ is homeomorphic to the Cantor
space. ∎

######  Theorem 4.3.

Let $X\subseteq[0,1]$, $X\neq\emptyset$, and suppose $F$ is an $X$-EPG smooth
fan. If $0\notin X$, then $E(F)$ is uncountable.

###### Proof.

Using Theorem 3.4, let $C$ be a comb such that $F\approx F_{C}$. By the
previous lemma, $\pi_{1}[\mathcal{E}(C)]$ is homeomorphic to the Cantor space,
which means $\mathcal{E}(C)$ is uncountable. But $E(F_{C})=\mathcal{E}(C)$, so
$E(F)$ is uncountable. ∎

######  Corollary 4.4.

If $X\subseteq[0,1]$, and there is an $X$-EPG smooth fan, then either
$X=\emptyset$, or else $0\in X$ or $1\in X$ (or both).

###### Proof.

By the preceding lemmas, if $F$ were an $X$-EPG smooth fan for any nonempty
$X\not\ni 0,1$, then $E(F)$ would be both countable and uncountable. ∎

This theorem answers Problem 3.41 from [12], which asked whether a smooth fan
can be $\\{\nicefrac{{1}}{{2}}\\}$-endpoint-generated.

Note that there are examples of $X$-EPG fans for sets $X$ satisfying each of
the four possibilities from Corollary 4.4:

  1. (1)

$X=\emptyset$: The simple $n$-od, for any $n\geq 3$, is $\emptyset$-EPG.

  2. (2)

$0,1\in X$: The Lelek fan is $[0,1]$-EPG.

  3. (3)

$1\in X$: The Cantor fan is $\\{1\\}$-EPG.

  4. (4)

$0\in X$: The star (i.e., the one point compactification of countably many
copies of $(0,1]$, drawn below) is $\\{0\\}$-EPG.

...

######  Theorem 4.5.

Suppose $X\subseteq[0,1]$ and there is an $X$-EPG smooth fan. If $0\notin X$
and $1\in X$, then $X=\\{1\\}$.

###### Proof.

Aiming for a contradiction, suppose that $F$ is an $X$-EPG smooth fan for some
$X$ with $0\notin X$, $1\in X$, and $X\neq\\{1\\}$. Applying Theorem 3.4, let
$C$ be a comb such that $F_{C}\approx F$.

For each $x$ such that $C$ has a blade at $x$, define

$\displaystyle e_{x}$ $\displaystyle\,=\,\max\left\\{y\colon(x,y)\in
B_{x}\cap\overline{\mathcal{E}(C)\setminus B_{x}}\right\\}$ $\displaystyle
m_{x}$ $\displaystyle\,=\,\min\left\\{y\colon(x,y)\in
B_{x}\cap\overline{\mathcal{E}(C)\setminus B_{x}}\right\\}.$

Because $1\in X$, the highest point at which
$\overline{\mathcal{E}(C)\setminus B}$ hits a blade $B$ is the tip of $B$. In
other words, $e_{x}$ is the $Y$-coordinate of the tip of $B_{x}$, and
$(x,e_{x})\in\mathcal{E}(C)$ (hence the choice of the letter $e$).

For each $n\in\mathbb{N}$, let

$L_{\nicefrac{{1}}{{n}}}\,=\,\left\\{x\colon
e_{x}-m_{x}\geq\nicefrac{{1}}{{n}}\right\\}.$

Lemma 3.8 implies that for each $B\in\mathcal{B}(C)$,
$B\cap\overline{\mathcal{E}(C)\setminus B}$ contains at least two points: the
tip of $B$ (because $1\in X$) and at least one other point of $B$ (because
$X\neq\\{1\\}$). Thus $e_{x}-m_{x}>0$ whenever $C$ has a blade at $x$.
Therefore

$\textstyle\bigcup_{n\in\mathbb{N}}L_{\nicefrac{{1}}{{n}}}\,=\,\left\\{x\colon
C\text{ has a blade at }x\right\\}\,=\,\pi_{1}[\mathcal{E}(C)].$

Because $0\notin X$ and $X\neq\emptyset$, Lemma 4.2 asserts that
$\pi_{1}[\mathcal{E}(C)]$ is homeomorphic to the Cantor space.

We claim the sets $L_{\nicefrac{{1}}{{n}}}$ are all closed. To see this, fix
$n\in\mathbb{N}$ and suppose $\left\langle x_{k}\colon
k\in\mathbb{N}\right\rangle$ is a sequence of points in
$L_{\nicefrac{{1}}{{n}}}$ converging to some point $x$; we wish to show $x\in
L_{\nicefrac{{1}}{{n}}}$. Because $\left\langle(x_{k},e_{x_{k}})\colon
k\in\mathbb{N}\right\rangle$ is a sequence of points in the compact set
$\overline{\mathcal{E}(C)}$, some subsequence
$\left\langle(x_{k_{i}},e_{x_{k_{i}}})\colon i\in\mathbb{N}\right\rangle$ of
this sequence converges to a point $(x,e)\in\overline{\mathcal{E}(C)}$. (Note
that this point must have first coordinate $x$, because
$x=\lim_{k\to\infty}x_{k}$.) Again,
$\left\langle(x_{k_{i}},m_{x_{k_{i}}})\colon i\in\mathbb{N}\right\rangle$ is a
sequence of points in $\overline{\mathcal{E}(C)}$, so some subsequence of this
sequence converges to a point $(x,m)\in\overline{\mathcal{E}(C)}$. Because
$e_{x_{k}}-m_{x_{k}}\geq\nicefrac{{1}}{{n}}$ for all $k$, we have
$e-m\geq\nicefrac{{1}}{{n}}$ as well. Because
$(x,m),(x,e)\in\overline{\mathcal{E}(C)}$, this implies $x\in
L_{\nicefrac{{1}}{{n}}}$ as claimed.

Because $\pi_{1}[\mathcal{E}(C)]$ is homeomorphic to the Cantor space, the
Baire Category Theorem asserts that it is not a countable union of closed sets
nowhere dense in $\pi_{1}[\mathcal{E}(C)]$. But
$\pi_{1}[\mathcal{E}(C)]=\bigcup_{n\in\mathbb{N}}L_{\nicefrac{{1}}{{n}}}$, and
each $L_{\nicefrac{{1}}{{n}}}$ is closed by the previous paragraph. It follows
that there is some $n\in\mathbb{N}$ such that $L_{\nicefrac{{1}}{{n}}}$ is not
nowhere dense in $\pi_{1}[\mathcal{E}(C)]$. Because $L_{\nicefrac{{1}}{{n}}}$
is closed in $\pi_{1}[\mathcal{E}(C)]$, this means $L_{\nicefrac{{1}}{{n}}}$
must have nonempty interior in $\pi_{1}[\mathcal{E}(C)]$.

Fix some nonempty, relatively open $U\subseteq\pi_{1}[\mathcal{E}(C)]$ such
that $U\subset L_{\nicefrac{{1}}{{n}}}$. Let $m=\inf\left\\{m_{x}\colon x\in
U\right\\}$, and fix some particular $x\in U$ such that
$m_{x}<m+\nicefrac{{1}}{{3n}}$. On the one hand, because
$(x,m_{x})\in\overline{\mathcal{E}(C)}$, there is some $x^{\prime}\in U$ such
that $(x^{\prime},e_{x^{\prime}})$ is within $\nicefrac{{1}}{{3n}}$ of
$(x,m_{x})$; in particular, $e_{x^{\prime}}\leq
m_{x}+\nicefrac{{1}}{{3n}}<m+\nicefrac{{2}}{{3n}}$. But $x^{\prime}\in
U\subseteq L_{\nicefrac{{1}}{{n}}}$, so this implies $m_{x^{\prime}}\leq
e_{x^{\prime}}-\nicefrac{{1}}{{n}}<m-\nicefrac{{1}}{{3n}},$ which contradicts
the definition of $m$. ∎

######  Corollary 4.6.

Up to homeomorphism, the Cantor fan is the only $\\{1\\}$-EPG smooth fan.
Furthermore, the Cantor fan is the only $X$-EPG smooth fan for a set $X$ with
$0\notin X$ and $1\in X$.

###### Proof.

If $F$ is a $\\{1\\}$-EPG smooth fan, then $E(F)$ is a closed subset of $F$
with no isolated points, and containing no arcs. By Brouwer’s characterization
of the Cantor space, this implies $E(F)$ is homeomorphic to the Cantor space.
By [13, Theorem 2.2], the Cantor fan is, up to homeomorphism, the unique
smooth fan in which $E(F)$ is homeomorphic to the Cantor space. This proves
the first statement of the corollary, and the second statement follows by
combining this with the previous theorem. ∎

The proof of the following theorem is similar to the proof of Theorem 4.5,
insofar as it employs the Baire Category Theorem in a similar way.

######  Theorem 4.7.

Suppose $X\subseteq[0,1]$ with $1\in X$, and there is an $X$-EPG smooth fan.
Then either $1$ is an isolated point of $X$, or $X=[0,1]$.

###### Proof.

Aiming for a contradiction, suppose that $F$ is an $X$-EPG smooth fan for some
$X\subseteq[0,1]$ such that $1\in X$, but $X\neq[0,1]$ and $1$ is not an
isolated point of $X$. By the previous theorem, if $0\notin X$ then
$X=\\{1\\}$ and we are done. Thus we may and do assume that $0\in X$. Applying
Theorem 3.4, let $C$ be a comb such that $F_{C}\approx F$.

Fix some $x_{0}$ such that $C$ has a blade at $x_{0}$. By Lemma 3.8, there is
a homeomorphism $h:I\to B_{x_{0}}$ mapping $X\setminus\\{0\\}$ onto
$(B_{x_{0}}\setminus\\{x_{0}\\})\cap\overline{\mathcal{E}(C)\setminus
B_{x_{0}}}$. In particular, because $0,1\in X$ but $X\neq[0,1]$, there is a
point of $(0,1)$ not in $X$, and the image of this point under $h$ is a point
of $B_{x_{0}}$ not in $\overline{\mathcal{E}(C)}$. Fix $y_{0}>0$ such that
$(x_{0},y_{0})\in B_{x_{0}}\setminus\overline{\mathcal{E}(C)}$.

Recall that $C$ is a comb over the Cantor set $K$, so $C\subseteq
I\times\\{0\\}\cup K\times I$. Because
$(x_{0},y_{0})\notin\overline{\mathcal{E}(C)}$, there is some basic clopen
$V\subseteq K$ with $x_{0}\in V$ and there is some $\varepsilon>0$ such that

$\big{(}V\times(y_{0}-\varepsilon,y_{0}+\varepsilon)\big{)}\cap\overline{\mathcal{E}(C)}\,=\,\emptyset.$

$x_{0}$$(x_{0},y_{0})$

Because $(x_{0},y_{0})\in B_{x_{0}}$, the tip of $B_{x_{0}}$ has
$Y$-coordinate $\geq y_{0}$. But the tip of $B_{x_{0}}$ is in
$\mathcal{E}(C)$, so this means its $Y$-coordinate is actually $\geq
y_{0}+\varepsilon$; i.e., the tip of $B_{x_{0}}$ sits above this rectangle.
Let $D$ denote the set of all $x$ such that there is a blade at $x$ with its
tip sitting above this rectangle: i.e.,

$D\,=\,\left\\{x\in V\colon\text{there is some $y\geq y_{0}+\varepsilon$ such
that }(x,y)\in C\right\\}.$

If $(x,y_{0})\in C$ for some $x$, then the tip of $B_{x}$ has $Y$-coordinate
$\geq\\!y_{0}+\varepsilon$, because
$\big{(}V\times(y_{0}-\varepsilon,y_{0}+\varepsilon)\big{)}\cap\overline{\mathcal{E}(C)}=\emptyset$.
Thus

$D\,=\,\left\\{x\in V\colon(x,y_{0})\in C\right\\}\,=\,C_{y_{0}}\cap V.$

In particular, this makes it clear that $D$ is a compact subset of $K$. Let

$E\,=\,\left\\{(x,y)\in\mathcal{E}(C)\colon x\in
D\right\\}\,=\,\mathcal{E}(C)\cap\big{(}V\times[y_{0},1]\big{)}.$

Because $1\in X$, every member of $E$ is a limit of points in
$\mathcal{E}(C)$. Because $V$ is clopen and
$\big{(}V\times(y_{0}-\varepsilon,y_{0}+\varepsilon)\big{)}\cap\mathcal{E}(C)=\emptyset$,
this implies $E$ has no isolated points. Because $D=\pi_{1}[E]$ and $\pi_{1}$
restricts to an injection on $E$, this implies $D$ has no isolated points.
Thus $D$ is a closed subset of the Cantor set $K$ with no isolated points.
Hence $D$ is homeomorphic to the Cantor space.

For each $x\in D$, define

$\displaystyle e_{x}$ $\displaystyle\,=\,\max\left\\{y\colon(x,y)\in
B_{x}\cap\overline{\mathcal{E}(C)\setminus B_{x}}\right\\}$ $\displaystyle
m_{x}$ $\displaystyle\,=\,\min\left\\{y\geq y_{0}\colon(x,y)\in
B_{x}\cap\overline{\mathcal{E}(C)\setminus B_{x}}\right\\}.$

Because $1\in X$, $(x,e_{x})$ is the tip of $B_{x}$. Because $1$ is not
isolated in $X$, Lemma 3.8 says that for each $x\in D$, $(x,e_{x})$ is not
isolated in $B_{x}\cap\overline{\mathcal{E}(C)\setminus B_{x}}$. If $x\in D$,
then $e_{x}>y_{0}$ (by the definition of $D$), and this implies that the set
$\left\\{y\geq y_{0}\colon(x,y)\in B_{x}\cap\overline{\mathcal{E}(C)\setminus
B_{x}}\right\\}$ contains points other than $e_{x}$. Hence $m_{x}\neq e_{x}$,
or equivalently $e_{x}-m_{x}>0$, for all $x\in D$.

Following the proof of the previous theorem, define

$L_{\nicefrac{{1}}{{n}}}\,=\,\left\\{x\colon
e_{x}-m_{x}\geq\nicefrac{{1}}{{n}}\right\\}$

for each $n\in\mathbb{N}$. By the previous paragraph,
$D=\bigcup_{n\in\mathbb{N}}L_{\nicefrac{{1}}{{n}}}$.

We claim that each of the sets $L_{\nicefrac{{1}}{{n}}}$ is closed. To see
this, fix $n\in\mathbb{N}$ and suppose $\left\langle x_{k}\colon
k\in\mathbb{N}\right\rangle$ is a sequence of points in
$L_{\nicefrac{{1}}{{n}}}$ converging to some point $x$; we wish to show $x\in
L_{\nicefrac{{1}}{{n}}}$. Because $\left\langle(x_{k},e_{x_{k}})\colon
k\in\mathbb{N}\right\rangle$ is a sequence of points in the compact set
$\big{(}V\times[y_{0},1]\big{)}\cap\overline{\mathcal{E}(C)}$, some
subsequence $\left\langle(x_{k_{i}},e_{x_{k_{i}}})\colon
i\in\mathbb{N}\right\rangle$ of this sequence converges to a point
$(x,e)\in\big{(}V\times[y_{0},1]\big{)}\cap\overline{\mathcal{E}(C)}$. (Note
that this point must have first coordinate $x$, because
$x=\lim_{k\to\infty}x_{k}$.) Again,
$\left\langle(x_{k_{i}},m_{x_{k_{i}}})\colon i\in\mathbb{N}\right\rangle$ is a
sequence of points in
$\big{(}V\times[y_{0},1]\big{)}\cap\overline{\mathcal{E}(C)}$, so a
subsequence of this sequence converges to a point
$(x,m)\in(V\times[y_{0},1])\cap\overline{\mathcal{E}(C)}$. Because
$e_{x_{k}}-m_{x_{k}}\geq\nicefrac{{1}}{{n}}$ for all $k$, we have
$e-m\geq\nicefrac{{1}}{{n}}$ as well. Because
$(x,m),(x,e)\in\big{(}V\times[y_{0},1]\big{)}\cap\overline{\mathcal{E}(C)}$,
this implies $x\in L_{\nicefrac{{1}}{{n}}}$ as claimed.

Because $D$ is homeomorphic to the Cantor space, the Baire Category Theorem
asserts that it is not a countable union of closed sets nowhere dense in $D$.
It follows that there is some $n\in\mathbb{N}$ such that
$L_{\nicefrac{{1}}{{n}}}$ is not nowhere dense in $D$, and therefore (since it
is closed) must have nonempty interior in $D$.

Fix a nonempty open $U\subseteq D$ such that $U\subset
L_{\nicefrac{{1}}{{n}}}$. Let $m=\inf\left\\{m_{x}\colon x\in U\right\\}$, and
fix some particular $x\in U$ such that $m_{x}<m+\nicefrac{{1}}{{3n}}$. On the
one hand, because $(x,m_{x})\in\overline{\mathcal{E}(C)}$, there is some
$x^{\prime}\in U$ such that $(x^{\prime},e_{x^{\prime}})$ is within
$\nicefrac{{1}}{{3n}}$ of $(x,m_{x})$; in particular, $e_{x^{\prime}}\leq
m_{x}+\nicefrac{{1}}{{3n}}<m+\nicefrac{{2}}{{3n}}$. But $x^{\prime}\in
U\subseteq L_{\nicefrac{{1}}{{n}}}$, so this implies $m_{x^{\prime}}\leq
e_{x^{\prime}}-\nicefrac{{1}}{{n}}<m-\nicefrac{{1}}{{3n}},$ contradicting our
choice of $m$. ∎

The results of this section are summarized as follows:

######  Corollary 4.8.

Suppose $X\subseteq[0,1]$ and there is an $X$-EPG fan. Then one of the
following must hold:

  1. (1)

$X=\emptyset$, $\\{1\\}$, or $[0,1]$, or

  2. (2)

$X$ is a closed subset of $[0,1]$ with $0\in X$ and $1\notin X$, or

  3. (3)

$X$ is a closed subset of $[0,1]$ with $0,1\in X$, and $1$ is an isolated
point of $X$.

##  5\. Constructing endpoint-homogeneous fans

In the previous section, we proved that for certain $X\subseteq[0,1]$ there
can be no $X$-EPG smooth fan. In this section, we show that for all
$X\subseteq[0,1]$ not ruled out by the results of the previous section, there
is an $X$-EPG smooth fan. In other words, the goal of this section is to
construct examples of $X$-EPG fans for every $X\subseteq[0,1]$ listed in
Corollary 4.8. Furthermore, we prove that all the examples we construct are in
fact endpoint-homogeneous.

If $X=\emptyset$, $\\{1\\}$, or $[0,1]$, as in the first bullet point, then we
have examples of $X$-EPG fans already:

$[0,1]$-EPG: the Lelek fan$\\{1\\}$-EPG: the Cantor fan$\emptyset$-EPG: the
simple $n$-ods$\dots$

All of these examples are endpoint-homogeneous. In fact, as mentioned in the
introduction, they are all $\frac{1}{3}$-homogeneous [2]. These examples are
also unique up to homeomorphism:

######  Observation 5.1.

Up to homeomorphism, the Cantor fan is the only $\\{1\\}$-EPG smooth fan, the
Lelek fan is the only $[0,1]$-EPG smooth fan, and the simple $n$-ods are the
only $\emptyset$-EPG smooth fans.

###### Proof.

For the Cantor fan, this follows from Corollary 4.6. For the Lelek fan, this
follows from Charatonik’s characterization of the Lelek fan in [10]. For the
$n$-ods, note that if $F$ is a $\emptyset$-EPG fan, then $E(F)$ has no
accumulation points. Because $F$ is compact, this implies $E(F)$ is finite,
which implies $F$ is a simple $n$-od. ∎

It remains to find $X$-EPG, endpoint-homogeneous fans for sets of type $(2)$
and $(3)$ in the list from Corollary 4.8. Recall from Theorems 4.1 and 4.3
that the $X$-EPG fans with $X$ of type $(2)$ should have countably many
blades, while the $X$-EPG fans with $X$ of type $(3)$ should have uncountably
many blades. We cover the case of countably many blades first.

######  Theorem 5.2.

If $X$ is a closed subset of $[0,1]$ with $1\notin X$, then there is an
$X$-EPG, endpoint-homogeneous smooth fan.

###### Proof.

If $X=\emptyset$ or $X=\\{0\\}$, then we already have examples of $X$-EPG
smooth fans: the simple $n$-ods for $X=\emptyset$, and the star for
$X=\\{0\\}$ (see the comments following Corollary 4.4). Furthermore, it is
clear that these examples are endpoint-homogeneous. Thus, for the remainder of
the proof, fix some closed $X\subseteq[0,1]$ such that $0\in X$, $1\notin X$,
and $X\cap(0,1)\neq\emptyset$.

Let $D$ be a countable dense subset of $X\setminus\\{0\\}$ (which is nonempty,
by the assumption of the previous paragraph), and let $\left\langle
y_{n}\colon n\in\mathbb{N}\right\rangle$ be an infinite sequence such that
$y_{n}\in D$ for every $n$, and for every $d\in D$ there are infinitely many
$n$ with $y_{n}=d$.

We now describe the construction of a comb over $K$ such that the
corresponding fan will be $X$-EPG and endpoint-homogeneous. To begin, let
$L^{0}=\\{0\\}\times[0,1]$. Let $M=\max(X)$, and note that $M<1$. Then:

  1. (1)

For each $n_{1}\in\mathbb{N}$, let

$L_{n_{1}}^{1}=\left\\{\frac{2}{3^{n_{1}}}\right\\}\times\big{[}0,y_{n_{1}}\big{]}.$

  2. (2)

For each $n_{1}$, $n_{2}\in\mathbb{N}$, let

$L_{n_{1},n_{2}}^{2}=\left\\{\frac{2}{3^{n_{1}}}+\frac{2}{3^{n_{1}+n_{2}}}\right\\}\times\big{[}0,M^{n_{1}}y_{n_{1}}y_{n_{2}}\big{]}.$

  3. (3)

For each $n_{1}$, $n_{2}$, $n_{3}\in\mathbb{N}$, let

$L_{n_{1},n_{2},n_{3}}^{3}=\left\\{\frac{2}{3^{n_{1}}}+\frac{2}{3^{n_{1}+n_{2}}}+\frac{2}{3^{n_{1}+n_{2}+n_{3}}}\right\\}\times\big{[}0,M^{n_{1}+n_{2}}y_{n_{1}}y_{n_{2}}y_{n_{3}}\big{]}.$

  4. (4)

More generally, for any given $k\in\mathbb{N}$ and for $n_{1}$,
$n_{2},\ldots,n_{k}\in\mathbb{N}$, let

$L_{n_{1},n_{2},\ldots,n_{k}}^{k}=\left\\{2\cdot\sum_{i=1}^{k}\left(\frac{1}{3}\right)^{\sum_{j=1}^{i}n_{j}}\right\\}\times\left[0\,,\,M^{\sum_{i=1}^{k-1}n_{i}}\cdot\prod_{i=1}^{k}y_{n_{i}}\right].$

For each $N\in\mathbb{N}$, let $C^{N}$ be the result of the first $N$ stages
of this construction, together with $I\times\\{0\\}$: i.e.,

$C^{N}\,=\,I\times\\{0\\}\cup
L^{0}\cup\bigcup\left\\{L^{k}_{n_{1},n_{2},\dots,n_{k}}\colon k\leq N\text{
and }(n_{1},n_{2},\dots,n_{k})\in\mathbb{N}^{k}\right\\}.$

Finally, let $C=\bigcup_{N\in\mathbb{N}}C^{N}$. This is illustrated below for
$X=\\{0,\nicefrac{{1}}{{2}},\nicefrac{{3}}{{4}}\\}$.

$C^{1}$$C^{2}$$C^{3}$$C$

Observe that the set of all $x\in I$ where this construction places a vertical
arc is exactly

$\textstyle\left\\{\sum_{i\in F}\frac{2}{3^{i}}\colon F\text{ is a finite
subset of }\mathbb{N}\right\\}.$

To put it more simply, this is just the set of all left-hand endpoints of the
Cantor set $K$. This is a countable dense subset of $K$.

Let us check that $C$ is closed. Let $\left\langle z_{n}\colon
n\in\mathbb{N}\right\rangle$ be a sequence of points in $C$; we wish to find a
convergent subsequence. If $\liminf_{n\to\infty}\pi_{2}(z_{n})=0$, then it is
not difficult to see that some subsequence converges to a point of
$I\times\\{0\\}$. So let us assume
$\liminf_{n\to\infty}\pi_{2}(z_{n})=\varepsilon>0$. The exponent on $M$ in the
definition of $L^{k}_{n_{1},n_{2},\dots,n_{k}}$ is at least $k-1$. Because
$M<0$, this means there is some $N\in\mathbb{N}$ such that
$M^{N}<\varepsilon$, which implies $\left\\{z_{n}\colon
n\in\mathbb{N}\right\\}\subseteq C^{N}.$ But it is not difficult to see that
$C^{N}$ is compact. So in this case, some subsequence of the $z_{n}$ converges
to a point of $C^{N}$.

Thus $C$ is closed, and it is easy to see that $C$ satisfies every other part
of Definition 3.2. Hence $C$ is a comb, and collapsing the bottom line of $C$
to a point yields a smooth fan $F_{C}$. To finish the proof, we must show that
this fan is an endpoint-homogeneous $X$-EPG fan.

For each vertical arc $L^{k}_{n_{1},\dots,n_{k}}$ produced by this
construction, let

$x^{k}_{n_{1},\dots,n_{k}}\,=\,\textstyle
2\cdot\sum_{i=1}^{k}\left(\frac{1}{3}\right)^{\sum_{j=1}^{i}n_{j}},$

i.e., $x^{k}_{n_{1},\dots,n_{k}}$ is the $X$-coordinate of the vertical arc
$L^{k}_{n_{1},\dots,n_{k}}$. Let

$e_{n_{1},\ldots,n_{k}}^{k}\,=\,\textstyle
M^{\sum_{i=1}^{k-1}n_{i}}\cdot\prod_{i=1}^{k}y_{n_{i}},$

or in other words, $e^{k}_{n_{1},\dots,n_{k}}$ is the $Y$-coordinate of the
tip of the blade at $x^{k}_{n_{1},\dots,n_{k}}$. Note that
$(x^{k}_{n_{1},\dots,n_{k}},e^{k}_{n_{1},\dots,n_{k}})\in\mathcal{E}(C)$. Let

$U^{k}_{n_{1},\dots,n_{k}}\,=\,\textstyle\Big{[}x^{k}_{n_{1},\dots,n_{k}}\,,\,x^{k}_{n_{1},\dots,n_{k}}+\left(\frac{1}{3}\right)^{\sum_{i=1}^{k}n_{i}}\Big{]}\cap
K.$

Because $x^{k}_{n_{1},\dots,n_{k}}$ is the left-hand endpoint of an interval
in the Cantor set $K$, this is simply the basic clopen subset of $K$ with
left-hand endpoint $x^{k}_{n_{1},\dots,n_{k}}$ and with width
$\left(\frac{1}{3}\right)^{\sum_{i=1}^{k}n_{i}}$. This particular width for
$U^{k}_{n_{1},\dots,n_{k}}$ is chosen precisely so as to make the following
observation true.

Observation: Let $x^{k}_{n_{1},\dots,n_{k}}$ and
$x^{\ell}_{m_{1},\dots,m_{\ell}}$ be two points at which $C$ has a blade. Then
$x^{\ell}_{m_{1},\dots,m_{\ell}}\in U^{k}_{n_{1},\dots,n_{k}}$ if and only if
$\ell\geq k$ and $(n_{1},n_{2},\dots,n_{k})=(m_{1},m_{2},\dots,m_{k})$ (but
$m_{k+1},\dots,m_{\ell}$ may be any natural numbers).

Let $C_{>0}=\bigcup_{y>0}C_{y}=C\cap(0,1]$. For each point
$x^{k}_{n_{1},\dots,n_{k}}$ at which $C$ has a blade, let
$V^{k}_{n_{1},\dots,n_{k}}=\big{(}U^{k}_{n_{1},\dots,n_{k}}\times(0,1]\big{)}\cap
C_{>0}$.

Claim: Let $x^{k}_{n_{1},\dots,n_{k}}$ be a point at which $C$ has a blade.
There is a homeomorphism
$h^{k}_{n_{1},\dots,n_{k}}:V^{k}_{n_{1},\dots,n_{k}}\to C_{>0}$ mapping the
tip of $B_{x^{k}_{n_{1},\dots,n_{k}}}$ to $(1,0)$.

_Proof of Claim:_ This claim is proved by exploiting the fractal-like self-
similarities of $C_{>0}$. The idea is simply that $V^{k}_{n_{1},\dots,n_{k}}$
is just a translated, dilated copy of $C_{>0}$, but where the dilation is not
quite even: the left-most blade of $V^{k}_{n_{1},\dots,n_{k}}$ is dilated by a
larger factor than every other blade.

To make this precise, define

$\displaystyle
m^{k}_{n_{1},\dots,n_{k}}\,=\,\max\left\\{y\colon(x^{k}_{n_{1},\dots,n_{k}},y)\in\overline{\mathcal{E}(C)\setminus
B_{x^{k}_{n_{1},\dots,n_{k}}}}\right\\}\textstyle\,=\,M^{\sum_{i=1}^{k}n_{i}}\cdot
M\cdot\prod_{i=1}^{k}y_{n_{i}}.$

In other words, this is the $Y$-coordinate of the highest point on
$B_{x^{k}_{n_{1},\dots,n_{k}}}$ (the left-most blade in
$V^{k}_{n_{1},\dots,n_{k}}$) that is in the closure of the other blades.
Define

$\displaystyle h^{k}_{n_{1},\dots,n_{k}}(x,y)$
$\displaystyle\,=\,\textstyle\Bigg{(}3^{\sum_{i=1}^{k}n_{i}}\Big{(}x-x^{k}_{n_{1},\dots,n_{k}}\Big{)}\,,\,\frac{1}{M^{\sum_{i=1}^{k}n_{i}}\cdot\prod_{i=1}^{k}y_{n_{i}}}y\Bigg{)}$
$\displaystyle\,=\,\textstyle\left(3^{\sum_{i=1}^{k}n_{i}}\Big{(}x-x^{k}_{n_{1},\dots,n_{k}}\Big{)}\,,\,\frac{M}{m^{k}_{n_{1},\dots,n_{k}}}y\right)$

for all $(x,y)\in V^{k}_{n_{1},\dots,n_{k}}$ such that $y\leq
m^{k}_{n_{1},\dots,n_{k}}$. This condition, that $y\leq
m^{k}_{n_{1},\dots,n_{k}}$, is true of every point of
$V^{k}_{n_{1},\dots,n_{k}}$ except for those in the topmost part of the blade
$B_{x^{k}_{n_{1},\dots,n_{k}}}$, specifically the points in
$\\{x^{k}_{n_{1},\dots,n_{k}}\\}\times(m^{k}_{n_{1},\dots,n_{k}},e^{k}_{n_{1},\dots,n_{k}}]$.
Define $h^{k}_{n_{1},\dots,n_{k}}$ on this part of this blade simply by
mapping it linearly onto $\\{0\\}\times(M,1]$. That is, define

$h^{k}_{n_{1},\dots,n_{k}}(x,y)\,=\,\textstyle\left(0\,,\,M+\frac{1-M}{e^{k}_{n_{1},\dots,n_{k}}-m^{k}_{n_{1},\dots,n_{k}}}\Big{(}y-m^{k}_{n_{1},\dots,n_{k}}\Big{)}\right)$

if $x=x^{k}_{n_{1},\dots,n_{k}}$ and $y>m^{k}_{n_{1},\dots,n_{k}}$.

$C_{>0}$$V^{k}_{n_{1},\dots,n_{k}}$

Observe that $h_{x^{k}_{n_{1},\dots,n_{k}}}$ maps the leftmost blade in in
$V^{k}_{n_{1},\dots,n_{k}}$, namely $B_{x^{k}_{n_{1},\dots,n_{k}}}$,
homeomorphically onto $\\{0\\}\times[0,1]$ via a piecewise-linear
homeomorphism.

If $B_{x^{\ell}_{m_{1},\dots,m_{\ell}}}$ is any other blade in
$V^{k}_{n_{1},\dots,n_{k}}$, then as observed above, $k\leq\ell$ and
$(n_{1},n_{2},\dots,n_{k})=(m_{1},m_{2},\dots,m_{k})$. But then

$\displaystyle
h^{k}_{n_{1},\dots,n_{k}}(x^{\ell}_{m_{1},\dots,m_{\ell}},e^{\ell}_{m_{1},\dots,m_{\ell}})$
$\displaystyle\,=\,\textstyle\Bigg{(}3^{\sum_{i=1}^{k}n_{i}}\Big{(}x^{\ell}_{m_{1},\dots,m_{\ell}}-x^{k}_{n_{1},\dots,n_{k}}\Big{)}\,,\,\frac{M}{m^{k}_{n_{1},\dots,n_{k}}}e^{\ell}_{m_{1},\dots,m_{k}}\Bigg{)}$
$\displaystyle\,=\,\textstyle\Bigg{(}3^{\sum_{i=1}^{k}n_{i}}\bigg{(}2\cdot\sum_{i=1}^{\ell}\left(\frac{1}{3}\right)^{\sum_{j=1}^{i}m_{j}}\,-\,2\cdot\sum_{i=1}^{k}\left(\frac{1}{3}\right)^{\sum_{j=1}^{i}n_{j}}\bigg{)}\,,\,$
$\displaystyle\textstyle\ \
\qquad\qquad\qquad\frac{M}{M^{\sum_{i=1}^{k}n_{i}}\cdot
M\cdot\prod_{i=1}^{k}y_{n_{i}}}M^{\sum_{i=1}^{\ell-1}m_{i}}\cdot\prod_{i=1}^{\ell}y_{m_{i}}\Bigg{)}$
$\displaystyle\,=\,\textstyle\Bigg{(}2\cdot\sum_{i=k+1}^{\ell}\left(\frac{1}{3}\right)^{\sum_{j=k+1}^{i}m_{j}}\,,\,M^{\sum_{i=k+1}^{\ell-1}m_{i}}\cdot\prod_{i=k+1}^{\ell}y_{m_{i}}\Bigg{)}$
$\displaystyle\,=\,\Big{(}x^{\ell-k}_{m_{k+1},\dots,m_{\ell}},e^{\ell-k}_{m_{k+1},\dots,m_{\ell}}\Big{)}.$

In other words, $h^{k}_{n_{1},\dots,n_{k}}$ maps the tip of the blade at
$x^{\ell}_{m_{1},\dots,m_{\ell}}$ to the tip of the blade at
$x^{\ell-k}_{m_{k+1},\dots,m_{\ell}}$. Because $h^{k}_{n_{1},\dots,n_{k}}$ is
just a scaling in the $Y$-coordinate, this means that
$h^{k}_{n_{1},\dots,n_{k}}$ maps the blade
$B_{x^{\ell}_{m_{1},\dots,m_{\ell}}}$ in $V^{k}_{n_{1},\dots,n_{k}}$ to the
blade $B_{x^{\ell-k}_{m_{k+1},\dots,m_{\ell}}}$ in $C_{>0}$. Thus every blade
of $V^{k}_{n_{1},\dots,n_{k}}$ maps to some blade of $C$, no two blades of
$V^{k}_{n_{1},\dots,n_{k}}$ map to the same blade of $C$, and every blade of
$C$ has some blade of $V^{k}_{n_{1},\dots,n_{k}}$ mapping onto it. Therefore
$h^{k}_{n_{1},\dots,n_{k}}$ is a bijection. And $h^{k}_{n_{1},\dots,n_{k}}$ is
clearly continuous in both directions, so $h^{k}_{n_{1},\dots,n_{k}}$ is a
homeomorphism as claimed. Furthermore, $h^{k}_{n_{1},\dots,n_{k}}$ maps the
endpoint of $B_{x^{k}_{n_{1},\dots,n_{k}}}$ to $(0,1)$. $\dashv$

Using this claim, we now show $F_{C}$ is endpoint-homogeneous. Suppose
$\bar{x}_{1}=x^{k}_{n_{1},\dots,n_{k}}$ and
$\bar{x}_{2}=x^{\ell}_{m_{1},\dots,m_{\ell}}$ are two different points at
which $C$ has a blade. By the claim above, there is a homeomorphism

$h\,=\,\big{(}h^{\ell}_{m_{1},\dots,m_{\ell}}\big{)}^{-1}\circ
h^{k}_{n_{1},\dots,n_{k}}:\,V^{k}_{n_{1},\dots,n_{k}}\to
V^{\ell}_{m_{1},\dots,m_{\ell}}$

that maps the tip of $B_{\bar{x}_{1}}$ to the tip of $B_{\bar{x}_{2}}$.
Because $h$ is a homeomorphism and $\bar{x}_{1}\neq\bar{x}_{2}$, there is a
clopen neighborhood $W$ of $B_{\bar{x}_{1}}$ such that $W\cap h[W]=\emptyset$.
Shrinking $W$ if necessary, we may (and do) assume that $W$ has the form
$V\times(0,1]$ for some basic clopen $V\subseteq K$. This implies $h[W]$ is
also a clopen subset of $K$. Now define a homeomorphism $H:F_{C}\to F_{C}$ as
follows:

$H(z)\,=\,\begin{cases}h(z)&\text{ if }z\in W,\\\ h^{-1}(z)&\text{ if }z\in
h[W],\\\ z&\text{ otherwise.}\end{cases}$

Because $W$ and $h[W]$ are both relatively clopen subsets of
$F_{C}\setminus\\{t\\}$, it is clear that
$H\\!\restriction\\!(F_{C}\setminus\\{t\\})$ is a homeomorphism
$F_{C}\setminus\\{t\\}\to F_{C}\setminus\\{t\\}$. But then this homeomorphism
extends (uniquely) to the one-point compactification of
$F_{C}\setminus\\{t\\}$, namely $F_{C}$, by mapping $t$ to $t$; this extension
is $H$. So $H$ is a homeomorphism $F_{C}\to F_{C}$ mapping the tip of the
blade at $\bar{x}_{1}$ to the tip of the blade at $\bar{x}_{2}$. Thus, given
two arbitrary points $\bar{x}_{1}$ and $\bar{x}_{2}$ at which $C$ has a blade,
we have found a homeomorphism $F_{C}\to F_{C}$ mapping the corresponding
endpoints to one another. Hence $F_{C}$ is endpoint-homogeneous.

It remains only to check that $F_{C}$ is $X$-EPG. Because we already know
$F_{C}$ is endpoint-homogeneous (so all blades are topologically
indistinguishable in the sense of Lemma 2.5), it suffices to show for a single
$B\in\mathcal{B}(F_{C})$ that there is a homeomorphism $\phi:I\to B$ such that
$\phi(0)=t$ and $\phi[X]=\overline{E(F_{C})\setminus B}$. Let $B=\\{0\\}\times
I$, the leftmost blade of $C$, and let $\phi$ be the natural homeomorphism
$\phi(z)=(0,z)$.

Because the tip of the blade at $x^{k}_{n_{1},\dots,n_{k}}$ has height
$\leq\\!M^{k}$, it is clear that $\overline{\mathcal{E}(C)}$ contains points
of $I\times\\{0\\}$ (in $C$), which means $t\in\overline{E(F_{c})\setminus B}$
(in $F_{C}$). Hence $\phi(0)=t\in\overline{E(F_{c})\setminus B}$.

Next let $x\in X\setminus\\{0\\}$. By our choice of the sequence $\left\langle
y_{n}\colon n\in\mathbb{N}\right\rangle$ in the construction, there is a
subsequence $\left\langle y_{n_{k}}\colon k\in\mathbb{N}\right\rangle$ such
that $\lim_{k\to\infty}y_{n_{k}}=x$. But then
$\left\langle(x^{1}_{n_{k}},e^{1}_{n_{k}})\colon
k\in\mathbb{N}\right\rangle=\left\langle(x^{1}_{n_{k}},y_{n_{k}})\colon
k\in\mathbb{N}\right\rangle$ is a sequence of points in
$\mathcal{E}(C)\setminus B_{0}$ converging to $(0,x)=\phi(x)$. Combined with
the previous paragraph, this shows
$\phi[X]\subseteq\overline{E(F_{C})\setminus B}$.

For the reverse inclusion, fix some $z\in B\cap\overline{E(F_{C})\setminus
B}$. Either $z=t$, or else $z=(0,y_{z})$ for some $y_{z}\in[0,1]$. Let
$\left\langle(\bar{x}_{k},\bar{e}_{k})\colon k\in\mathbb{N}\right\rangle$ be a
sequence in $E(F_{C})\setminus B=\mathcal{E}(C)\setminus B$ converging to $z$.
This implies $\lim_{k\to\infty}\bar{x}_{k}=0$, and we would like to show
$\lim_{k\to\infty}\bar{e}_{k}\in X$, since this would mean
$z=(0,\lim_{k\to\infty}\bar{e}_{k})\in\phi[X]$. We consider two cases.

For the first case, suppose there are infinitely many $k$ such that
$\bar{x}_{k}=x^{1}_{n_{k}}$ for some $n_{k}$. Then by construction,
$(\bar{x}_{k},\bar{e}_{k})=(x^{1}_{n_{k}},y_{n_{k}})$ for infinitely many $k$.
But then $\lim_{k\to\infty}\bar{y}_{n}\in\overline{\left\\{y_{n}\colon
n\in\mathbb{N}\right\\}}=X$.

For the second case, suppose there are only finitely many $k$ such that
$\bar{x}_{k}=x^{1}_{n_{k}}$. So, for all sufficiently large $k$,
$\bar{x}_{k}=x^{\ell}_{n^{k}_{1},n^{k}_{2},\dots,n^{k}_{\ell}}$ for some
$\ell>1$ and some $n^{k}_{1},n^{k}_{2},\dots,n^{k}_{\ell}\in\mathbb{N}$. But
$\bar{x}_{k}=x^{\ell}_{n^{k}_{1},n^{k}_{2},\dots,n^{k}_{\ell}}\geq\frac{2}{3^{n^{k}_{1}}}$
by construction. Because $\lim_{k\to\infty}\bar{x}_{k}=0$, this implies
$\lim_{k\to\infty}n^{k}_{1}=\infty$. But also $\bar{e}_{k}\leq M^{n^{1}_{k}}$
by construction, so this implies $\lim_{k\to\infty}\bar{e}_{k}=0\in X$.

In either case $z\in\phi[X]$, so $\phi[X]\supseteq
B\cap\overline{E(F_{C})\setminus B}$ as claimed. Hence
$\phi[X]=B\cap\overline{E(F_{C})\setminus B}$, and $F_{C}$ is an $X$-EPG fan.
∎

This concludes the proof that for sets $X$ of type $(1)$ or $(2)$ in Corollary
4.8, there is an endpoint-homogeneous, $X$-EPG smooth fan. It remains to show
that the same is true for $X$ of type $(3)$. As it turns out, this is a
relatively easy consequence of the previous theorem.

######  Theorem 5.3.

Suppose $X$ is a closed subset of $[0,1]$ with $0,1\in X$, and $1$ is an
isolated point of $X$. Then there is an endpoint-homogeneous, $X$-EPG smooth
fan.

###### Proof.

Because $1$ is an isolated point of $X$, $Y=X\setminus\\{1\\}$ is a closed
subset of $[0,1]$ with $1\notin Y$. By the previous theorem, there is an
endpoint-homogeneous, $Y$-EPG smooth fan $F_{Y}$. Let $t_{0}$ denote the top
of $F_{Y}$.

Let $F$ denote the quotient space obtained from $F_{Y}\times K$ (where $K$
denotes the Cantor set) by collapsing $\\{t_{0}\\}\times K$ to a single point.
(Equivalently, one may view $F$ as the one-point compactification of
$(F_{Y}\setminus\\{t_{0}\\})\times K$.) Let $t$ denote the top of $F$. For
convenience, we identify $F\setminus\\{t\\}$ with
$(F_{Y}\setminus\\{t_{0}\\})\times K$.

Because $F_{Y}$ is a smooth fan, it is a subcontinuum of the Cantor fan
$F_{K}$ (with top $t_{K}$). This means $F$ is a subset of the quotient space
obtained from $F_{K}\times K$ by collapsing $\\{t_{K}\\}\times K$ to a single
point (or, equivalently, the one-point compactification of
$(F_{K}\setminus\\{t_{K}\\})\times K$). This latter space is homeomorphic to
$F_{K}$, because $K\times K\approx K$. Thus $F$ is a subset of a Cantor fan.

It is straightforward to check that $F$ is path-connected, compact, and
contains a copy of $F_{Y}$. Combined with the previous paragraph, this means
$F$ is a smooth fan. Furthermore, it is clear that $E(F)=E(F_{Y})\times K$.

Let $e,e^{\prime}\in E(F)$. Then $e=(e_{1},e_{2})$ and
$e^{\prime}=(e_{1}^{\prime},e_{2}^{\prime})$, where $e_{1},e_{1}^{\prime}\in
E(F_{Y})$ and $e_{2},e_{2}^{\prime}\in K$. Because $F_{Y}$ is endpoint-
homogeneous, there is a homeomorphism $h_{1}:F_{Y}\to F_{Y}$ with
$h_{1}(e_{1})=e_{1}^{\prime}$. Because $K$ is homogeneous, there is a
homeomorphism $h_{2}:K\to K$ with $h_{2}(e_{2})=e_{2}^{\prime}$. But then

$H(z)\,=\,\begin{cases}(h_{1}(x),h_{2}(y))&\text{ if
}z=(x,y)\in(F_{Y}\setminus\\{t_{0}\\})\times K,\\\ t&\text{ if
}z=t\end{cases}$

is a homeomorphism $F\to F$ with $H(e)=e^{\prime}$. So $F$ is endpoint-
homogeneous.

Lastly, we must show that $F$ is an $X$-EPG fan. Fix $B\in\mathcal{B}(F)$.
Observe that $B=B_{0}\times\\{c\\}$ for some $B_{0}\in\mathcal{B}(F_{Y})$ and
$c\in K$, provided that we are willing to abuse notation slightly and identify
$(t_{0},c)$ with $t$. Because $F_{Y}$ is $Y$-EPG, there is a homeomorphism
$\phi:I\to B$ with $\phi(0)=t$ such that
$\phi[Y]=B\cap\overline{E(F_{Y})\setminus B}$. We claim that
$\phi[Y]=B\cap\overline{E(F)\setminus B}$.

On the one hand, $F_{Y}\times\\{c\\}$ is a homeomorphic copy of $F_{Y}$
sitting inside of $F$, and the endpoints of this copy of $F_{Y}$ are all
endpoints of $F$. Therefore

$B\cap\overline{E(F)\setminus
B}\,\supseteq\,B\cap\overline{E(F_{Y}\times\\{c\\})\setminus B}\,=\,\phi[Y].$

But also $\phi(1)$, the endpoint of $B$, is in $\overline{E(F)\setminus B}$,
because it is a limit point of $\\{\phi(1)\\}\times K\subseteq E(F)$. Hence
$B\cap\overline{E(F)\setminus B}\supseteq\phi[Y]\cup\\{\phi(1)\\}=\phi[X]$.

On the other hand, suppose $x\in B$ and $x\notin\phi[X]$. Fix $x_{0}\in B_{0}$
such that $x=(x_{0},c)$. Then $x_{0}\notin\phi[Y]$, and $x_{0}$ is not the
endpoint of $B_{0}$ (because $x\notin\phi[X]$), so there is an open
neighborhood $U\subseteq F_{Y}$ with $x_{0}\in U$ such that $U\cap
E(F_{Y})=\emptyset$. But then $U\times K$ is an open neighborhood of
$(x_{0},c)$ in $F$ such that $(U\times K)\cap E(F)=(U\times
K)\cap(E(F_{Y})\times K)=\emptyset$. Hence $x\notin\overline{E(F)\setminus
B}$. ∎

######  Corollary 5.4.

Let $X\subseteq[0,1]$. The following are equivalent:

  1. (1)

There is an endpoint-homogeneous, $X$-EPG smooth fan.

  2. (2)

There is an $X$-EPG smooth fan.

  3. (3)

$X$ satisfies one of the conditions listed in Corollary 4.8.

######  Theorem 5.5.

There are uncountably many distinct homeomorphism types of endpoint-
homogeneous smooth fans.

###### Proof.

In light of the previous Corollary 5.4 and Observation 2.11, it suffices to
show that there is an uncountable family of subsets of $[0,1]$ such that every
member of the family satisfies the three statements in Corollary 5.4, but no
two of them are equivalently embedded in $[0,1]$.

For each $n\in\mathbb{N}$, fix some $a_{n},b_{n}$ such that
$\frac{1}{n+1}<a_{n}<b_{n}<\frac{1}{n}$. Then, for each
$A\subseteq\mathbb{N}$, let

$X_{A}\,=\,\textstyle\\{0\\}\cup\left\\{\frac{1}{n}\colon
n\in\mathbb{N}\right\\}\cup\left\\{[a_{n},b_{n}]\vphantom{f^{f}}\colon n\in
A\right\\}.$

The family of sets $\left\\{X_{A}\colon A\subseteq\mathbb{N}\right\\}$ has the
required properties. ∎

##  6\. $\frac{1}{n}$-homogeneity

Recall from the introduction that a space $X$ is _$\frac{1}{n}$ -homogeneous_
if there is a partition of $X$ into $n$ sets (but no fewer) such that if
$x,y\in X$ lie in the same partition piece, then there is a homeomorphism
$h:X\to X$ with $h(x)=y$.

In [2], Acosta, Hoehn, and Juárez completely classify all
$\frac{1}{3}$-homogeneous smooth fans: they are the simple $n$-ods, the Cantor
fan, the Lelek fan, and the $\\{0\\}$-EPG and $\\{0,1\\}$-EPG fans constructed
in the previous section.

Cantor fanLelek fansimple $n$-ods$\dots$...the $\\{0\\}$-EPG smooth fan...the
$\\{0,1\\}$-EPG smooth fan

Acosta, Hoehn, and Juárez also prove in [2] (see Theorem 4.1) that no smooth
fan is $\frac{1}{4}$-homogeneous. The main result of this section is that for
every $n\geq 5$, there is an endpoint-homogeneous, $\frac{1}{n}$-homogeneous
smooth fan. No new fans are constructed though: rather, we show that for
certain choices of $X$ (depending on $n$), the $X$-EPG fans constructed in the
proof of Theorem 5.2 are $\frac{1}{n}$-homogeneous. The sets $X$ used for odd
$n$ and for even $n$ are rather different. The odd case is simpler, and we
handle it first.

######  Theorem 6.1.

There is an endpoint-homogeneous, $\frac{1}{n}$-homogeneous smooth fan for all
odd $n\geq 5$.

###### Proof.

Let $n$ be an odd number $\geq\\!5$, and let $m=\frac{n-3}{2}$. Fix some
$a_{1},a_{2},\dots,a_{m}$ such that $0<a_{1}<\dots<a_{m}<1$, and let
$X=\\{0,a_{1},\dots,a_{m}\\}$. We claim that the $X$-EPG fan constructed in
the proof of Theorem 5.2 is $\frac{1}{n}$-homogeneous. Let $C$ denote the comb
of this fan, as constructed in the proof of Theorem 5.2, and let $F_{C}$
denote the fan itself. We know already that $F_{C}$ is an endpoint-homogeneous
smooth fan, so we need only show it is $\frac{1}{n}$-homogeneous.

Note that $X$ naturally divides $[0,1]$ into $n$ sets:

$\displaystyle T$ $\displaystyle\,=\,\\{0\\},$ $\displaystyle E$
$\displaystyle\,=\,\\{1\\},$ $\displaystyle A_{1}$
$\displaystyle\,=\,\\{a_{1}\\},\ A_{2}\,=\,\\{a_{2}\\},\dots,\
A_{m}\,=\,\\{a_{m}\\},$ $\displaystyle D_{1}$ $\displaystyle\,=\,(0,a_{1}),\
D_{2}\,=\,(a_{1},a_{2}),\dots,\ D_{m}\,=\,(a_{m-1,}a_{m}),\
D_{m+1}\,=\,(a_{m},1).$

$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$0$$a_{1}$$a_{2}$$\cdots$$a_{m-1}$$a_{m}$$T$$A_{1}$$A_{2}$$\cdots$$A_{m-1}$$A_{m}$$D_{1}$$D_{2}$$D_{m}$$D_{m+1}$$E$

For every $B\in\mathcal{B}(F_{C})$, there is a homeomorphism $\phi_{B}:I\to B$
mapping $1$ to the tip of $B$ such that
$\phi_{B}[X]=B\cap\overline{E(F_{C})\setminus B}$. Partition $F_{C}$ into $n$
classes of points by taking

$P_{Z}\,=\,\bigcup_{B\in\mathcal{B}(F_{C})}\phi_{B}^{-1}[Z]$

where $Z$ is any of the classes listed above, $T$, $E$, $A_{1},\dots,A_{m}$,
$D_{1},\dots,D_{m+1}$. It is clear that the $P_{Z}$ partition $F_{C}$ into $n$
sets, and we claim that this partition witnesses the $\frac{1}{n}$-homogeneity
of $F_{C}$.

First, it is not difficult to see that these classes can be defined
topologically. For example, $T=\\{t\\}$ (the top of $F_{C}$) and $E=E(F_{C})$.
Given $i\leq m$, $A_{i}$ consists of all those points $x\in F_{C}$ that are
limits of endpoints, and such that the arc $[t,x]$ contains $i+1$ limits of
endpoints (counting $t$ and $x$). Given $i\leq m+1$, $D_{i}$ consists of all
those points $x\in F_{C}$ that are not limits of endpoints, and such that the
arc $[t,x]$ contains $i$ limits of endpoints.

Because the $P_{Z}$ admit this kind of topological definition, every
homeomorphism $F_{C}\to F_{C}$ preserves each of the $P_{Z}$. It remains to
show that if $x,y\in P_{Z}$ for some particular $Z$, then there is a
homeomorphism $h:F_{C}\to F_{C}$ with $h(x)=y$.

This is obvious for $Z=T$, and it is true for $Z=E$ because $F_{C}$ is
endpoint-homogeneous.

Fix $i\leq m$, and let $x,y\in A_{i}$. Because $F_{C}$ is endpoint-
homogeneous, there is a homeomorphism $h:F_{C}\to F_{C}$ mapping the blade
containing $x$ to the blade containing $y$. But $P_{A_{i}}$ contains exactly
one point of every blade, and $h$ preserves $P_{A_{i}}$, so this implies
$h(x)=y$.

It remains to consider the $D_{i}$. Fix $i\leq m+1$. Instead of considering
two arbitrary members of $D_{i}$, we begin by consider two members of
$P_{D_{i}}$ on $B_{0}$, the leftmost blade of $C$. Fix $z_{1}=(0,y_{1})$ and
$z_{2}=(0,y_{2})$ in $P_{D_{i}}\cap B_{0}$. Because $[z_{1},z_{2}]\subseteq
P_{D_{i}}$ and $P_{D_{i}}\cap\overline{\mathcal{E}(C)}=\emptyset$, there is an
open $U\supseteq[z_{1},z_{2}]$ such that $U\cap\mathcal{E}(C)=\emptyset$.
Shrinking $U$ if necessary (and relabelling $z_{1}$ and $z_{2}$ so that
$z_{1}$ is between $t$ and $z_{2}$), we may (and do) assume that

$U\,=\,V\times(y_{1}-\varepsilon\,,\,y_{2}+\varepsilon)$

for some clopen $V\subseteq K$ containing $0$. There is a homeomorphism
$h_{0}:I\to I$ such that $h_{0}(y_{1})=y_{2}$ and $h_{0}(y)=y$ for all
$y\notin(y_{1}-\varepsilon,y_{2}+\varepsilon)$. But then

$h(z)\,=\,\begin{cases}\big{(}x,h_{0}(y)\big{)}&\text{ if }z=(x,y)\in U,\\\
z&\text{ if not}\end{cases}$

is a homeomorphism $F_{C}\to F_{C}$ mapping $z_{1}$ to $z_{2}$.

Thus for any two points of $P_{D_{i}}\cap B_{0}$, there is a homeomorphism
$F_{C}\to F_{C}$ mapping one to the other. But also, because $F_{C}$ is
endpoint-homogeneous and $P_{D_{i}}$ is homeomorphism-invariant, for every
$x\in P_{D_{i}}$ there is a homeomorphism $F_{C}\to F_{C}$ mapping it to a
point of $B_{0}$. By composing homeomorphisms, it follows that if $x,y\in
P_{D_{i}}$ then there is a homeomorphism $F_{C}\to F_{C}$ mapping $x$ to $y$.
∎

Observe that not all $\frac{1}{n}$-homogeneous fans need be smooth, and not
all $\frac{1}{n}$-homogeneous smooth fans need be endpoint-homogeneous. The
picture following Definition 2.6 shows two non-smooth fans; the one on the
left is $\frac{1}{7}$-homogeneous and the one on the right is
$\frac{1}{5}$-homogeneous. Drawn below is a $\frac{1}{5}$-homogeneous smooth
fan that is not endpoint-homogeneous.

......

######  Theorem 6.2.

There is an endpoint-homogeneous, $\frac{1}{n}$-homogeneous smooth fan for all
$n\geq 5$.

###### Proof.

In light of the previous theorem, it suffices to show that there is an
endpoint-homogeneous, $\frac{1}{n}$-homogeneous smooth fan for all even $n\geq
6$.

Let $\left\\{p_{i}\colon i\in\mathbb{Z}\right\\}$ be a subset of $[0,1]$ such
that $p_{i}<p_{j}$ whenever $i<j$, and $\lim_{i\to-\infty}p_{i}=0$, and
$\lim_{i\to\infty}p_{i}<1$. Let $a_{1}=\lim_{i\to\infty}p_{i}$, and if $n\geq
8$ fix some $a_{2}<\dots<a_{m}<1$, where $m=\frac{n-4}{2}$. Let

$X\,=\,\left\\{p_{i}\colon
i\in\mathbb{Z}\right\\}\cup\\{0,a_{1}\\}\cup\\{a_{2},\dots,a_{m}\\}.$

This set $X$ is illustrated below for $n=10$.

$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$\bullet$$p_{{}_{0}}$$p_{{}_{-1}}$$p_{{}_{1}}$$p_{{}_{-2}}$$p_{{}_{2}}$$a_{1}$$0$$\cdots$$\cdots$$\bullet$$a_{2}$$\bullet$$a_{3}$

We claim that the $X$-EPG fan constructed in the proof of Theorem 5.2 is
$\frac{1}{n}$-homogeneous. Let $C$ denote the comb of this fan, as constructed
in the proof of Theorem 5.2, and let $F_{C}$ denote the fan itself. We know
already that $F_{C}$ is an endpoint-homogeneous smooth fan, so we need only
show it is $\frac{1}{n}$-homogeneous.

Not unlike in the previous proof, $X$ naturally divides $[0,1]$ into $n$ sets:

$\displaystyle T$ $\displaystyle\,=\,\\{0\\},$ $\displaystyle E$
$\displaystyle\,=\,\\{1\\},$ $\displaystyle L$
$\displaystyle\,=\,\left\\{p_{i}\colon i\in\mathbb{Z}\right\\},$
$\displaystyle G$
$\displaystyle\,=\,\textstyle\bigcup\left\\{(p_{i},p_{i+1})\colon
i\in\mathbb{Z}\right\\}$ $\displaystyle A_{1}$
$\displaystyle\,=\,\\{a_{1}\\},\ A_{2}\,=\,\\{a_{2}\\},\ \dots,\
A_{m-1}\,=\,\\{a_{m-1}\\},\ A_{m}\,=\,\\{a_{m}\\},$ $\displaystyle D_{1}$
$\displaystyle\,=\,(a_{1},a_{2}),\ D_{2}\,=\,(a_{2},a_{3}),\ \dots,\
D_{m-1}\,=\,(a_{m-1},a_{m}),\ D_{m}\,=\,(a_{m},1).$

For every $B\in\mathcal{B}(F_{C})$, there is a homeomorphism $\phi_{B}:I\to B$
mapping $1$ to the tip of $B$ such that
$\phi_{B}[X]=B\cap\overline{E(F_{C})\setminus B}$. Partition $F_{C}$ into $n$
classes of points by taking

$P_{Z}\,=\,\bigcup_{B\in\mathcal{B}(F_{C})}\phi_{B}^{-1}[Z],$

where $Z$ is any of the classes listed above, $T$, $E$, $P$, $I$,
$A_{1},\dots,A_{m}$, $D_{1},\dots,D_{m}$. It is clear that these sets $P_{Z}$
form a partition $F_{C}$ into $n$ sets, and we claim that this partition
witnesses the $\frac{1}{n}$-homogeneity of $F_{C}$.

As in the previous proof, it is not difficult to see that each of the
partition pieces $P_{Z}$ admits a purely topological definition. Explicitly:
$T=\\{t\\}$ is just the top of $F_{C}$; $E=E(F_{C})$; $L$ is the set of all
those $x$ that are a limit of endpoints, and such that
$[t,x]\cap\overline{\mathcal{E}(C)\setminus[t,x]}$ is a decreasing convergent
sequence; $G$ is defined similarly, except that $x$ is not a limit of
endpoints; $A_{i}$ is all those $x$ that are a limit of endpoints, and such
that, if $e$ denotes the endpoint on the blade of $x$, then $[x,e]$ contains
$m-i$ limits of endpoints; $D_{i}$ is defined similarly, except that $x$ is
not a limit of endpoints.

Because each of the $P_{Z}$ admits a purely topological definition, every
homeomorphism $F_{C}\to F_{C}$ preserves each of the $P_{Z}$. It remains to
show that if $x,y\in P_{Z}$ for some particular $Z$, then there is a
homeomorphism $h:F_{C}\to F_{C}$ with $h(x)=y$.

This is obvious for $Z=T$, and it is true for $Z=E$ because $F_{C}$ is
endpoint-homogeneous. The case $Z=A_{i}$ (for any $i\leq m$) follows from the
endpoint-homogeneity of $F_{C}$ as in the previous proof. To be explicit, fix
$i\leq m$ and let $x,y\in P_{A_{i}}$. Because $F_{C}$ is endpoint-homogeneous,
there is a homeomorphism $F_{C}\to F_{C}$ mapping the blade containing $x$ to
the blade containing $y$. Because $P_{A_{i}}$ contains exactly one point of
every blade, and $h$ preserves $P_{A_{i}}$, this homeomorphism must map $x$ to
$y$.

Like with the $A_{i}$, the argument for the $D_{i}$ is essentially the same as
in the proof of the previous theorem. Just as we did there, we begin here by
consider two members of $P_{D_{i}}$ on $B_{0}$, the leftmost blade of $C$. Fix
$z_{1}=(0,y_{1})$ and $z_{2}=(0,y_{2})$ in $P_{D_{i}}\cap B_{0}$. Because
$[z_{1},z_{2}]\subseteq P_{D_{i}}$ and
$P_{D_{i}}\cap\overline{\mathcal{E}(C)}=\emptyset$, there is an open
neighborhood $U\subseteq C$ of $[z_{1},z_{2}]$ such that
$U\cap\mathcal{E}(C)=\emptyset$. Shrinking $U$ if necessary (and relabelling
$z_{1}$ and $z_{2}$ so that $z_{1}$ is between $t$ and $z_{2}$), we may (and
do) assume that

$U\,=\,V\times(y_{1}-\varepsilon\,,\,y_{2}+\varepsilon)$

for some clopen $V\subseteq K$ containing $0$. There is a homeomorphism
$h_{0}:I\to I$ such that $h_{0}(y_{1})=y_{2}$ and $h_{0}(y)=y$ for all
$y\notin(y_{1}-\varepsilon,y_{2}+\varepsilon)$. But then

$h(z)\,=\,\begin{cases}\big{(}x,h_{0}(y)\big{)}&\text{ if }z=(x,y)\in U,\\\
z&\text{ if not}\end{cases}$

is a homeomorphism $F_{C}\to F_{C}$ mapping $z_{1}$ to $z_{2}$.

Thus for any two points of $P_{D_{i}}\cap B_{0}$, there is a homeomorphism
$F_{C}\to F_{C}$ mapping one to the other. But also, because $F_{C}$ is
endpoint-homogeneous and $P_{D_{i}}$ is homeomorphism-invariant, for every
$x\in P_{D_{i}}$ there is a homeomorphism $F_{C}\to F_{C}$ mapping $x$ to a
point of $B_{0}$. By composing homeomorphisms, it follows that if $x,y\in
P_{D_{i}}$ then there is a homeomorphism $F_{C}\to F_{C}$ mapping $x$ to $y$.

Observe that the first part of this argument for the $P_{D_{i}}$ (not the
previous paragraph, but the one before it) shows a little more than what is
claimed. An essentially identical argument shows:

  * $(*)$

If $i\in\mathbb{Z}$ and $z_{1},z_{2}\in\\{0\\}\times(p_{i},p_{i+1})$, then
there is a homeomorphism $F_{C}\to F_{C}$ mapping $z_{1}$ to $z_{2}$.

This fact will be used to examine $P_{G}$ later, but next we look at $P_{L}$.

Like with the $D_{i}$, we begin here by consider two members of $P_{L}$ on
$B_{0}$. Fix $(0,p_{i})$ and $(0,p_{j})$ in $P_{L}\cap B_{0}$, with $i<j$. Let
$\phi:I\to I$ be the homeomorphism such that

  * $\circ$

$\phi(0)=0$ and $\phi(z)=z$ for all $z\geq a_{1}$.

  * $\circ$

$\phi(p_{k})=p_{k+j-i}$ for all $k\in\mathbb{Z}$, and

  * $\circ$

$\phi$ maps $(p_{k},p_{k+1})$ linearly onto $(p_{k+j-i},p_{k+j-i+1})$ for all
$k\in\mathbb{Z}$.

Note, in particular, that $\phi[X]=X$ and $\phi(p_{i})=p_{j}$. We aim to find
a homeomorphism $h:F_{C}\to F_{C}$ such that $h(0,y)=(0,\phi(y))$ for all
$y>0$ (in other words, $h$ restricts to $\phi$ on $B_{0}\setminus\\{t\\}$).

In the proof of Theorem 5.2, our construction begins with a countable dense
$D\subseteq X$. Because each of the $p_{k}$ is isolated in $X$,
$\left\\{p_{k}\colon k\in\mathbb{Z}\right\\}\subseteq D$. The construction
proceeds by fixing an infinite sequence $\left\langle y_{n}\colon
n\in\mathbb{N}\right\rangle$ such that each member of $D$ appears infinitely
often in the sequence. In particular,

$Y_{k}\,=\,\left\\{n\colon y_{n}=p_{k}\right\\}$

is infinite for every $k\in\mathbb{N}$. Let $\psi:\mathbb{N}\to\mathbb{N}$ be
a permutation of $\mathbb{N}$ such that

  * $\circ$

$\psi[Y_{k}]=Y_{k+j-i}$ for all $k\in\mathbb{Z}$, and

  * $\circ$

$\psi$ acts as the identity map on
$\mathbb{N}\setminus\bigcup_{k\in\mathbb{Z}}Y_{k}$.

Observe that $\phi(y_{n})=y_{\psi(n)}$ for all $n\in\mathbb{N}$.

Recalling more notation from the proof of Theorem 5.2, $C_{>0}=C\cap(0,1]$,
and for each $n\in\mathbb{N}$ there is a relatively clopen subset $V^{1}_{n}$
of $C_{>0}$ containing the blade at $x^{1}_{n}=\nicefrac{{2}}{{3^{n}}}$.
Furtheremore, the $V^{1}_{n}$ form a partition of $C_{>0}\setminus B_{0}$.

Recall that $M$ denotes the largest member of $X$, and $M^{n+1}y_{n}$ is the
$Y$-coordinate of the highest point on the leftmost blade of $V^{1}_{n}$ that
is a limit of endpoints. We divide $V^{1}_{n}$ into two pieces overlapping at
one point:

$V^{1+}_{n}\,=\,\left\\{(x,y)\in V^{1}_{n}\colon y\geq
M^{n+1}y_{n}\right\\},$$V^{1-}_{n}\,=\,\left\\{(x,y)\in V^{1}_{n}\colon y\leq
M^{n+1}y_{n}\right\\}.$

In other words, $V^{1+}_{n}$ is just the top part of the leftmost blade of
$V^{1}_{n}$, and $V^{1-}_{n}$ is everything else. In particular, note that
$V^{1+}_{n}$ is a vertical line segment.

In the proof of the Theorem 5.2, we show that for any $m,n\in\mathbb{N}$ there
is a homeomorphism $V^{1}_{n}\to V^{1}_{m}$, which in the proof is denoted
$\big{(}h^{1}_{m}\big{)}^{-1}\circ h^{1}_{n}$. This homeomorphism scales and
translates $V^{1+}_{n}$ onto $V^{1+}_{m}$ and (with a different scaling
factor) $V^{1-}_{n}$ onto $V^{1-}_{m}$. For the present proof, we need a
homeomorphism $V_{n}^{1}\to V_{\psi(n)}^{1}$, defined by modifying
$\big{(}h^{1}_{\psi(n)}\big{)}^{-1}\circ h^{1}_{n}$ as follows.

The idea is that we would like to adjust the map
$\big{(}h^{1}_{\psi(n)}\big{)}^{-1}\circ h^{1}_{n}:V_{n}^{1}\to
V_{\psi(n)}^{1}$ on the vertical segment $V_{n}^{1+}$ only, so that instead of
a simple linear mapping onto $V_{\psi(n)}^{1+}$, it mimics
$\phi\\!\restriction\\![0,y_{n}]$. More specifically, for each $n$ let
$\ell_{n}$ denote the linear mapping $V^{1+}_{n}\to[0,y_{n}]$ sending the
topmost point of $V^{1+}_{n}$ to $y_{n}$:

$\ell_{n}(\nicefrac{{2}}{{3^{n}}},y)\,=\,\textstyle\frac{y_{n}}{y_{n}-M^{n+1}y_{n}}\big{(}y-M^{n+1}y_{n}\big{)}.$

Observe that these mappings are invertible, and
$\big{(}\ell_{\psi(n)}\big{)}^{-1}$ is the linear mapping $[0,y_{\psi(n)}]\to
V^{1+}_{\psi(n)}$ that sends $y_{\psi(n)}$ to the topmost point of
$V^{1+}_{\psi(n)}$:

$\big{(}\ell_{\psi(n)}\big{)}^{-1}(y)\,=\,\textstyle\Big{(}\frac{2}{3^{\psi(n)}}\,,\,\frac{y_{\psi(n)}-M^{\psi(n)+1}y_{\psi(n)}}{y_{\psi(n)}}y+M^{\psi(n)+1}y_{\psi(n)}\Big{)}.$

Define $h_{n}:V^{1}_{n}\to V^{1}_{\psi(n)}$ by setting

$h_{n}(x)\,=\,\begin{cases}\big{(}h^{1}_{\psi(n)}\big{)}^{-1}\circ
h^{1}_{n}(z)&\text{ if }z\in V^{1-}_{n},\\\
\big{(}\ell_{\psi(n)}\big{)}^{-1}\circ\phi\circ\ell_{n}(z)&\text{ if }z\in
V^{1+}_{n}.\end{cases}$

Note that $\ell_{n}$ is a linear mapping $V^{1+}_{n}\to[0,y_{n}]$, $\phi$ maps
$[0,y_{n}]$ homeomorphically to $[0,y_{\psi(n)}]$, and
$\big{(}\ell_{\psi(n)}\big{)}^{-1}$ is a linear mapping $[0,y_{\psi(n)}]\to
V^{1+}_{\psi(n)}$. Thus the composition
$\big{(}\ell_{\psi(n)}\big{)}^{-1}\circ\phi\circ\ell_{n}$ is a homeomorphism
$V^{1+}_{n}\to V^{1+}_{\psi(n)}$. Furthermore,
$\big{(}h^{1}_{\psi(n)}\big{)}^{-1}\circ h^{1}_{n}(z)$ is a homeomorphism
$V^{1-}_{n}\to V^{1-}_{\psi(n)}$, and it is straightforward to check that each
of these homeomorphisms sends the common point of $V^{1-}_{n}\cap V^{1+}_{n}$
to the common point of $V^{1-}_{\psi(n)}\cap V^{1+}_{\psi(n)}$. It follows
that $h_{n}$ is a homeomorphism $V^{1}_{n}\to V^{1}_{\psi(n)}$.

Define a mapping $h:F_{C}\to F_{C}$ as follows:

$h(z)\,=\,\begin{cases}t&\text{ if }z=t,\\\ \big{(}0,\phi(y)\big{)}&\text{ if
}z=(0,y)\in B_{0}\text{ and }y>0,\\\ h_{n}(z)&\text{ if }z\in V^{1}_{n}\text{
for some }n.\end{cases}$

Observe that $F_{C}=\\{t\\}\cup C_{>0}=\\{t\\}\cup
B_{0}\cup\bigcup_{n\in\mathbb{N}}V^{1}_{n}$, so $h$ is defined on all of
$F_{C}$. Because $\psi$ is a permutation of $\mathbb{N}$, and each
$h_{n,\psi(n)}$ is a bijection $V^{1}_{n}\to V^{1}_{\psi(n)}$, and because
$\phi:I\to I$ is a bijection with $\phi(0)=0$, $h$ is a bijection. We claim
that $h$ is also continuous.

Because the $V^{1}_{n}$ are all clopen and the $h_{n,\psi(n)}$ are all
homeomorphisms, it is clear that $h$ is continuous at every
$z\in\bigcup_{n\in\mathbb{N}}V^{1}_{n}$.

Let $z=(0,y)\in B_{0}\setminus\\{t\\}$ (so $y\in(0,1]$). We show next that $h$
is continuous at $z$ in $\\{z\\}\cup\bigcup_{n\in\mathbb{N}}V^{1}_{n}$.

To this end, let $\left\langle z_{n}\colon n\in\mathbb{N}\right\rangle$ be a
sequence of points in $\bigcup_{n\in\mathbb{N}}V^{1}_{n}$ converging to $z$.
For each $n$ let $\bar{x}_{n}=\pi_{1}(z_{n})$ and
$\bar{y}_{n}=\pi_{2}(z_{n})$, so that $z_{n}=(\bar{x}_{n},\bar{y}_{n})$.

For each $n$, let $k_{n}$ denote the integer such that $z_{n}\in V^{1}_{k}$.
Because each $V^{1}_{k}$ is clopen in $C$ and
$\lim_{n\to\infty}z_{n}=(0,y)\notin\bigcup_{k\in\mathbb{N}}V^{1}_{k}$, no
single $V^{1}_{k}$ can contain infinitely many of the $z_{n}$. Consequently,
$\lim_{n\to\infty}k_{n}=\infty$.

Next, observe that the maximum $Y$-coordinate of $V^{1-}_{k}$ is
$M^{k+1}y_{n}$. Because $M=a_{m}<1$, this goes to $0$ as $k$ increases.
Because $\lim_{n\to\infty}k_{n}=\infty$ and
$\lim_{n\to\infty}\bar{y}_{n}=y>0$, it follows that $z_{n}\notin
V^{1-}_{k_{n}}$ for sufficiently large values of $n$. Consequently, $z_{n}\in
V^{1+}_{k_{n}}$ for all sufficiently large $n$.

Thus, for sufficiently large $n$, we have

$\bar{x}_{n}\,=\,\textstyle\frac{2}{3^{k_{n}}}\qquad\text{and}\qquad\bar{y}_{n}\in\big{[}M^{k_{n}+1}y_{n}\,,\,y_{n}\big{]}.$

Plugging these values into the formulas above, we get
$h(\bar{x}_{n},\bar{y}_{n})\,=\,$

$\textstyle\Bigg{(}\frac{2}{3^{{\psi(k_{n})}}}\,,\,\frac{y_{\psi(k_{n})}-M^{\psi(k_{n})+1}y_{\psi(k_{n})}}{y_{\psi(k_{n})}}\phi\bigg{(}\frac{y_{k_{n}}}{y_{k_{n}}-M^{k_{n}+1}y_{k_{n}}}\big{(}y-M^{k_{n}+1}y_{k_{n}}\big{)}\bigg{)}+M^{\psi(k_{n})+1}y_{\psi(k_{n})}\Bigg{)}$

for all sufficiently large $n$. Recall that $\lim_{n\to\infty}k_{n}=\infty$,
and (because $\psi$ is a bijection) this also implies
$\lim_{n\to\infty}\psi(k_{n})=\infty$. In particular,
$\lim_{n\to\infty}\frac{2}{3^{\psi(k_{n})}}=0$ and, because $M<1$,
$\lim_{n\to\infty}M^{k_{n}+1}=0$ and $\lim_{n\to\infty}M^{\psi(k_{n})+1}=0$.
Therefore, taking the limit of the above expression for
$h(\bar{x}_{n},\bar{y}_{n})$ as $n\to\infty$ gives

$\textstyle\lim_{n\to\infty}h(\bar{x}_{n},\bar{y}_{n})\,=\,(0,\phi(y))\,=\,\phi(z).$

Thus $h$ is continuous at $z$ in the space
$\\{z\\}\cup\bigcup_{n\in\mathbb{N}}V^{1}_{n}$.

But now observe that $h$ is clearly continuous at $z$ in
$F_{C}\setminus\bigcup_{n\in\mathbb{N}}V^{1}_{n}$, because this space is just
a copy of $[0,1]$ on which $h$ acts like $\phi$.

Because $h$ is continuous at $z$ both in
$\\{z\\}\cup\bigcup_{n\in\mathbb{N}}V^{1}_{n}$ and in
$F_{C}\setminus\bigcup_{n\in\mathbb{N}}V^{1}_{n}$, it follows that $h$ is
continuous at $z$ in $F_{C}$. Because $z$ was an arbitrary point of
$B_{0}\setminus\\{t\\}$, and because we already observed $h$ is continuous on
$F_{C}\setminus B_{0}=\bigcup_{n\in\mathbb{N}}V^{1}_{n}$, this shows that $h$
is continuous on $F_{C}\setminus\\{t\\}$.

But if $h$ is continuous on $F_{C}\setminus\\{t\\}$ then it is continuous.
Generally speaking, any continuous bijection $F_{C}\setminus\\{t\\}\to
F_{C}\setminus\\{t\\}$ extends to a continuous bijection $F_{C}\to F_{C}$ by
mapping $t$ to $t$, because $F_{C}$ is the one-point compactification of
$F_{C}\setminus\\{t\\}$.

Hence $h$ is a continuous bijection $F_{C}\to F_{C}$. Because $F_{C}$ is
compact, $h$ is a homeomorphism. Furthermore,
$h(0,p_{i})=(0,\phi(p_{i}))=(0,p_{j})$ and so, as desired, we have found a
homeomorphism sending one arbitrary point of $B_{0}\cap P_{L}$ to another.

But, because $F_{C}$ is endpoint-homogeneous and $P_{L}$ is homeomorphism-
invariant, for every $x\in P_{L}$ there is a homeomorphism $F_{C}\to F_{C}$
mapping it to a point of $B_{0}$. By composing homeomorphisms, it follows that
if $x,y\in P_{L}$ then there is a homeomorphism $F_{C}\to F_{C}$ mapping $x$
to $y$.

To complete the proof of the theorem, it now remains only to show that the
same is true for the final piece of our partition, $P_{G}$. Thankfully, this
follows relatively easily from the work we have already done for $P_{L}$, plus
the statement $(*)$ from earlier in the proof.

Observe that with $p_{i},p_{j}$ as before, the homeomorphism $h$ described
above does not only map $(0,p_{i})$ to $(0,p_{j})$; it maps the entire
vertical line segment $\\{0\\}\times(p_{i},p_{i+1})$ onto
$\\{0\\}\times(\phi(p_{i}),\phi(p_{i+1}))=\\{0\\}\times(p_{j},p_{j+1})$.

Fix $x,y\in B_{0}\cap P_{G}$. There are some $i,j\in\mathbb{Z}$ such that
$x\in\\{0\\}\times(p_{i},p_{i+1})$ and $y\in\\{0\\}\times(p_{j},p_{j+1})$. As
mentioned in the previous paragraph, there is a homeomorphism $F_{C}\to F_{C}$
that sends $\\{0\\}\times(p_{i},p_{i+1})$ to $\\{0\\}\times(p_{j},p_{j+1})$.
But also, by the statement $(*)$, there is a homeomorphism mapping any given
point of $\\{0\\}\times(p_{i},p_{i+1})$ to any other point of
$\\{0\\}\times(p_{i},p_{i+1})$. By composing homeomorphisms, it follows that
there is a homeomorphism $F_{C}\to F_{C}$ that maps $x$ to $y$.

Thus for any two points of $B_{0}\cap P_{G}$, there is a homeomorphism
$F_{C}\to F_{C}$ mapping one to the other. But also, because $F_{C}$ is
endpoint-homogeneous and $P_{G}$ is homeomorphism-invariant, for every $x\in
P_{G}$ there is a homeomorphism $F_{C}\to F_{C}$ mapping it to a point of
$B_{0}$. By composing homeomorphisms, it follows that if $x,y\in P_{G}$ then
there is a homeomorphism $F_{C}\to F_{C}$ mapping $x$ to $y$. ∎

##  7\. Open problems, and some non-smooth fans

In this final section of the paper we state some open questions relating to
endpoing-homogeneous and endpoint-generated fans. We also describe some
examples of non-smooth fans relevant to these questions. These examples are
meant more as illustrations than formal theorems. Accordingly, we do not
include full proofs of some of our claims about these examples.

######  Problem 7.1.

Suppose $F$ and $G$ are smooth, endpoint-homogeneous fans with
$EPG(F)=EPG(G)$. Are they homeomorphic?

Conjecture: Yes, except for the degenerate case of $\emptyset$-EPG fans (the
$n$-ods).

Some partial results in this direction are known already:

  * $\circ$

The Cantor fan is the only $\\{1\\}$-EPG smooth fan (Corollary 4.6).

  * $\circ$

The Lelek fan is the only $[0,1]$-EPG smooth fan (Charatonik).

  * $\circ$

The star is the only $\\{0\\}$-EPG smooth fan. (implicit in [2])

  * $\circ$

The shrinking Cantor fan (i.e., the $\\{0,1\\}$-EPG fan constructed in Section
5; see the picture at the beginning of Section 6) is the only $\\{0,1\\}$-EPG
smooth fan. (implicit in [2, Theorem 2.2(2)])

  * $\circ$

The $\left\\{\nicefrac{{1}}{{n}}\colon n\geq 1\right\\}$-EPG smooth fans
construction in Section 5 are the only $\left\\{\nicefrac{{1}}{{n}}\colon
n\in\mathbb{N}\right\\}$-EPG smooth fans. (see [6])

  * $\circ$

The $\left\\{\nicefrac{{1}}{{n}}\colon n\geq 2\right\\}$-EPG smooth fans
construction in Section 5 are the only $\left\\{\nicefrac{{1}}{{n}}\colon
n\in\mathbb{N}\right\\}$-EPG smooth fans. (see [6])

It is worth pointing out that the above conjecture becomes false if the
smoothness assumption is removed. The following example shows that if we allow
non-smooth fans, then there are distinct homeomorphism types of
$\\{0,\nicefrac{{1}}{{2}}\\}$-EPG fans, and there are even distinct
homeomorphism types of non-smooth $\\{0,\nicefrac{{1}}{{2}}\\}$-EPG endpoint-
homogeneous fans.

######  Example 7.2.

Consider the following picture, which shows our construction of the comb for a
$\\{0,\nicefrac{{1}}{{2}}\\}$-EPG endpoint-homogeneous smooth fan:

Here are two ways of modifying this picture to obtain planar dendroids that,
while no longer combs over $K$, still have the property that collapsing the
bottom line to a point produces a (non-smooth) fan:

These fans are still endpoint-homogeneous, with essentially the same proof as
for the original version. (All the pieces of this picture are non-uniformly
scaled versions of the whole, after making a topological adjustment to the
leftmost blade.) This shows that there are at least two homeomorphism types of
non-smooth, endpoint-homogeneous, $\\{\nicefrac{{1}}{{2}}\\}$-EPG fans.
$\dashv$

It should be clear that the two pictures shown here can be modified in many
other ways, producing a zoo of non-smooth, endpoint-homogeneous,
$\\{0,\nicefrac{{1}}{{2}}\\}$-EPG fans. And there is nothing special about the
set $\\{0,\nicefrac{{1}}{{2}}\\}$: the same thing works for any of the sets
considered in the proof of Theorem 5.2. In light of this, we do not expect a
nice classification of endpoint-homogeneous non-smooth fans, along the lines
of what the conjecture following Problem 7.1 proposes for smooth fans.

Considering non-smooth fans raises further questions. The results in Sections
4 and 5 characterize precisely those $X$ such that there is a smooth $X$-EPG
fan. But what if we remove the smoothness requirement?

######  Problem 7.3.

For which $X\subseteq[0,1]$ is there an $X$-EPG fan?

The following example shows that it is possible to have a non-smooth
$[\nicefrac{{1}}{{2}},1]$-EPG fan, despite the fact that there are no smooth
$[\nicefrac{{1}}{{2}},1]$-EPG fans.

######  Example 7.4.

The Cantor space $K$ has a basis of clopen sets, each homeomorphic to $K$, and
also contains many nowhere dense sets homeomorphic to $K$ (because every
nonempty closed subset of $K$ is homeomorphic to $K$). Using these two facts,
a straightforward recursive argument can be used to find a sequence
$\left\langle K_{n}\colon n\in\mathbb{N}\right\rangle$ of nowhere dense
subsets of $K$, each homeomorphic to $K$, such that for any sequence of points
$\left\langle d_{n}\colon n\in\mathbb{N}\right\rangle$ such that $d_{n}\in
K_{n}$ for all $n$, the set $\left\\{d_{n}\colon n\in\mathbb{N}\right\\}$ is
dense in $K$.

Let $C_{K}$ denote the Cantor comb, viewed as a subset of the $X$-$Y$ plane in
$3$-dimensional space: i.e.,
$C_{K}=(K\times[0,1]\times\\{0\\})\cup([0,1]\times\\{0\\}\times\\{0\\})$.

Let $C_{L}$ be a comb whose corresponding fan is the Lelek fan (that is,
$E(C_{L})$ is dense in $C_{L}$). For each $n\in\mathbb{N}$, define
$C^{n}_{L}\subseteq\mathbb{R}^{3}$ as follows. First, fix a homeomorphism
$\phi_{n}:K\to K_{n}$. Next, define a function $C_{L}\to\mathbb{R}^{3}$ by

$(x,y)\textstyle\,\mapsto\,\Big{(}\phi_{n}(x)\,,\,1-\frac{y}{2}\,,\,\frac{y}{2n}\Big{)}.$

Finally, let $C_{L}^{n}$ be the image of $C_{L}$ under this mapping. Roughly
speaking, $C_{L}^{n}$ is a copy of the Lelek comb, with the bottom line
removed, but (1) situated over $K_{n}$ instead of over $K$, and (2) beginning
at the top of the Cantor comb $C_{K}$, instead of on the $X$-axis, and $(3)$
tilted down towards the Cantor comb, so that $C^{n}_{L}$ approaches $C_{K}$ as
$n$ increases. Let

$C\,=\,C_{K}\cup\textstyle\bigcup_{n\in\mathbb{N}}C^{n}_{L}.$

Here we have illustrated $C_{K}\cup\bigcup_{n=1}^{5}C^{n}_{L}$ (but with the
copies of the Lelek fan drawn more like the Cantor fan, so that they show up
better; and we have made no attempt to show that the $K_{n}$ are nowhere
dense).

It is not difficult to check that $C$ is a dendroid. Furthermore (like with
combs) collapsing $[0,1]\times\\{0\\}\times\\{0\\}$ to a point yields a (non-
smooth) fan. We claim it is a $[\nicefrac{{1}}{{2}},1]$-EPG fan.

Now let $B$ be a blade of the fan. These come in two types: some lie flat in
the $X$-$Y$ plane (for example, the co-meagerly many blades above points in
$K\setminus\bigcup_{n=1}^{\infty}K_{n}$ in $C_{K}$), and some have a piece
angled out of the $X$-$Y$ plane. Let us called these the type I and type II
blades, respectively.

Because of how we chose the $K_{n}$, and because the endpoints of the Lelek
comb are dense in the Lelek comb, the closure of the endpoints of the type II
blades is all of $K\times[\nicefrac{{1}}{{2}},1]\times\\{0\\}$, i.e., the top
half of the comb $C_{K}$.

If $B$ is a type I blade, then the top half of $B$ is contained in
$\overline{E(C)\setminus B}$, by the previous paragraph. But no end points of
$C$ have $X$-coordinate below $\nicefrac{{1}}{{2}}$, so
$B\cap\overline{E(C)\setminus B}$ is precisely the top half of $B$.

If $B$ is a type II blade, then $B=B^{\prime}\cup B^{\prime\prime}$, where
$B^{\prime}$ lies flat in the $X$-$Y$ plane and $B^{\prime\prime}$ is angled
up from the $X$-$Y$ plane. As in the previous paragraph,
$B^{\prime}\cap\overline{E(C)\setminus B}$ is precisely the top half of
$B^{\prime}$. But observe that $B^{\prime\prime}$ is a blade in a copy of the
Lelek fan attached to $C_{K}$, which means that all of $B^{\prime\prime}$ is
contained in $\overline{E(C)\setminus B}$. Hence $B\cap\overline{E(C)\setminus
B}$ is the top half of $B^{\prime}$ plus all of $B^{\prime\prime}$, which is a
closed final segment of $B$. $\dashv$

######  Problem 7.5.

Is the fan from Example 7.4 planar?

Observe that this example of a non-smooth $[\nicefrac{{1}}{{2}}]$-EPG fan is
not endpoint-homogeneous. (This is because only some of the blades of the fan
witness its non-smoothness. More precisely, say $B\in\mathcal{B}(F)$ is
_smooth_ if for every sequence $\left\langle x_{n}\colon
n\in\mathbb{N}\right\rangle$ in $F\setminus B$ converging to a point $x\in B$,
we have $\lim_{n\to\infty}[t,x_{n}]=[t,x]$. Some of the blades in our
$[\nicefrac{{1}}{{2}}]$-EPG fan are smooth (the type II blades), while some
are not (the type I blades), and no homeomorphism of the fan can map a smooth
blade onto a non-smooth blade.)

######  Problem 7.6.

For which $X\subseteq[0,1]$ is there an endpoint-homogeneous $X$-EPG (but not
necessarily smooth) fan?

While the previous example shows there is an endpoint-generated fan that is
not endpoint-homogeneous, we do not have a smooth example.

######  Problem 7.7.

Is every $X$-EPG smooth fan endpoint-homogeneous?

If the answer to this question is yes, then our conjecture above (if true)
would provide a classification of all endpoint-generated smooth fans.

######  Problem 7.8.

Up to homeomorphism, is there only one $\frac{1}{6}$-homogeneous smooth fan?

As mentioned already, Acosta, Hoehn, and Juárez prove in [2] that there is no
$\frac{1}{4}$-homogeneous smooth fan. They do not rule out the possibility of
a $\frac{1}{4}$-homogeneous non-smooth fan, however; they even prove some
results about what such a fan would look like (if it exists).

######  Problem 7.9.

Is there a $\frac{1}{4}$-homogeneous (non-smooth) fan?

For the next question, recall that $X,Y\subseteq[0,1]$ are _equivalently
embedded_ if there is an order-preserving homeomorphism $[0,1]\to[0,1]$
mapping $X$ onto $Y$. Given $X\subseteq[0,1]$, let us say $x,x^{\prime}\in X$
are _equivalently embedded_ if there is an order-preserving homeomorphism
$\phi:[0,1]\to[0,1]$ mapping $X$ onto $X$ such that $\phi(x)=x^{\prime}$.

######  Problem 7.10.

Suppose $X\subseteq[0,1]$ has $n$ equivalence classes of points with respect
to the “equivalently embedded” relation described above. If there is an
$X$-EPG fan, must it be $\frac{1}{n}$-homogeneous? More specifically, is this
true for the smooth fans constructed in Section 5?

We end with a question less directly related to the material of the paper.
Endpoint-homogeneous fans have rich automorphism groups in one sense. But
there are other senses in which a space can have a rich automorphism group,
for example, the automorphism group’s universal minimal flow can be extremely
amenable. The Lelek fan has this property [5], and the fans constructed in
Section 5 are, in a sense, generalizations of the Lelek fan. This suggests the
following question.

######  Problem 7.11.

Except for the degenerate cases of the simple $n$-ods and the star, is the
universal minimal flow of automorphism group of an endpoint-homogeneous fan
extremely amenable? More specifically, is this true for the smooth fans
constructed in Section 5?

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