# List chromatic numbers and singular compactness

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

Let $G=(V,E)$ be a graph of size $\mu$ where $\mu>{\rm cf}(\mu)$. Assume that
$\aleph_{0}\leq\lambda<\mu$ and $\eta^{\lambda}<\mu$ for every $\eta\in\mu$.
We prove that if ${\rm List}(H)\leq\lambda$ for every $H\leq G$ such that
$|H|<\mu$ then ${\rm List}(G)\leq\lambda$. In particular, if $\mu$ is a strong
limit singular cardinal then singular compactness holds at $\mu$ with respect
to the list-chromatic number for every $\lambda\in\mu$. We also use the list-
chromatic number in order to prove a statement about the club principle.

###### Key words and phrases:

List chromatic number, bipartite graphs, singular cardinals, tiltan,
supercompact cardinal

###### 2010 Mathematics Subject Classification:

05C15, 05C63

##  0\. Introduction

An abelian group $A$ is called almost free if every subgroup $B$ of $A$ of
size less than the size of $A$ is free. An interesting problem is whether
there is an almost free non-free abelian group or whether almost free implies
free. An important parameter here is the cardinality of $A$. In many cases
this question is independent of ZFC. This includes the case of a successor of
a singular cardinal as proved in [6]. But if the cardinality of $A$ is a
singular cardinal $\mu$ then almost free abelian groups of size $\mu$ are
free.

This phenomenon has been studied by Shelah in [9] and it is called _singular
compactness_. The range of singular compactness is much wider than free
abelian groups, and it is known to exist in many mathematical areas including
graph theory. In the general setting we have a family of structures, a
property $\mathcal{P}$ of these structures and a cardinal $\lambda<\mu$ where
$\mu$ is a singular cardinal. We shall say that the family satisfies singular
compactness with respect to $\mathcal{P}$ iff given any $\mathfrak{A}$ of size
$\mu$ it is true that $\mathcal{P}(\mathfrak{A})\leq\lambda$ whenever
$\mathcal{P}(\mathfrak{B})\leq\lambda$ for every substructure $\mathfrak{B}$
of $\mathfrak{A}$ of size less than $\mu$.

In this paper we deal with singular compactness applied to graph theory. More
specifically, we prove a statement about the list chromatic number in infinite
graphs. Let $G=(V,E)$ be a graph. A function $f:V\rightarrow{\rm Ord}$ is a
_good coloring_ iff $f(x)\neq f(y)$ whenever $\\{x,y\\}\in E$. If $\kappa$ is
a cardinal then a $\kappa$-assignment for $G$ is a function
$F:V\rightarrow[{\rm Ord}]^{\kappa}$. The list-chromatic number of $G$,
denoted ${\rm List}(G)$, is the minimal $\kappa$ such that for every
$\kappa$-assignment $F:V^{G}\rightarrow[{\rm Ord}]^{\kappa}$ one can find a
good coloring $f:V^{G}\rightarrow{\rm Ord}$ so that $f(x)\in F(x)$ whenever
$x\in V^{G}$.

The list chromatic number was defined, independently, by Vizing in [13] and by
Erdős, Rubin and Taylor in [2]. Singular compactness with respect to the list
chromatic number is the following statement. Suppose that $G=(V,E)$ is a graph
where $|V|=\mu$ and $\mu>{\rm cf}(\mu)$. If $\lambda<\mu$ and ${\rm
List}(H)\leq\lambda$ for every $H\leq G$ of size less than $\mu$ then ${\rm
List}(G)\leq\lambda$ as well. It is known that the list chromatic number fails
to satisfy singular compcatness without additional assumptions, and a simple
example for this failure is given by Komjáth in [4]. Komjáth proved that if
$G$ is a bipartite graph with $V^{G}=A\cup B,|A|=\aleph_{0}$ and
$|B|<2^{\aleph_{0}}$ then ${\rm List}(G)\leq\aleph_{0}$. On the other hand,
there are many bipartite graphs $G$ such that $V^{G}=A\cup B,|A|=\aleph_{0}$
and $|B|=2^{\aleph_{0}}$ and ${\rm List}(G)>\aleph_{0}$. Hence if one forces
$2^{\aleph_{0}}=\mu$ where $\mu>{\rm cf}(\mu)$ then singular compactness fails
with respect to ${\rm List}(G)$.

The example of Komjáth generalizes to larger cardinalities, and one concludes
that ${\rm List}(G)$ does not satisfy singular compactness without further
assumptions. However, Usuba proved in [12] that under some cardinal arithmetic
assumptions on the singular cardinal $\mu$ and the power-set operation below
$\mu$ one can prove singular compacntess for ${\rm List}(G)$ where $|G|=\mu$.
He asked whether the assumption $\eta^{\lambda}<\mu$ for every $\eta\in\mu$
will be sufficient, and gave some partial results, see [12, Question 3.8]. Our
goal is to give a positive answer to this question.

In the last section of the paper we deal with reflection principles which
emerge from the list chromatic number (and from another invariant of graphs,
the coloring number). We observe that an interesting theorem of Usuba from
[12] which assumes diamonds can be proved with the club principle instead of
diamond. We conclude from this fact that any $ccc$ forcing notion which forces
$2^{\omega}>\omega_{1}$ must destroy the club principle at some stationary
set.

Our notation is standard. We denote cardinals by
$\theta,\kappa,\lambda,\mu,\chi$ and ordinals which are not cardinals by
$\alpha,\beta,\gamma,\delta$ and also by $i,j$. If $\lambda={\rm
cf}(\lambda)<\kappa$ then $S^{\kappa}_{\lambda}=\\{\alpha\in\kappa:{\rm
cf}(\alpha)=\lambda\\}$. Graph theoretical notation is usually consistent with
[1].

##  1\. Singular compactness

In this section we give sufficient conditions for singular compactness with
respect to the list chromatic number in terms of cardinal arithmetic. As a
first step we introduce a slight improvement upon [12, Fact 2.4] by proving a
similar statement but weakening one of the assumptions. Given a graph
$G=(V,E)$ and $x\in V$ let $E^{x}=\\{y\in V:\\{x,y\\}\in E\\}$.

######  Lemma 1.

Let $G=(V,E)$ be a graph and suppose that $\lambda\leq\tau$ are infinite
cardinals. Assume that there are $W_{0},W_{1}\subseteq V$ such that
$|W_{0}|=\tau,|W_{1}|\geq\tau^{\lambda},W_{0}\cap W_{1}=\varnothing$ and
$|E^{x}\cap W_{0}|\geq\lambda$ for every $x\in W_{1}$. Then ${\rm
List}(G)>\lambda$.

###### Proof.

Since $H\leq G$ implies ${\rm List}(H)\leq{\rm List}(G)$ it is sufficient to
find some $H\leq G$ for which ${\rm List}(H)>\lambda$. Hence we assume that
$|E^{x}\cap W_{0}|=\lambda$ for every $x\in W_{1}$ by eliminating some edges
from $E$ if needed.

Fix a disjoint family $\\{A_{y}:y\in W_{0}\\}\subseteq[\tau]^{\lambda}$. The
cardinality of the product $\prod_{y\in W_{0}}A_{y}$ is $\tau^{\lambda}$. Let
$H$ be the graph induced by $W_{0}\cup W_{1}$. We shall define a
$\lambda$-assignment for $H$ so that no function $f:V^{H}\rightarrow{\rm
Ord}$, amenable to this assignment, will be good. This will show that ${\rm
List}(H)>\lambda$ and hence ${\rm List}(G)>\lambda$ as desired.

Enumerate the elements of $\prod_{y\in W_{0}}A_{y}$ by $(g_{z}:z\in W_{1})$.
If $|W_{1}|>\tau^{\lambda}$ then either take a subset of $W_{1}$ of size
$\tau^{\lambda}$ or use repetitions. For every $y\in W_{0}$ let $F(y)=A_{y}$
and for every $z\in W_{1}$ let $F(z)=\\{g_{z}(y):y\in W_{0},\\{y,z\\}\in
E\\}$. Observe that $F$ is a $\lambda$-assignment since $|E^{x}\cap
W_{0}|=\lambda$ for every $x\in W_{1}$ and since $A_{y}\cap A_{z}=\varnothing$
whenever $y,z\in W_{0}$ and $y\neq z$.

Suppose that $f$ is a coloring of $H$ for which $f(x)\in F(x)$ whenever $x\in
W_{0}\cup W_{1}=V^{H}$. Notice that $f\upharpoonright W_{0}\in\prod_{y\in
W_{0}}A_{y}$ and choose $z\in W_{1}$ such that $f\upharpoonright W_{0}=g_{z}$.
By definition, $F(z)=\\{g_{z}(y):y\in E^{z}\cap W_{0}\\}=\\{f(y):y\in
E^{z}\cap W_{0}\\}$. But $f(z)\in F(z)$ so $f(z)=f(y)$ for some $y\in
E^{z}\cap W_{0}$. This means that $f$ is not a good coloring, so we are done.
∎

The above lemma supplies a condition under which ${\rm List}(G)$ is large,
namely larger than $\lambda$. The next lemma goes in the opposite direction
and bounds ${\rm List}(G)$ by $\lambda$. This lemma is a quotation from [12].
Recall that if $A$ is a set and $\delta$ is an ordinal then a
$\delta$-filtration of $A$ is a $\subseteq$-increasing and continuous sequence
$(A_{\alpha}:\alpha\in\delta)$ such that $|A_{\alpha}|<|A|$ for every
$\alpha\in\delta$ and $A=\bigcup_{\alpha\in\delta}A_{\alpha}$. If $G=(V,E)$
and $A_{\alpha}\subseteq V$ then ${\rm List}(A_{\alpha})$ is the list-
chromatic number of the subgraph of $G$ induced by $A_{\alpha}$. The following
is [12, Lemma 3.5].

######  Lemma 2.

Let $G=(V,E)$ be a graph and let $\lambda$ be an infinite cardinal. Let
$(A_{\alpha}:\alpha\in\delta)$ be a filtration of $V$ such that:

  1. $(\aleph)$

${\rm List}(A_{\alpha})\leq\lambda$ for every $\alpha\in\delta$.

  2. $(\beth)$

$|E^{x}\cap A_{\alpha}|<\lambda$ for every $\alpha\in\delta$ and every $x\in
V-A_{\alpha}$.

Then ${\rm List}(G)\leq\lambda$.

We can prove now our main result in a similar way to that of [12, Proposition
3.6], but using Lemma 1 we can relax the assumptions needed for singular
compactness. Notice that if our singular cardinal is strong limit then
singular compactness holds at $\mu$ for the list chromatic number with no
additional parameters. That is, for every $\lambda<\mu$ if ${\rm
List}(H)\leq\lambda$ whenever $H\leq G$ is of size less than $\mu$ then ${\rm
List}(G)\leq\lambda$ as well.

######  Theorem 3.

Assume that:

  1. $(a)$

$\mu>{\rm cf}(\mu)=\theta$ and $\aleph_{0}\leq\lambda<\mu$.

  2. $(b)$

$\eta^{\lambda}<\mu$ for every $\eta<\mu$.

  3. $(c)$

$G=(V,E)$ is a graph of size $\mu$.

If ${\rm List}(H)\leq\lambda$ for every $H\leq G$ of size less than $\mu$ then
${\rm List}(G)\leq\lambda$. Hence if $\mu$ is strong limit then the statement
holds for every $\lambda<\mu$.

###### Proof.

Let $(\mu_{i}:i\in\theta)$ be an increasing continuous sequence of cardinals
such that $\theta,\lambda<\mu_{0}$ and $\mu=\bigcup_{i\in\theta}\mu_{i}$. For
every $i\in\theta$ let $\nu_{i}=\mu_{i}^{\lambda}$, so $\nu_{i}<\mu$ for every
$i\in\theta$ by virtue of $(b)$.

The following observation is central to the proof. If $U\subseteq V$ and
$|U|=\mu_{i}$ then $|Y_{U}|<\nu_{i}$ where $Y_{U}=\\{x\in V:|E^{x}\cap
U|\geq\lambda\\}$. Indeed, if $U$ forms a counterexample then one can apply
Lemma 1 to the subgraph of $G$ induced by $U\cup Y_{U}$ and conclude that
${\rm List}(U\cup Y_{U})>\lambda$ contrary to the assumption that ${\rm
List}(H)\leq\lambda$ whenever $H\leq G$ and $|H|<\mu$.

Bearing in mind the above observation we choose an array
$(M^{\alpha}_{i}:i\in\theta,\alpha\in\lambda^{+})$ of elementary submodels of
$\mathcal{H}(\chi)$ where $\chi$ is a sufficiently large regular cardinal,
such that the following requirements are met:

  1. $(\alpha)$

$|M^{\alpha}_{i}|=\nu_{i}$ and $\nu_{i}+1\subseteq M^{\alpha}_{i}$ for every
$i\in\theta,\alpha\in\lambda^{+}$.

  2. $(\beta)$

$(M^{\alpha}_{i}:i\in\theta)$ is $\subseteq$-increasing and continuous for
every $\alpha\in\lambda^{+}$.

  3. $(\gamma)$

If $\alpha\in\lambda^{+}$ is a limit ordinal then
$M^{\alpha}_{i}=\bigcup_{\beta\in\alpha}M^{\beta}_{i}$ for every $i\in\theta$.

  4. $(\delta)$

$(M^{\alpha}_{i}:i\in\theta)\in M^{\alpha+1}_{0}$ for every
$\alpha\in\lambda^{+}$.

  5. $(\varepsilon)$

$\mu,G\in M^{\alpha}_{i}$ for every $i\in\theta,\alpha\in\lambda^{+}$.

We may assume, without loss of generality, that $V=\mu$ so $G=(\mu,E)$. For
every $i\in\theta$ let $M_{i}=\bigcup_{\alpha\in\lambda^{+}}M^{\alpha}_{i}$
and let $A_{i}=M_{i}\cap\mu$. Notice that $M^{\alpha}_{i}\in M_{i}$ for every
$i\in\theta,\alpha\in\lambda^{+}$ and $\mu\subseteq\bigcup_{i\in\theta}M_{i}$.
Also, $|A_{i}|=\nu_{i}<\mu$ and hence $(A_{i}:i\in\theta)$ is a filtration of
$V$. By the assumptions of the theorem it is true that ${\rm
List}(A_{i})\leq\lambda$ for every $i\in\theta$, so our filtrafion satisfies
the first assumption of Lemma 2.

We claim that for every $i\in\theta$ and every $x\in V-A_{i}$ it is true that
$|E^{x}\cap A_{i}|<\lambda$. If we prove this claim then from Lemma 2 we will
infer that ${\rm List}(G)\leq\lambda$ and the proof of the theorem will be
accomplished. Assume, therefore, that the claim fails and fix $i\in\theta$ and
$x\in\mu-A_{i}$ such that $|E^{x}\cap A_{i}|\geq\lambda$. Since
$A_{i}=M_{i}\cap\mu=\bigcup_{\alpha\in\lambda^{+}}M^{\alpha}_{i}\cap\mu$ one
can choose $\alpha\in\lambda^{+}$ for which $|E^{x}\cap
M^{\alpha}_{i}|\geq\lambda$.

Define $Y=\\{y\in\mu:|E^{y}\cap M^{\alpha}_{i}|\geq\lambda\\}$. The set $Y$ is
definable in $M_{i}$ since $M^{\alpha}_{i}\in M_{i}$ as well as the other
parameters, so $Y\in M_{i}$. By the observation at the beginning of the proof
we have $|Y|<\nu_{i}$ and hence $Y\subseteq M_{i}$. But $x$ satisfies
$|E^{x}\cap M^{\alpha}_{i}|\geq\lambda$ and hence $x\in Y\subseteq M_{i}$.
This means that $x\in M_{i}\cap\mu=A_{i}$, which is impossible as $x\in\mu-
A_{i}$, so we are done. ∎

As mentioned in the introduction, the invariant of list chromatic numbers does
not satisfy singular compactness without further assumptions. One may wonder
whether $\eta^{\lambda}<\mu$ for every $\eta<\mu$ is the optimal assumption
for singular compactness. There is no much room for improving this assumption
since $2^{\omega}=\mu$ already implies the failure of singular compactness
with respect to the list chromatic number at $\mu$, and $\omega$ can be
replaced by every strongly inaccessible cardinal $\kappa$.

Pondering the possibility that $2^{\lambda}<\mu$ is insufficient for singular
compactness at $\mu$ (with $\lambda$ as a parameter) we conclude from the
above theorem that in such a case there must be $\eta<\mu$ so that
$\eta^{\lambda}\geq\mu$. The following claim shows that the first such
cardinal $\eta$ has some interesting properties.

######  Claim 4.

Suppose that $2^{\lambda}<\mu,\mu$ is a singular cardinal and singular
compactness for the list chromatic number fails at $\mu$ with respect to
$\lambda$. Then there is a singular cardinal $\eta<\mu$ which satisfies ${\rm
cf}(\eta)\leq\lambda,\eta^{\lambda}\geq\mu$ and
$\theta<\eta\Rightarrow\theta^{\lambda}<\eta$.

###### Proof.

Let $\eta$ be the first cardinal below $\mu$ for which
$\eta^{\lambda}\geq\mu$. Such a cardinal exists by virtue of Theorem 3 and the
assumptions of the claim. If $\theta<\eta$ and $\theta^{\lambda}\geq\eta$ then
$\theta^{\lambda}=(\theta^{\lambda})^{\lambda}\geq\eta^{\lambda}\geq\mu$, and
this is impossible since $\eta$ is the first cardinal which satisfies
$\eta^{\lambda}\geq\mu$, hence $\theta<\eta\Rightarrow\theta^{\lambda}<\eta$.
Thus necessarily ${\rm cf}(\eta)\leq\lambda$. Indeed, if ${\rm
cf}(\eta)>\lambda$ then
$\eta^{\lambda}=\bigcup_{\theta<\eta}\theta^{\lambda}=\eta<\mu$. The proof of
the claim is accomplished. ∎

There are some interesting upshots which follow from this claim. Call $\mu$ a
_weird_ cardinal iff $\mu$ is singular and there is some $\lambda<\mu$ such
that $2^{\lambda}<\mu$ and singular compactness for list chromatic numbers
fails at $\mu$ with respect to $\lambda$. If $\lambda$ is specified then we
shall say that $\mu$ is $\lambda$-weird. Let us show that in some sense the
collection of weirdos is small. Fix a cardinal $\kappa$ of uncountable
cofinality. Suppose that $\mu<\kappa$ is $\lambda$-weird for some
$\lambda<\mu$. From Claim 4 we see that there is some $\eta<\mu$ so that
$\eta^{\lambda}\geq\mu$. If $\eta^{\lambda}<\kappa$ for every $\lambda$-weird
cardinal $\mu<\kappa$ then $\kappa$ will be called a _sane_ cardinal.

######  Corollary 5.

If $\kappa$ is sane then the set of weird cardinals below $\kappa$ is not
stationary in $\kappa$.

###### Proof.

As a first step, we fix an infinite cardinal $\lambda$ and show that the set
of $\lambda$-weird cardinals below $\kappa$ is not stationary. Call this set
$S$, and assume towards contradiction that $S$ is stationary in $\kappa$.
Applying Claim 4 to each element of $S$, let $\eta_{\mu}<\mu$ be the pertinent
cardinal, and since $\kappa$ is sane we see that $\eta_{\mu}^{\lambda}<\kappa$
for every $\mu\in S$.

The function $h(\mu)=\eta_{\mu}$ is regressive on $S$, hence there is a
stationary subset $T\subseteq S$ and a fixed $\eta$ such that $\mu\in
T\Rightarrow\eta_{\mu}=\eta$. Since $\kappa$ is sane we see that
$\eta^{\lambda}<\kappa$. On the other hand, since $T$ is unbounded in $\kappa$
one can pick some $\mu\in T$ so that $\eta^{\lambda}<\mu$ and this is
impossible since $\eta^{\lambda}=\eta_{\mu}^{\lambda}\geq\mu$.

For the second step, let $S_{0}$ be the set of weird cardinals below $\kappa$.
For every $\mu\in S_{0}$ there exists some $\lambda_{\mu}<\mu$ such that $\mu$
is $\lambda_{\mu}$-weird. If $S_{0}$ is stationary then there are a fixed
$\lambda$ and a stationary subset $S_{1}$ of $S_{0}$ so that $\mu\in
S_{1}\Rightarrow\lambda_{\mu}=\lambda$. Apply the first step to $S_{1}$ in
order to obtain the desired contradiction. ∎

If $\kappa$ is a strong limit cardinal then the above corollary follows simply
from the fact that the set of strong limit singular cardinals below $\kappa$
is a club, and every weird cardinal is not strong limit. However, the above
corollary applies to cases in which $\kappa$ is not strong limit, and it shows
that the set of weirdos is _topologically_ small, in the sense that it is not
stationary. This leads us to the possibility of an unbounded subset of weirdos
below a sane cardinal.

In the following theorem we employ a reflection argument with respect to a
supercompact cardinal. A similar argument works at many local properties, e.g.
the fact that if there is a huge cardinal above a supercompact cardinal then
there are unboundedly many huge cardinals below the supercompact, see [7].

######  Theorem 6.

Let $\kappa$ be a supercompact cardinal. If $\kappa<\lambda<\mu$ and $\mu$ is
$\lambda$-weird then the set of weird cardinals below $\kappa$ is unbounded in
$\kappa$.

###### Proof.

Let $\mu$ be $\lambda$-weird and notice that necessarily $\lambda>\kappa$.
Recall that $\mu$ is a singular cardinal, $2^{\lambda}<\mu$ and singular
compactness fails at $\mu$ and $\lambda$ with respect to the list chromatic
number. Let $G$ be a graph of size $\mu$ which witnesses this fact in $V$.
Namely, if $H\leq G$ and $|H|<\mu$ then ${\rm List}(H)\leq\lambda$ despite the
fact that ${\rm List}(G)>\lambda$.

Choose a sufficiently large $\delta>\mu$ (the cardinal $\delta=(2^{\mu})^{+}$
is oso ondo) and an elementary embedding $\jmath:V\rightarrow M$ so that
$\kappa={\rm crit}(\jmath),\jmath(\kappa)>\lambda$ and ${}^{\delta}M\subseteq
M$. By the closure properties of $M$ we see that $G\in M$ and it witnesses the
fact that $\mu$ is $\lambda$-weird. But $\mu<\jmath(\kappa)$ and hence $M$
knows that there exists a $\lambda$-weird cardinal below $\jmath(\kappa)$.
Apply elementarity and conclude that there exists a weird cardinal $\theta$
below $\kappa$ in $V$. Moreover, if $\gamma\in\kappa$ then
$\jmath(\gamma)=\gamma$ since $\kappa={\rm crit}(\jmath)$. Hence for every
$\gamma\in\kappa$ we know, in $M$, that there is a weird cardinal
$\mu<\jmath(\kappa)$ such that $\mu>\jmath(\gamma)$. Therefore, in $V$ there
is a weird cardinal $\theta<\kappa$ such that $\theta>\gamma$, and since
$\gamma\in\kappa$ was arbitrary we are done. ∎

As we have seen already, the set of weirdos below $\kappa$ will be unbounded
(under the assumption of the theorem) but it will never be stationary. We do
not know if the existence of a weird cardinal, especially above a supercompact
cardinal, is consistent at all. Actually, one may wonder whether singular
compactness for the list chromatic number can fail above a supercompact
cardinal. The answer is positive, since the counterexample of [4] with
$\omega$ works at every strongly inaccessible cardinal $\lambda$ instead of
$\omega$. For completeness, we unfold the details.

######  Proposition 7.

Let $\kappa$ be a supercompact cardinal. Let $\lambda$ be a strongly
inaccessible cardinal greater than $\kappa$. One can force the existence of a
singular cardinal $\mu>\lambda$ such that singular compactness fails at
$\lambda$ and $\mu$ with respect to the list chromatic number, while
preserving the supercompactness of $\kappa$.

###### Proof.

Based on [5] we may assume that $\kappa$ is Laver-indestructible. Let
$\mathbb{P}$ be a $\kappa$-directed closed forcing notion which forces
$2^{\lambda}=\mu$ for some singular cardinal $\mu>\lambda$. Notice that
$\kappa$ remains supercompact in the generic extension. Let us show that $\mu$
and $\lambda$ satisfy the statement of the proposition.

Let $G$ be the complete bipartite graph over $V=A\cup B$, where
$A=\\{a_{\alpha}:\alpha\in\lambda\\}$ and $B=\\{b_{\beta}:\beta\in\mu\\}$. Fix
a family of disjoint sets of ordinals $\\{S_{\alpha}:\alpha\in\lambda\\}$ so
that $|S_{\alpha}|=\lambda$ for every $\alpha\in\lambda$. We define an
assignment $F$ which exemplifies the fact that ${\rm List}(G)>\lambda$.
Firstly, let $F(a_{\alpha})=S_{\alpha}$ for each $\alpha\in\lambda$. Secondly,
fix a bijection $h$ between $B$ and $\prod_{a_{\alpha}\in A}S_{\alpha}$ which
maps $b_{\beta}$ to $g_{\beta}$, and for each $\beta\in\mu$ let
$F(b_{\beta})=\\{g_{\beta}(a_{\alpha}):\alpha\in\lambda\\}$.

Suppose that $f$ is a good coloring of $G$ and $f(x)\in F(x)$ for every $x\in
V$. In particular, if $g=f\upharpoonright A$ then $g\in\prod_{a_{\alpha}\in
A}S_{\alpha}$. Let $\beta\in\mu$ be such that $h(b_{\beta})=g$. It follows
that $f(b_{\beta})\in
F(b_{\beta})=\\{g_{\beta}(a_{\alpha}):\alpha\in\lambda\\}$ and hence
$f(b_{\beta})=g(a_{\alpha})$ for some $\alpha\in\lambda$. But
$a_{\alpha}Eb_{\beta}$ and $f(b_{\beta})=g(a_{\alpha})=f(a_{\alpha})$ as
$g=f\upharpoonright A$, so $f$ is not good after all.

Suppose now that $H\leq G$ and $|H|<\mu$. Let $V_{H}=A_{H}\cup B_{H}$ and
without loss of generality $A_{H}=A$. Suppose that
$|B_{H}|=\theta<\mu=2^{\lambda}$, let
$A_{H}=\\{a_{\alpha}:\alpha\in\lambda\\}$ and let
$B_{H}=\\{b_{\beta}:\beta\in\theta\\}$. Given $F:A_{H}\cup
B_{H}\rightarrow[{\rm Ord}]^{\lambda}$ we wish to find an $F$-amenable good
coloring.

For every $\alpha\in\lambda$, the set of functions ${}^{\alpha}2$ is of size
less than $\lambda$, so one can choose for every $\varphi\in{}^{\alpha}2$ an
element $c_{\varphi}\in F(a_{\alpha})$ such that
$\varphi,\psi\in{}^{\alpha}2\wedge\varphi\neq\psi\Rightarrow c_{\varphi}\neq
c_{\psi}$. We do this for every $\alpha\in\lambda$. Now for each
$h\in{}^{\lambda}2$ we define the set
$T_{h}=\\{c_{h\upharpoonright\alpha}:\alpha\in\lambda\\}$. Notice that
$|T_{h}|=\lambda$ whenever $h\in{}^{\lambda}2$. Likewise, if $h\neq
h^{\prime}$ then $|T_{h}\cap T_{h^{\prime}}|<\lambda$ since
$h\upharpoonright\alpha\neq h^{\prime}\upharpoonright\alpha$ over an end-
segment of $\alpha\in\lambda$.

Therefore, if $\beta\in\theta$ then $F(b_{\beta})\subseteq T_{h}$ is possible
for at most one $h\in{}^{\lambda}2$. Since $\theta<\mu=2^{\lambda}$ one can
choose an element $h\in{}^{\lambda}2$ such that $\neg(F(b_{\beta})\subseteq
T_{h})$ for every $\beta\in\theta$. Fix such a function $h$ and define a good
coloring $f$ by the following procedure. If $\alpha\in\lambda$ then
$f(a_{\alpha})=c_{h\upharpoonright\alpha}$ which belongs to $F(a_{\alpha})$ by
definition. If $\beta\in\theta$ then $f(b_{\beta})$ is any element from
$F(b_{\beta})-T_{h}$. One can verify that $f$ is a good coloring so we are
done. ∎

One may wonder whether the failure of singular compactness with respect to the
list-chromatic number is limited to one cardinality. Namely, if $\mu>{\rm
cf}(\mu),|G|=\mu,\lambda<\mu$ and ${\rm List}(H)\leq\lambda$ for every small
subgraph $H$ of $G$, is it true that ${\rm List}(G)\leq\lambda^{+}$ (or a
similar bound which depends on $\lambda$)? It turns out that ${\rm List}(G)$
can be arbitrarily large from the $\aleph$-scale point of view. For instance,
if $2^{\omega}=2^{\omega_{1}}=\mu$ and $\mu>{\rm cf}(\mu)$ then Komjath’
example of the complete bipartite graph $G$ of size $\omega\times 2^{\omega}$
satisfies ${\rm List}(H)\leq\aleph_{0}$ for every $H\leq G$ of size less than
$\mu$, but ${\rm List}(G)>\aleph_{1}$. This can be generalized as in
Proposition 7. It seems, however, that the cofinality of $\mu$ is restricted
in such examples.

We indicate that the situation of $\kappa<\lambda<\eta<\mu$ with $\kappa$
being supercompact, $2^{\lambda}<\eta$ and $\eta^{\lambda}\geq\mu$ is
forceable. Begin with a Laver-indestructible supercompact cardinal $\kappa$,
choose $\lambda<\eta<\mu$ so that $\kappa<\lambda$ and add many Cohen subsets
to $\eta$ of size $\lambda$ (the supercompactness will be preserved). As in
the proof of Theorem 6, unboundedly many triples
$\lambda^{\prime}<\eta^{\prime}<\mu^{\prime}$ below $\kappa$ will satisfy
$2^{\lambda^{\prime}}<\eta^{\prime}$ and
$\eta^{\prime\lambda^{\prime}}\geq\mu^{\prime}$. We do not know, however, if
this gives weird cardinals.

######  Question 8.

Is it consistent that there exists a weird cardinal?

##  2\. Tiltan and reflection

Suppose that $\kappa={\rm cf}(\kappa)>\aleph_{0}$ and $S\subseteq\kappa$ is
stationary. The club principle $\clubsuit_{S}$ is a prediction principle
introduced by Ostaszewski in [8]. It says that there exists a sequence
$(T_{\alpha}:\alpha\in{S})$ where each $T_{\alpha}$ is an unbounded subset of
$\alpha$, and for every unbounded $A\subseteq\kappa$ the set
$S_{A}=\\{\alpha\in{S}:T_{\alpha}\subseteq{A}\\}$ is stationary. Henceforth we
call the club principle _tiltan_ , to avoid double meaning and overuse of the
term club.

It is clear that $\Diamond_{S}$ implies $\clubsuit_{S}$ and this is true, in
particular, where $S=\omega_{1}$. For the converse,
$\clubsuit_{\omega_{1}}\wedge 2^{\omega}=\omega_{1}$ implies
$\Diamond_{\omega_{1}}$. However, one can force tiltan at $\omega_{1}$ along
with $2^{\omega}>\omega_{1}$, in which case diamond fails at $\omega_{1}$.
Hence tiltan is strictly weaker than diamond.

The methods for obtaining tiltan with a large value of $2^{\omega}$ divide
into two groups. One approach is to use collapses. Shelah proved in [10] that
if one begins with diamond at every stationary subset of $\omega_{1}$ and
$\omega_{2}$, increases $2^{\omega}$ and collapses with
$L\acute{e}vy(\aleph_{0},\aleph_{1})$ then tiltan holds at every stationary
subset of $\omega_{1}$ in the generic extension. Of course,
$2^{\omega}>\omega_{1}$ and hence diamond fails despite the fact that tiltan
holds. Todorčević, [11], employs a similar idea. He iterates proper forcing
notions with countable support where the length of the iteration is more than
$\omega_{2}$.

Another approach is taken in [3], and here no collapse is involved. The
forcing there is not $\aleph_{1}$-cc but a special argument proves that
$\aleph_{1}$ is preserved. Clearly, if one collapses $\aleph_{1}$ as in the
first approach then the forcing is not $\aleph_{1}$-cc. One may wonder,
therefore, if this is always the case. That is, whether a $ccc$ forcing notion
which increases $2^{\omega}$ necessarily destroys tiltan at some stationary
set. The main result of this section is a positive statement in this
direction, provided that there exists a supercompact cardinal in $V$.

Let us consider another phenomenon related to compactness with respect to
list-chromatic numbers. This is a reflection principle discussed in [12] and
denoted by $\textsf{RP}({\rm List},\lambda)$. It says that for every graph $G$
such that $|G|\leq\lambda$, if ${\rm List}(G)>\aleph_{0}$ then there exists
$H\leq{G}$ such that $|H|=\aleph_{1}$ and ${\rm List}(H)>\aleph_{0}$.

Another invariant of graphs is called the coloring number. Given a graph $G$,
the coloring number ${\rm Col}(G)$ is the minimal $\kappa$ such that there
exists a well-ordering $<^{*}$ of $V_{G}$ which satisfies $|\\{u\in
E^{v}:u<^{*}v\\}|<\kappa$ for every $v\in V_{G}$. Notice that ${\rm
List}(G)\leq{\rm Col}(G)$ for every graph $G$. The corresponding reflection
principle is defined in the same way. Thus $\textsf{RP}({\rm Col},\lambda)$ is
the statement that for every graph $G$ with $|G|\leq\lambda$, if ${\rm
Col}(G)>\aleph_{0}$ then there exists $H\leq{G}$ such that $|H|=\aleph_{1}$
and ${\rm Col}(H)>\aleph_{0}$. By writing $\textsf{RP}({\rm Col})$ or
$\textsf{RP}({\rm List})$ we assert that these principles hold at every
$\lambda$. By [12, Fact 2.12] the principle $\textsf{RP}({\rm Col})$ is
preserved in $ccc$ forcing extensions.

The following is a modification of a theorem from [12] in which $\Diamond_{S}$
is replaced by $\clubsuit_{S}$. The proof is essentially the same proof, but
there are some points in which one has to change the argument slightly. We
spell-out the details for the sake of completeness.

######  Theorem 9.

Assume that $\lambda={\rm cf}(\lambda)<\kappa={\rm cf}(\kappa)$. Suppose that
$\clubsuit_{S}$ holds at every stationary $S\subseteq S^{\kappa}_{\lambda}$.
Let $G$ be a graph with $V_{G}=\kappa$. Suppose that ${\rm Col}(G)>\lambda$
and ${\rm Col}(H)\leq\lambda$ for every $H\leq{G}$ of size less than $\kappa$.
Then ${\rm List}(G)>\lambda$.

###### Proof.

Let $S=\\{\alpha\in S^{\kappa}_{\lambda}:\exists\beta\geq\alpha,|\alpha\cap
E^{\beta}|\geq\lambda\\}$. By [12, Fact 2.7] one knows that $S$ is a
stationary subset of $\kappa$. For every $\alpha\in{S}$ choose
$\beta(\alpha)\geq\alpha$ such that $|\alpha\cap
E^{\beta(\alpha)}|\geq\lambda$. Without loss of generality, if
$\alpha_{0},\alpha_{1}\in{S}$ and $\alpha_{0}<\alpha_{1}$ then
$\beta(\alpha_{0})<\beta(\alpha_{1})$.

Let $(r_{\alpha},t_{\alpha}:\alpha\in{S})$ be a sequence of functions where
both $r_{\alpha}$ and $t_{\alpha}$ are partial functions from $\alpha$ into
$\alpha$ and if $f,g:\kappa\rightarrow\kappa$ then
$r_{\alpha}\subseteq{f}\wedge t_{\alpha}\subseteq{g}$ for stationarily many
$\alpha\in{S}$. We may assume that the domain of each $r_{\alpha}$ and each
$t_{\alpha}$ is unbounded in $\alpha$.

Let $(A_{\beta}:\beta\in\kappa)$ be a sequence of pairwise disjoint sets where
$A_{\beta}\in[\kappa]^{\lambda}$ for every $\beta\in\kappa$. Recall that
$|\alpha\cap E^{\beta(\alpha)}|\geq\lambda$ for every $\alpha\in{S}$, and fix
a set $x_{\alpha}\subseteq\alpha\cap E^{\beta(\alpha)}$ of size $\lambda$ for
each $\alpha\in{S}$. Based on the proof of [12, Proposition 3.1] we define two
$\lambda$-assignments $F_{0},F_{1}$ and we show that (at least) one of them
witnesses ${\rm List}(G)>\lambda$.

Suppose that $\beta\in\kappa$. The substantial part of the definition lies on
the cases in which $\beta=\beta(\alpha)$ for some $\alpha\in{S}$, so if
$\beta\neq\beta(\alpha)$ for every $\alpha\in{S}$ then simply let
$F_{0}(\beta)=F_{1}(\beta)=A_{\beta}$. If $\beta=\beta(\alpha)$ for some
$\alpha\in{S}$ then $\alpha$ is unique, and we distinguish three cases:

  1. $(a)$

If $|r^{\prime\prime}_{\alpha}x_{\alpha}|=\lambda$ then
$F_{0}(\beta)=r^{\prime\prime}_{\alpha}x_{\alpha}$ and
$F_{1}(\beta)=A_{\beta}$.

  2. $(b)$

If $|r^{\prime\prime}_{\alpha}x_{\alpha}|<\lambda$ and
$|t^{\prime\prime}_{\alpha}x_{\alpha}|=\lambda$ then $F_{0}(\beta)=A_{\beta}$
and $F_{1}(\beta)=t^{\prime\prime}_{\alpha}x_{\alpha}$.

  3. $(c)$

If both $|r^{\prime\prime}_{\alpha}x_{\alpha}|<\lambda$ and
$|t^{\prime\prime}_{\alpha}x_{\alpha}|<\lambda$ then
$F_{0}(\beta)=F_{1}(\beta)=A_{\beta}$.

Assume that $f:\kappa\rightarrow\kappa$ is $F_{0}$-amenable and
$g:\kappa\rightarrow\kappa$ is $F_{1}$-amenable. Our goal is to show that (at
least) one of them is not good. Suppose, therefore, that $f$ is good. Choose
$\alpha\in{S}$ such that $r_{\alpha}\subseteq{f}\wedge
t_{\alpha}\subseteq{g}$. In particular,
$r_{\alpha}\upharpoonright{x_{\alpha}}\subseteq f\upharpoonright{x_{\alpha}}$
and $t_{\alpha}\upharpoonright{x_{\alpha}}\subseteq
g\upharpoonright{x_{\alpha}}$ since $x_{\alpha}\subseteq\alpha$.

Let $\beta=\beta(\alpha)$. We consider the above three possibilities. If $(a)$
holds, namely $|r^{\prime\prime}_{\alpha}x_{\alpha}|=\lambda$ then
$F_{0}(\beta)=r^{\prime\prime}_{\alpha}x_{\alpha}=f^{\prime\prime}x_{\alpha}$.
Since $f$ is $F_{0}$-amenable we see that $f(\beta)\in
f^{\prime\prime}x_{\alpha}$. Choose $\eta\in{x_{\alpha}}$ such that
$f(\beta)=f(\eta)$ and notice that $\eta\neq\beta$ since
$\eta\in{x_{\alpha}}\subseteq\alpha$ and $\beta=\beta(\alpha)\geq\alpha$.
However, $x_{\alpha}\subseteq E^{\beta}$ and hence $(\eta,\beta)\in{E}$ yet
$f(\eta)=f(\beta)$. This is impossible since $f$ is good.

We conclude, therefore, that $(a)$ fails, that is to say
$|r^{\prime\prime}_{\alpha}x_{\alpha}|<\lambda$. Recall that we fixed an
ordinal $\alpha\in{S}$ such that $r_{\alpha}\subseteq{f}$ and
$t_{\alpha}\subseteq{g}$. We may assume, without loss of generality, that both
$r_{\alpha}$ and $t_{\alpha}$ are defined on every $\gamma\in{x_{\alpha}}$.
Indeed, define $x=\bigcup_{\alpha\in{S}}x_{\alpha}$ and consider all the
ordinals $\alpha$ for which $r_{\alpha}\subseteq f\upharpoonright{x}$ and
$t_{\alpha}\subseteq g\upharpoonright{x}$. By Fodor’s lemma there must be some
$\alpha\in{S}$ and $y_{\alpha}\in[x_{\alpha}]^{\lambda}$ for which both
$r_{\alpha}$ and $t_{\alpha}$ are defined on every $\gamma\in{y_{\alpha}}$.
Since all we need in the following argument is just a set of size $\lambda$ we
may assume that $y_{\alpha}=x_{\alpha}$.

Let $A=\\{\gamma\in{x_{\alpha}}:F_{0}(\gamma)=A_{\gamma}\\}$. We claim that
$|A|<\lambda$. To see this, notice that $f^{\prime\prime}A$ is one-to-one
since $f(\gamma)\in A_{\gamma}=F_{0}(\gamma)$ for each $\gamma\in{A}$ and
$A_{\gamma}\cap A_{\delta}=\varnothing$ whenever $\gamma\neq\delta$. Hence if
$|A|=\lambda$ then $|f^{\prime\prime}A|=\lambda$ as well. But then
$|r^{\prime\prime}_{\alpha}{A}|=|f^{\prime\prime}A|=\lambda$ since
$A\subseteq{x_{\alpha}}$ and then
$|r^{\prime\prime}_{\alpha}x_{\alpha}|=\lambda$ contradicting the fact that
$|r^{\prime\prime}_{\alpha}x_{\alpha}|<\lambda$.

Now if $\gamma\in{x_{\alpha}}-A$ (and there are $\lambda$-many such ordinals)
then $F_{0}(\gamma)\neq{A_{\gamma}}$. By definition,
$F_{1}(\gamma)=A_{\gamma}$ for each such ordinal $\gamma$, and this will
happen at $\lambda$-many $\gamma$s. Since $g$ is $F_{1}$-amenable, if
$\gamma,\delta\in{x_{\alpha}}-A$ and $\gamma\neq\delta$ then $g(\gamma)\neq
g(\delta)$. Hence
$|t^{\prime\prime}_{\alpha}x_{\alpha}|=|g^{\prime\prime}x_{\alpha}|=\lambda$
and one concludes that necessarily $(b)$ holds.

Let us show that $(b)$ implies that $g$ is not good, and this will accomplish
the proof. Recall that $\beta=\beta(\alpha)$ and $(b)$ asserts that
$F_{1}(\beta)=t^{\prime\prime}_{\alpha}x_{\alpha}$. Pick $\eta\in\alpha$ such
that $g(\beta)=t_{\alpha}(\eta)$. Since
$t^{\prime\prime}_{\alpha}x_{\alpha}\subseteq g^{\prime\prime}x_{\alpha}$ we
see that $g(\beta)=g(\eta)$. But $\eta\neq\beta$ since
$\eta\in{x_{\alpha}}\subseteq\alpha\leq\beta(\alpha)=\beta$, so
$(\eta,\beta)\in{E}$ and hence $g$ is not good as required. ∎

From the above theorem one can derive an interesting conclusion with regard to
the club principle. We are interested in the properties of forcing notions
which increase the size of the continuum but preserve tiltan at various
stationary sets.

######  Theorem 10.

Let $\kappa$ be supercompact and let $G\subseteq{\rm
L\acute{e}vy}(\omega_{1},<\kappa)$ be $V$-generic. Let $\mathbb{P}$ be a
forcing notion in $V[G]$ and let $H\subseteq\mathbb{P}$ be $V[G]$-generic. If
the continuum hypothesis fails in $V[G][H]$ then for some stationary
$S\subseteq\omega_{1}$ one has $V[G][H]\models\neg\clubsuit_{S}$.

###### Proof.

We commence with the following general statement. Assume that $\lambda={\rm
cf}(\lambda)<\kappa$. Assume further that $\clubsuit_{S}$ holds whenever
$S\subseteq S^{\theta}_{\lambda}$ for every $\theta\in{\rm
Reg}\cap(\lambda,\kappa)$. Then for every graph $G$ of size less than $\kappa$
it is true that ${\rm Col}(G)>\lambda$ iff ${\rm List}(G)>\lambda$. Of course,
one direction is immediate since the inequality ${\rm List}(G)\leq{\rm
Col}(G)$ is always true. For the opposite direction suppose that ${\rm
Col}(G)>\lambda$. Let $H\leq{G}$ be such that ${\rm Col}(H)>\lambda$ and for
every $I\leq{H}$ if $|I|<|H|$ then ${\rm Col}(I)\leq\lambda$. Shelah proved in
[9] that the coloring number satisfies singular compactness. Hence if $H$ is
of minimal cardinality then necessarily $|H|$ is a regular cardinal which
falls within the scope of Theorem 9. Applying Theorem 9 to the graph $H$ one
concludes that ${\rm List}(H)>\lambda$ and hence ${\rm List}(G)>\lambda$.

In particular, if $\clubsuit_{S}$ holds at every stationary
$S\subseteq\omega_{1}$ and $G$ is a graph of size $\aleph_{1}$ such that ${\rm
Col}(G)=\aleph_{1}$ then ${\rm List}(G)=\aleph_{1}$. Therefore, if
$\clubsuit_{S}$ holds at every stationary $S\subseteq\omega_{1}$ and
$\textsf{RP}({\rm Col})$ is true then $\textsf{RP}({\rm List})$ is true as
well. Indeed, let $G$ be any graph of size $\kappa>\aleph_{1}$ such that ${\rm
List}(G)>\aleph_{0}$. We know that ${\rm Col}(G)\geq{\rm List}(G)>\aleph_{0}$
so from $\textsf{RP}({\rm Col})$ we choose $H\leq{G},|H|=\aleph_{1}$ such that
${\rm Col}(H)>\aleph_{0}$. From the above argument we conclude that ${\rm
List}(H)>\aleph_{0}$, that is to say $\textsf{RP}({\rm List})$.

From [12] we know that if $\kappa$ is supercompact and $G$ is generic for the
collapse of the predecessors of $\kappa$ then $\textsf{RP}({\rm Col})$ holds
in $V[G]$. Notice further that $\Diamond_{S}$ holds at every stationary
$S\subseteq\omega_{1}$ and, in particular, $2^{\omega}=\omega_{1}$. Since
$\mathbb{P}$ is $ccc$ we know that $\textsf{RP}({\rm Col})$ holds in
$V[G][H]$.

Assume towards a contradiction that $\clubsuit_{S}$ holds in $V[G][H]$
whenever $S\subseteq\omega_{1}$ is stationary. By the above considerations
$\textsf{RP}({\rm List})$ is true, and in particular $\textsf{RP}({\rm
List},2^{\omega})$ holds in $V[G][H]$. But $\textsf{RP}({\rm
List},2^{\omega})$ implies $2^{\omega}=\omega_{1}$ as proved in [12, Lemma
4.1]. We are assuming that $\mathbb{P}$ forces $2^{\omega}>\omega_{1}$, thus
we arrived at a contradiction. ∎

We believe that the limitation on forcing notions which increase the continuum
and preserve tiltan as expressed in the above theorem does not depend on the
existence of large cardinals in the ground model. Thus we phrase the
following:

######  Conjecture 11.

Suppose that $V=L$. Let $\mathbb{P}$ be a $ccc$ forcing notion and let
$G\subseteq\mathbb{P}$ be generic. If $2^{\omega}>\omega_{1}$ in $V[G]$ then
$\clubsuit_{S}$ fails in $V[G]$ for some stationary set
$S\subseteq\omega_{1}$.

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