Let $X \neq \emptyset$ be a set. We say that ${\cal F} \subseteq {\cal P}(X)$ is a *down-set* if ${\cal F}$ is closed under taking subsets. Whenever $a \in X$, we let ${\cal F}\_a = \{ S \in F : a \in S\}$.


We say ${\cal G} \subseteq {\cal P}(X)$ is *intersecting* if $S \cap T$ is non-empty whenever $S, T \in {\cal G}$.


Consider the following statement:



> 
> (C): Whenever $X$ is a non-empty set, ${\cal F} \subseteq {\cal P}(X)$ is a down-set, and ${\cal G} \subseteq {\cal F}$ is an intersecting family, then there is $a \in X$, and an injection $\iota: {\cal G} \to {\cal F}\_a$.
> 
> 
> 


It is not known whether (C) is true for finite non-empty sets, and this open problem is called Chvatal's conjecture. I was thinking that for $X = \omega$ maybe a family with the finite intersection property could be used to show that (C) is false for infinite ground sets $X$, but I failed to come up with an argument.


**Question.** What is a counterexample to (C) in the infinite case?



Counterexample for Chvatal's conjecture in an infinite set
Let $\kappa$ be a cardinal of
countable cofinality that is strictly larger than the continuum.
I will construct a counterexample to (C) on
$X=(\kappa\times \omega)\cup \omega$.



Since $\kappa$ has countable cofinality, there exists
a countable increasing sequence of cardinals
$(\kappa\_i)\_{i\in\omega}$ with limit $\kappa$.
For each ordinal $\mu<\kappa$, let $n\_{\mu}$
be the least $n\in\omega$ such that $\mu\in \kappa\_n$.
For each $\mu<\kappa$ let
$A\_{\mu}\subseteq X$ be the countable set
$$
A\_{\mu} = (\{\mu\}\times \omega)\cup \{2^c\cdot 3^d\;|\; c\leq n\_{\mu} \leq d\}.
$$
Let $\mathcal G = \{A\_{\mu}\;|\;\mu<\kappa\}$.
Let $\mathcal F$ be the down-closure of ${\mathcal G}$ in ${\mathcal P}(X)$.
Necessarily $\mathcal F$ is a down-set in ${\mathcal P}(X)$ containing $\mathcal G$.
I claim that


* $\mathcal G$ is intersecting.
* $\mathcal G$ has cardinality $\kappa$.
* Each ${\mathcal F}\_a$ has size $<\kappa$.


The last two items show that there is no injection from $\mathcal G$
to any ${\mathcal F}\_a$.



To show that $\mathcal G$ is intersecting, note that
if $\mu \leq \nu$, then $2^{n\_{\mu}}3^{n\_{\nu}}\in A\_{\mu}\cap A\_{\nu}$.



To see that $|{\mathcal G}| = \kappa$, it is enough to
observe that the $A\_{\mu}$'s are indexed by $\kappa$
and that they are distinct from each other since
$(\mu,0)\in A\_{\mu}-A\_{\nu}$
when $\mu\neq \nu$.



It remains to show that
each ${\mathcal F}\_a$ has size $<\kappa$.
Here there are cases depending on whether
(i) $a \in \kappa\times \omega$,
(ii) $a$ has the form $2^c\cdot 3^d$ for some $c\leq d$,
or (iii)
$a$ does not belong to any of the sets $A\_{\mu}$.


In case (i) $a=(\lambda,k)\in \kappa\times \omega$.
There is exactly one set $A\_{\mu}$
containing $a$, namely $A\_{\lambda}$.
Since ${\mathcal F}$ is the order ideal
generated by $\mathcal G$, the set
${\mathcal F}\_a$ is ${\mathcal P}(A\_{\lambda})\_a$.
Since $A\_{\lambda}$ is countably infinite,
${\mathcal P}(A\_{\lambda})\_a$ has size continuum, which is $<\kappa$.


In case (ii) we have $a=2^c\cdot 3^d$ for some $c\leq d$.
If $a\in A\_{\mu}$ for some $\mu$, then
$c\leq n\_{\mu}\leq d$. There are at most finitely
many $n\_{\mu}$'s that can satisfy this, let $N$ be the largest such one.
The elements of ${\mathcal F}\_a$ are subsets
of those $A\_{\mu}$'s where $c\leq n\_{\mu}\leq d$.
Hence
${\mathcal F}\_a\subseteq \bigcup\_{\mu\leq \kappa\_N} {\mathcal P}(A\_{\mu})\_a$.
Since each $A\_{\mu}$ is countably infinite
${\mathcal P}(A\_{\mu})\_a$ has size $2^{\aleph\_0}$.
Therefore $|{\mathcal F}\_a|\leq 2^{\aleph\_0}\cdot \kappa\_{N} = \max(2^{\aleph\_0},\kappa\_N)<\kappa$.


In case (iii) ${\mathcal F}\_a$ is empty, so has size $<\kappa$. \\\










