Counterexample for Chvatal's conjecture in an infinite set 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?
 A: The usual definition of ideal also includes the requirement that $F$ is closed under unions.
But in this case, I claim that Chvatal's property holds for all infinite ideals.
First, if $a\in Y\in F$, then $\{a\}\in F$. And in this case, we may associate any $Z\in F$ that omits $a$ with the set $Z\cup\{a\}$, and this makes a bijection between $F-F_a$ and $F_a$.
Thus, if $F$ is infinite, we have split it in half, and these halves must therefore have the same size as $F$ itself. So $F_a$ is bijective with $F$. And so $G$ will inject into such an $F_a$ as desired.
So perhaps you don't intend to use ideals, but merely downsets?
A: 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 $<\kappa$. \\\
