TITLE: Frequently in Nets
QUESTION [6 upvotes]: Let $(x_\alpha)_{\alpha \in I}$ be a net with elements $x_{\alpha}$ in some countable set $A$. Suppose that no element of $A$ is frequently in $(x_\alpha)_{\alpha \in I}$. Is it true that there is some infinite subset $B \subseteq A$ so that for any infinite subset $C \subseteq B$, $C$ is frequently in $(x_\alpha)_{\alpha \in I}$? That is, $\forall \alpha \in I$, $\exists \beta \gtrsim \alpha$ so that $x_\beta \in C$.
I think this is true but can't find a proof. It's hard because nets are very different from sequences.

REPLY [8 votes]: It's not true, in general.
Equip $A$ with the discrete topology and let $\beta A$ be its Stone–Čech compactification.  Fix some $p \in \beta A \setminus A$.  Since $A$ is dense in $\beta A$, there is a net $(x_\alpha)_{\alpha \in I} \subset A$ which converges to $p$.
Suppose $x \in A$.  Since $\beta A$ is Hausdorff, $\{x\}$ is closed in $\beta A$.  So $\beta A \setminus \{x\}$ is an open neighborhood of $p$.  As such, $(x_\alpha)$ is eventually in $\beta A \setminus \{x\}$, so it is not the case that $x$ is frequently in $(x_\alpha)$.  Thus $(x_\alpha)$ satisfies the desired hypothesis.
Let $B$ be any infinite subset of $A$, and let $C_0, C_1$ be any two disjoint infinite subsets of $B$.  Let $f : A \to \{0,1\}$ be any function which maps $C_0$ to $0$ and $C_1$ to $1$; note $f$ is continuous since $A$ is discrete.  By the universal property of $\beta A$, $f$ has a continuous extension $g : \beta A \to \{0,1\}$.  
Suppose first that $g(p) = 0$.  By continuity we have $g(x_\alpha) \to 0$ which means that $g(x_\alpha) = 0$ eventually.  Since $g(C_1) = 1$ it is not the case that $x_\alpha$ is frequently in $C_1$.  
Similarly, if we have $g(p)=1$ then it is not the case that $x_\alpha$ is frequently in $C_0$.

Alternate phrasing:
Let $B$ be any infinite subset of $A$.  By the lemma below, $B$ has at least two limit points, so in particular it has a limit point $q \ne p$.  Since $\beta A$ is Hausdorff, $p,q$ have disjoint open neighborhoods $U,V$.  Let $C = B \cap V$, which is infinite because $q$ is a limit point of $B$.  As before, $(x_\alpha)$ is eventually in $U$, which is disjoint from $C$.  So it is not the case that $C$ is frequently in $(x_\alpha)$.
Lemma. Every infinite subset $B$ of $A$ has at least two limit points in $\beta A$.  (Actually it can be shown that it has $2^{\mathfrak{c}}$ many limit points, see Result 4 here.)
Proof. Partition $B$ into two infinite subsets $B_0, B_1$.  Consider the continuous function $f : A \to \{0,1\}$ which maps $B_0$ to $0$ and $A \setminus B_0$ to $1$.  By the universal property of $\beta A$, $f$ has a continuous extension $g : \beta A \to \{0,1\}$.  Let $E_i = g^{-1}(\{i\})$ for $i=0,1$, so that each $E_i$ is compact and $B_i \subset E_i$.  Hence each $B_i$ has a limit point $q_i$ in $E_i$.  Necessarily $q_1, q_2$ are distinct, since $E_1, E_2$ are disjoint, and $q_1, q_2$ are both limit points of $B$.