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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.

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  • $\begingroup$ If I'm reading your question right, a net that maps everything to a constant value is a counterexample. Any infinite subset B of A will itself have an infinite subset C that doesn't have that constant value in it. $\endgroup$ – btilly Mar 24 '17 at 23:10
  • $\begingroup$ Thanks! I added the needed assumption. $\endgroup$ – mathworker21 Mar 24 '17 at 23:15
  • $\begingroup$ Why do you write "$C$ is frequently in $(x_\alpha)_{\alpha \in I}$" when you mean "$(x_\alpha)_{\alpha \in I}$ is frequently in $C$"? $\endgroup$ – bof Mar 25 '17 at 2:56
  • $\begingroup$ Because I'm stupid $\endgroup$ – mathworker21 Mar 25 '17 at 3:13
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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 $B$.

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  • $\begingroup$ Amazing! Deserves many upvotes $\endgroup$ – mathworker21 Mar 25 '17 at 3:07
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I don't have an answer, but I can get a limited version of the result. And that may give others insight into why it might or might not be true.

Your result is true if the domain $D$ of the net $\Phi$ is a countable set $d_1, d_2, \ldots$. (Those order has nothing to do with the counting.)

We will have $B = \Phi(E)$ where $E$ is to be figured out. Its first element is $d_1$. If we have already chosen $e_1, \ldots, e_n$ then we pick $e_{n+1}$ to be the first element in the counting of $D$ that is $\ge$ to every element up to and including $e_n$, and is mapped to something different than $\Phi(e_1), \ldots, \Phi(e_n)$. This always exists because none of that finite number of points is an accumulation point, and we can always find something in the domain that is bigger than any finite set of points.

Now $B$ is an infinite set. And for any infinite subset $C$ of $B$, and any element $d$ of the domain, all but a finite number of elements of $C$ are the image of something that is $\ge d$.

(Of course the case you're probably interested in is an uncountable domain $D$, but this is the argument that I came up with...)

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  • $\begingroup$ Hi, thanks for your answer. And yea, i'm mostly worried about the case when the net is over an uncountable indexed set $I$. $\endgroup$ – mathworker21 Mar 25 '17 at 2:27
  • $\begingroup$ I'm thinking maybe some Zorn's Lemma argument would work, I don't know. $\endgroup$ – mathworker21 Mar 25 '17 at 2:28

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