Definition: A net in a topological space $X$ is universal if for any $A\subseteq X$, either the net is eventually in $A$ or it is eventually in $X\setminus A$.

Problem: A sequence is a universal net if and only if it is eventually constant.

I've shown that if a sequence is eventually constant, then it is universal. I'm trying to figure out the other direction.

My attempt: Let $X$ be a topological space, $\Phi \colon \mathbb{Z}^+ \to X$ be a sequence, and suppose $\Phi$ is not eventually constant. Consider two cases.

Case 1: There is an element $x_0\in X$ such that $\Phi(n) = x_0$ for infinitely many $n$. In this case, let $A = \{x_0\}$. Then given any $n$, there exists $m\geq n$ and $k\geq n$ such that $\Phi(m) = x_0\in A$ and $\Phi(k) \neq x_0$ so $\Phi(k) \notin A$. Thus $\Phi$ cannot eventually be in $A$ nor can it eventually be in $X\setminus A$, so $\Phi$ is not universal.

Case 2: There are infinitely many elements in $\Phi(\mathbb{Z}^+)$, each appearing finitely many times in the sequence.

To finish case 2, I would now like to define two disjoint subsequences (in that they share no common elements), which I think should be possible. This would then finish the proof. However, I cannot think of a way to do this precisely.

Question: Is this proof going in the right direction?

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    $\begingroup$ Do you have a copy of the ebook of Wilde, entitled "Basic Analysis-Gently Done Topological Vector Spaces"? There you may look at page 22. $\endgroup$ – Juniven Aug 6 '13 at 2:33
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    $\begingroup$ You can write $\Phi(\mathbb{Z}^+)$ as a disjoint union of two infinite sets $S$ and $T$. Then you can define two disjoint subsequences using $\Phi^{-1}(S)$ and $\Phi^{-1}(T)$ (any infinite subset of $\mathbb{Z}^+$ is order-bijective with $\mathbb{Z}^+$). To finish the proof you just need to find the right choice of $A$. $\endgroup$ – Kevin Ventullo Aug 6 '13 at 3:19

The idea indeed is to split it into two subsequences.

Let $\Phi: \mathbb{Z}^+ \to X$ be your sequence, which is a universal net. Suppose that $\Phi$ is not eventually constant. We will get a contradiction:

Let $x$ be any value of the sequence, so a point in $\Phi[\mathbb{Z}^+]$. As $\Phi$ is not eventually constant, it is not eventually in $\{x\}$, so by being a universal net, it must be eventually in $X \setminus \{x\}$, so after some initial finite segment, all values are unequal to $x$. This shows that $x$ appears at most finitely many times as a value. As $x$ was arbitrary, all values of the sequence appear finitely many times, and hence so $\Phi[\mathbb{Z}+]$ is infinite.

Now split that image set into two disjoint infinite sets $A$ and $B$. Now $\Phi$ cannot be eventually in $A$ (as a finite start segment cannot exhaust all values that are in $B$, so there are always points beyond it with values in $B$, which are not in $A$), and symmetrically, it cannot be eventually in $X\setminus A$ as well. This contradicts the universality of the sequence.

So the assumption led to a contradiction, and $\Phi$ is eventually constant.

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  • $\begingroup$ Since I struggled a little to understand Henno Brandsa's parenthetical remark, let me add a few clarifying words. We claim $\Phi$ cannot be eventually in $A$. If that were the case, then $\Phi$ visited $B$ only a finite number of times. But this cannot happen, since $B$ is infinite, and $\Phi$ must visit every element in it, because $B\subset \Phi[\mathbb{Z}^+]$. $\endgroup$ – Teddy Sep 14 '17 at 12:56

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