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If $x:\Lambda \rightarrow X$ is a net in a topological space $X$ and $\Lambda '\subseteq \Lambda$ is a cofinal subset of the directed set $\Lambda$, then $x|_{\Lambda '}$ is a subnet of $x$. We call subnets of this form strict subnets.

Of course, not all subnets are of this form, and indeed, the definition of a subnet is chosen the way it is in part so that certain theorems we would like to be true are in fact true. I am in particular curious about the following standard theorem.

$X$ is quasicompact iff every net in $X$ has a convergent subnet.

Looking at the proof, I definitely see where it would no go through if we were restricted to strict subnets, and it seems as if there would be no easy way to fix this with this restriction. Of course, this doesn't mean that it can't be proven---maybe we just haven't been clever enough to figure out how to do it with only strict subnets? I am quite confident this is not actually the case, however, and that there is indeed a counter-example.

Kelley presents an example (pg. 77 of his General Topology, attributed to Arens) that is meant to demonstrate why strict subnets are not enough. He presents an example of a net with a cluster point to which no subnet converges. Unfortunately, the example he gives is Lindelöf, but not quasicompact (and personally, I could live with nets not having subnets converging to cluster points (this is already the case for limit points, for example)).

So then, is there a well-known example of a quasicompact space and a net with no convergent strict subnet?

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This phenomenon is very typical in compact spaces whose construction involves taking an uncountable product. For instance, consider $X=\{0,1\}^{\mathcal{P}(\mathbb{N})}$ with the product topology. Consider the sequence $(x_n)$ in $X$ where $x_n(A)=1$ iff $n\in A$. If a subsequence were to converge, that would mean there is an infinite subset $B\subseteq\mathbb{N}$ such that for all $A\subseteq\mathbb{N}$, $A\cap B$ is either finite or cofinite in $B$. Obviously no such subset exists. But $X$ is compact by Tychonoff's theorem.

In fact, this example sits inside the closed subspace $\beta\mathbb{N}\subset \{0,1\}^{\mathcal{P}(\mathbb{N})}$ of (characteristic functions of) ultrafilters on $\mathbb{N}$, and each $x_n$ is the principal ultrafilter corresponding to $n$. This subspace is known to be the Stone-Cech compactification of $\mathbb{N}$; that is, the free compact Hausdorff space on the set $\mathbb{N}$. So this is actually the universal example of a sequence in a compact Hausdorff space. It is thus unsurprising that this sequence has no convergent subsequence: if there existed any sequence in a compact Hausdorff space with no convergent subsequence, then by universality this sequence would have to be such a sequence.

This example has obvious generalizations: whenever you have different nets with the same index set on a bunch of compact spaces which converge on different strict subnets, you can form the product space, and corresponding net in the product will have a convergent strict subnet only if there was a common strict subnet of all the strict subnets on which the different original nets converged.

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    $\begingroup$ In other words...compact does not imply sequentially compact. $\endgroup$ – Pete L. Clark Jul 7 '15 at 2:49

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