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I am reading Kelley’s book on general topology. There are a few statements on nets there (chapter 2), but the characterization of compact sets in the language of nets is not given. How should we prove the following

Theorem: A topological space X is compact iff every net has a convergent subnet.

Stefan Hamcke
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superAnnoyingUser
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3 Answers3

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See Theorem $15.3$ in this excellent PDF, Translating Between Nets and Filters, by Saitulaa Naranong; it’s well worth reading the whole thing.

Added 28 May 2024: The link has been updated to point to the copy archived at the Wayback Machine. As noted in the comments, $\Phi$ and $\Psi$ are interchanged in the displayed line in Definition $\mathbf{10.2}$, but the one-sentence paragraph immediately following Definition $\mathbf{11.1}$ is correct.

Brian M. Scott
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  • An interesting paper, but I was a bit confused, because they switched $\Psi$ and $\Phi$ in the definition of subnet on page 11. – Stefan Hamcke Mar 09 '13 at 18:29
  • @Stefan: I’m not sure what you mean: on that page he consistently uses $\Psi$ for the subnet and $\Phi$ for the original net. – Brian M. Scott Mar 09 '13 at 18:32
  • I mean $\Psi$ is subnet of $\Phi$ iff $\Psi$ is eventually in all the sets in which $\Phi$ is eventually. So in the second line of Definition 10.2 the arrow should point in the other direction. – Stefan Hamcke Mar 09 '13 at 18:39
  • @Stefan: No, the definition goes the other way: $\Phi$ is eventually in all of the sets in which $\Psi$ is eventually. The subnet may be in more sets eventually. – Brian M. Scott Mar 09 '13 at 18:41
  • Are you sure? I tried Theorem 11.3.(1), and it seems to work only if it's the other way. – Stefan Hamcke Mar 09 '13 at 18:43
  • For example, the sequence $(\frac1n)\mathbb N$ should rather be a subnet of $(n%2)\mathbb N$ than the other way. – Stefan Hamcke Mar 09 '13 at 18:49
  • @Stefan: No, it works only with the definition as given. The subnet is eventually in at least as many sets, so its associated filter has at least as many sets. – Brian M. Scott Mar 09 '13 at 18:49
  • @Stefan: What do you mean by $(n%2)_{\Bbb N}$? – Brian M. Scott Mar 09 '13 at 18:50
  • the remainder $n$ mod 2, the sequence alternating between $0$ and $1$. – Stefan Hamcke Mar 09 '13 at 18:51
  • If $\Psi$ is eventually in the sets of $\Phi$, then a tail of $\Phi$ contains a tail of $\Psi$ as a subset, hence Filter($\Phi$)$\subseteq$Filter($\Psi$). – Stefan Hamcke Mar 09 '13 at 18:55
  • @Stefan: Neither of those is a subnet of the other: the first is eventually in $(-1/2,1/2)$, and the second one isn’t, while the second is eventually in ${0,1}$, and the first one isn’t. – Brian M. Scott Mar 09 '13 at 18:58
  • Oops, you're right. Replace the first one by any sequence which is eventually $0$, but maybe differs at the first few terms. – Stefan Hamcke Mar 09 '13 at 19:01
  • @Stefan: I don’t understand what you mean by $\Psi$ being eventually in ‘the sets of $\Phi$’. If you mean that $\Phi$ eventually in $A$ implies that $\Psi$ is eventually in $A$, then you have the filter inclusion backwards. – Brian M. Scott Mar 09 '13 at 19:01
  • Yes, that is what I mean. Doesn't my second to last comment make sense? – Stefan Hamcke Mar 09 '13 at 19:02
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    @Stefan: Aaargghh! You’re right: I’m so used to the ideas that I kept reading what it meant instead of what it said. The one-sentence paragraph under Definition $11.1$ is right, and $\Phi$ and $\Psi$ are indeed reversed in the displayed line of Definition $10.2$. – Brian M. Scott Mar 09 '13 at 19:06
  • :-D Yes, everything they write makes sense except that particular line – Stefan Hamcke Mar 09 '13 at 19:09
  • @Stefan: So is everything okay now? – Brian M. Scott Mar 09 '13 at 19:10
  • Yes, of course, if both of us agree now, everthing is fine :-) – Stefan Hamcke Mar 09 '13 at 19:13
  • @Stefan: Great! whew :-) – Brian M. Scott Mar 09 '13 at 19:13
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    The link is broken but http://web.archive.org/web/20130308175220/http://www.math.tamu.edu/%7Esaichu/netsfilters.pdf still works. – Watson Mar 07 '17 at 12:24
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Suppose that $x_\alpha$ is a net in $X$ with no convergent subnet. Then $(x_\alpha)$ do not have accumulation point in $X$. For each $x\in X$, let $V_x$ be an open neighbourhood of $x$ that excludes all the part of the net from some term onward. Let $V=\{V_x:\ x\in X\}$ and note that $V$ is an open cover of $X$. Can you prove that it is impossible to find a finite subcover of $X$ in $V$?

On the other hand, let $V$ be a open cover of $X$ such that every finite subcover of $V$ do not cover $X$. Consider the open cover $U$ of $X$ consisting of finite unions of elements of $V$. If $A,B\in U$, we say that $A\leq B$ when $A\subseteq B$. With this relation, $V$ is an directed set. For $A\in V$, let $x_A\in X\setminus A$. Can you show that the net $x_A$ does not have any convergent subnet?

Tomás
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If you are studying Kelley, and you want to prove this subnet characterization of compactness, then a hint is Problem 2 J.

GEdgar
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