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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.

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3 Answers 3

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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.

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  • $\begingroup$ An interesting paper, but I was a bit confused, because they switched $\Psi$ and $\Phi$ in the definition of subnet on page 11. $\endgroup$ Commented Mar 9, 2013 at 18:29
  • $\begingroup$ @Stefan: I’m not sure what you mean: on that page he consistently uses $\Psi$ for the subnet and $\Phi$ for the original net. $\endgroup$ Commented Mar 9, 2013 at 18:32
  • $\begingroup$ 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. $\endgroup$ Commented Mar 9, 2013 at 18:39
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    $\begingroup$ @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 $10.2$. $\endgroup$ Commented Mar 9, 2013 at 19:06
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    $\begingroup$ The link is broken but web.archive.org/web/20130308175220/http://www.math.tamu.edu/… still works. $\endgroup$
    – Watson
    Commented Mar 7, 2017 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?

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  • $\begingroup$ I suppose you mean for me to prove those claims by contradiction. Can you give me a hint about where the contradictions stem from. $\endgroup$ Commented Mar 13, 2013 at 18:08
  • $\begingroup$ Where do you got stucked? $\endgroup$
    – Tomás
    Commented Mar 13, 2013 at 18:43
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    $\begingroup$ I'm about as far as nowhere. $\endgroup$ Commented Mar 13, 2013 at 19:41
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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.

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