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I'm reviewing some topology and have found myself at a bit of a loss after trying to prove the proposition below. Proposition. A space is compact iff every net has an accumulation point.

My problem is that I have produced a "proof" of the incorrect claim: "every cover has a finite subcover if and only if every net has an accumulation point." Of course this is not true since we can always cover a space by its singletons, and not every sequentially compact space is finite. Where is the necessary topological input that I have lost?

"Proof" of $\implies$:

Assume every cover has a finite subcover and towards a contradiction suppose $(x_\alpha)$ has no accumulation point. Then for every $x\in X$ there's a neighborhood not frequented. These neighborhoods yield a cover $(U_x)$, which has a finite subcover $(U_{x_i})$. That $U_{x_i}$ is not frequented means there's a $\beta_{x_i}$ in the directed set beyond which the net does not intersect $U_{x_i}$. By directedness, there's a $\beta$ greater than all the $\beta_{x_i}$, thus $(x_\alpha)_{\alpha\geq \beta}\cap X=\emptyset$. But the net is assumed in $X$ which is a contradiction.

"Proof" of $\impliedby$:

Assume every net has an accumulation point. Let $(B_i)$ be a family of subsets of $X$. To add direction let $(C_\alpha)$ be the family obtained by adjoining their finite intersections ordered by containment. This induces a direction on $A$. For each $C_\alpha$ choose an element $x_\alpha$ and define a net $\alpha\mapsto x_\alpha$. By assumption this net has an accumulation point $x$. Noting $(B_i)$ has the finite intersection property iff $(C_\alpha)$ does, it suffices to assume the latter and prove $\bigcap_\alpha C_\alpha\neq \emptyset$. We prove by contradiction the accumulation point $x$ is in this intersection. If not, it's in its complement and by de Morgan in some $C^c_\alpha$. By directedness, $x_\beta \in C^c_\alpha$ for some $\beta>\alpha$. Thus $x_\beta\in C^c_\alpha\cap V_\beta=\emptyset$ which is a contradiction.

What am I completely missing? Where is openness/closedness needed?

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    $\begingroup$ $\frac{1}{2n} \in \bigl(0,\frac{1}{n}\bigr)$, but $0 = \lim \frac{1}{2n} \notin \bigl(0,\frac{1}{k}\bigr)$ for any $k$. You need closed subsets to have the conclusion that the limit belongs to the sets. $\endgroup$ – Daniel Fischer Oct 25 '16 at 19:19
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The forward direction is fine. The problem with the other direction is the one noted by Daniel Fischer in the comments: you can’t prove that $x\in\bigcap_\alpha C_\alpha$ unless the sets $C_\alpha$ are closed.

We have a family $\mathscr{C}=\{C_\alpha:\alpha\in\Lambda\}$ of non-empty sets that is closed under finite intersections, so that $\langle\mathscr{C},\supseteq\rangle$ is a directed set, and we have a net $\langle x_\alpha:\alpha\in\Lambda\rangle$ such that $x_\alpha\in C_\alpha$ for each $\alpha\in\Lambda$. Finally, we have an accumulation point $x$ of this net.

Fix $\alpha\in\Lambda$. If $U$ is an open nbhd of $x$, there is a $\beta\in\Lambda$ such that $C_\beta\subseteq C_\alpha$ and $x_\beta\in U$. Evidently $x_\beta\in C_\alpha$, so $U\cap C_\alpha\ne\varnothing$, and it follows that $x\in\operatorname{cl}C_\alpha$. Since $\alpha$ was arbitrary, $x\in\bigcap_{\alpha\in\Lambda}\operatorname{cl}C_\alpha$. If the sets $C_\alpha$ are closed, we can conclude that $x\in\bigcap_{\alpha\in\Lambda}C_\alpha$ and hence that $X$ is compact.

If the sets $C_\alpha$ are not closed, however, $x$ need not be in any of them, and their intersection may indeed be empty even if $X$ is compact, as the following example shows.

Let $X=[0,1]$ with the usual topology, and let $C_n=(0,2^{-n})$ for $n\in\Bbb N$. The family $\mathscr{C}=\{C_n:n\in\Bbb N\}$ is closed under finite intersections and has the finite intersection property. Following the proof, let $x_n\in C_n$ for each $n\in\Bbb N$. The resulting net is actually just the sequence $\langle x_n:n\in\Bbb N\rangle$, which clearly converges to $0$ and therefore has $0$ as an accumulation point; indeed, $0$ is its only accumulation point. However,

$$\bigcap_{n\in\Bbb N}C_n=\bigcap_{n\in\Bbb N}(0,2^{-n})=\varnothing\;.$$

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  • $\begingroup$ Very clear, thank you. In the second line of the third paragraph, is $U_\alpha$ actually $C_\alpha$? $\endgroup$ – Arrow Oct 25 '16 at 22:46
  • $\begingroup$ @Arrow: You're welcome. It is indeed; thanks for catching that. $\endgroup$ – Brian M. Scott Oct 25 '16 at 22:53
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If every cover has a finite subcover, then the space is finite, because, as you observe, the family of singletons is a cover.

In a finite space every net must visit at least one point infinitely often, so there's an eventually constant subnet.

Your argument for $\impliedby$ fails at the start: the set of finite intersections is directed by cointainment, but why should you be able to pick a point in every $C_\alpha$? A finite intersection of members of the family $(B_i)$ can well be empty.

The proof that “if any net has a convergent subnet then any open cover has a finite subcover” is more involved.

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  • $\begingroup$ I picked an element of each $C_\alpha$ under the assumption $(B_i)$ has the finite intersection property. I think this does not need them to be closed sets yet. $\endgroup$ – Arrow Oct 26 '16 at 8:48
  • $\begingroup$ @Arrow I see; your $(B_i)$ is the family of complements of the elements of a cover? $\endgroup$ – egreg Oct 26 '16 at 10:42
  • $\begingroup$ Yep. Rereading though, what I wrote is unclear... $\endgroup$ – Arrow Oct 26 '16 at 10:51

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