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From the definition of compactness, I think one-point sets are always compact in any topological space. But, I am not sure about my judgement. Am I correct?

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Since I was asked to repost my comment as an answer:

Any open cover of any topological space is a subset of the power set of the underlying set, and power sets of finite sets are finite. So all open covers of finite spaces are already finite; in other words, finite spaces only have finitely many open sets.

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  • $\begingroup$ "So all open covers of finite spaces are already finite" That doesn't follow. An open cover is a set of elements from the power set of the topological space, not of the covered space. Consider the set {0}. This has an infinite cover of {(n-2,n+2)} in R. So your argument fails to establish that {0} is compact in every topological space. $\endgroup$ – Acccumulation Feb 26 '18 at 3:50
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    $\begingroup$ @Acccumulation If one considers a finite set $X$ lying inside an ambient space as a topological space with the subspace topology then the OP is correct, since $X$ will only have finitely many open sets to begin with $\endgroup$ – leibnewtz Feb 26 '18 at 7:50
  • $\begingroup$ Two caveats: Some (French) people include Hausdorff in their definition of compact. This certainly holds for singletons, but not for arbitrary finite sets. Second, I would rather say that a cover is a family of open sets and what you describe is merely the image of this map, but that does not effect your argument. $\endgroup$ – MaoWao Feb 26 '18 at 8:17
  • $\begingroup$ @Acccumulation Yes, as leibnewtz said, I was considering the subspace topology. Inside $\mathbb{R}$, the subset $\{0\}$ has lots of open neighbourhoods. However, the definition of "compactness" doesn't require us to cover $\{0\}$ with open sets of $\mathbb{R}$, but with open sets of $\{0\}$. There are far fewer of those. $\endgroup$ – Billy Feb 26 '18 at 9:31
  • $\begingroup$ @MaoWao Thanks for your comments! $\endgroup$ – Billy Feb 26 '18 at 9:32
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Every finite set is compact. This is because, one can always find a finite subcover, which can be proven inductively:

Let $X:=\{x_1, \dots x_n\}$ be a finite set. Suppose that $\{U_x\}$ covers $X$. Take any $U_{x_1}$ that covers $x_1$. Consider $\{U_x\}\setminus U_{x_1}$. Then pick one that covers $x_2$ if $U_{x_1}$ does not, etc.

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    $\begingroup$ This sounds very indirect to me. Surely the real point is that any open cover of any topological space is a subset of the power set of the underlying set, and power sets of finite sets are finite! $\endgroup$ – Billy Feb 25 '18 at 17:59
  • $\begingroup$ You must consider cases because, what happens if $x_2\in U_{x_1}$ and there is not another open set in $\{U_x\}$ such that contains to $x_2$? $\endgroup$ – Gödel Feb 25 '18 at 18:00
  • $\begingroup$ @Gödel sure, that is true. $\endgroup$ – Andres Mejia Feb 25 '18 at 18:03
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    $\begingroup$ @Billy post a new answer, I think that is a nice point as well. Mine was moreso not considering subspaces of potentially infinite spaces, but it is true that no matter which way, in the subspace topology your argument works. $\endgroup$ – Andres Mejia Feb 25 '18 at 18:04
  • $\begingroup$ @Gödel That is a simple modification. Instead of $x_2$, cover the first $x_i$ such that $x_i \notin U$ (assuming that all points are not already covered). We know that this will always be possible. $\endgroup$ – Matthew Feb 25 '18 at 21:37
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Yes, one point set is always compact in any topological space, because it will be contained in an open set of any cover and that is the finite one.

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Yes. Suppose the open sets $U_i$ with $i$ in some index set $I$ cover your one-point set $\{x\}$. Covering this set means that $x\in U_i$ for some $i\in I$. Therefore your one-point set is contained in a single open set of your open cover. This is certainly a finite subcover!

Notice that the argument did not use the fact that the sets $U_i$ are open. This is a symptom of generality: It is true for any sets $U_i$ and therefore for any topology of the ambient space at all.

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  • $\begingroup$ While your second paragraph is true, there are not too many topologies on a one-point set. ;) $\endgroup$ – MaoWao Feb 26 '18 at 8:19
  • $\begingroup$ @MaoWao I was referring to the topology of the ambient space, where there might be a bigger selection. :) I'll edit to clarify that... $\endgroup$ – Joonas Ilmavirta Feb 26 '18 at 8:20

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