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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?

Keith
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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.

Billy
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  • "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. – Acccumulation Feb 26 '18 at 03:50
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    @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 – Exit path Feb 26 '18 at 07:50
  • 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. – MaoWao Feb 26 '18 at 08:17
  • @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. – Billy Feb 26 '18 at 09:31
  • @MaoWao Thanks for your comments! – Billy Feb 26 '18 at 09:32
  • @leibnewtz By "OP", do you mean Keith or Billy? Keith did not ask whether a singleton has a topological space in which it is compact, but rather whether it is compact in every topological space. If someone were to ask "Given a continuous function, is the pre-image of an open set an open set?", responding "Sure! All sets are open in their own subspace topology" would not be an appropriate answer. – Acccumulation Feb 26 '18 at 15:39
  • @Acccumulation "Compactness" isn't relative to an ambient space, whereas "openness of a set" is. I define a subset U of a space X to be compact if U is itself a compact top. space in the subspace topology inherited from X. Maybe you define it differently, e.g. "any collection of open sets of X which covers U can be refined to a finite one". But these definitions are equivalent, and my answer remains essentially the same: take a cover of open sets of X, intersect them all with U, find a finite subcover there, as open sets of U, and then lift the finite subcover back up to X. – Billy Feb 26 '18 at 16:25
  • (As a linguistic point: this is why I called them "finite spaces", i.e. topological spaces on a finite set, rather than "finite subsets of topological spaces". :)) – Billy Feb 26 '18 at 16:31
  • @Acccumulation We first define compactness intrinsically, and then we say a subset $A \subseteq X$ is compact in some space $X$ if it's compact in the subspace topology – Exit path Feb 26 '18 at 19:08
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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.

Andres Mejia
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    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! – Billy Feb 25 '18 at 17:59
  • 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 $x_2$? – YCB Feb 25 '18 at 18:00
  • @Gödel sure, that is true. – Andres Mejia Feb 25 '18 at 18:03
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    @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. – Andres Mejia Feb 25 '18 at 18:04
  • @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. – Matthew Feb 25 '18 at 21:37
  • Your argument is rather vague. For starters, you should define what your terms refer to. – Acccumulation Feb 26 '18 at 03:55
  • @Acccumulation I fail to see anywhere where the terms are unclear. I also don’t know where your criticism is either – Andres Mejia Feb 26 '18 at 04:13
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    The notation {Ux} is vague. You call this "induction", but there is no base case, inductive hypothesis, or proof of the n+1 case. I guess your idea is to take a Ui for each xi, and then the collection of Ui is then a finite subcover, but you just leave that to the reader to figure out. – Acccumulation Feb 26 '18 at 04:24
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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.

Ri-Li
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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.

Joonas Ilmavirta
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  • While your second paragraph is true, there are not too many topologies on a one-point set. ;) – MaoWao Feb 26 '18 at 08:19
  • @MaoWao I was referring to the topology of the ambient space, where there might be a bigger selection. :) I'll edit to clarify that... – Joonas Ilmavirta Feb 26 '18 at 08:20