# Are one-point sets always compact in any topological spaces?

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?

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.

• "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 3:50
• @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 – leibnewtz Feb 26 '18 at 7:50
• 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 8:17
• @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 9:31
• @MaoWao Thanks for your comments! – Billy Feb 26 '18 at 9:32

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.

• 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 $x_2$? – Gödel Feb 25 '18 at 18:00
• @Gödel sure, that is true. – Andres Mejia Feb 25 '18 at 18:03
• @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

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.

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.

• While your second paragraph is true, there are not too many topologies on a one-point set. ;) – MaoWao Feb 26 '18 at 8:19
• @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 8:20