# How is every subset of the set of reals with the finite complement topology compact?

While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)?

What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology?

• I'm almost a newbie in topology, but isn't $[0,1]$ an infinite set, which doesn't have a finite complement a counterexample of the statement you want to prove? Aug 16, 2013 at 9:43
• @Dima: $[0,1]$ is not open. But it's still compact in this topology. Aug 16, 2013 at 9:53
• You could prove the following: If every proper closed subset of $X$ is compact, then $X$ is compact. This is true for every space $X$. Now in your case the proper closed subsets are all finite, thus compact. Aug 16, 2013 at 13:39

Let $X$ be any set endowed with the finite complement topology. Let $\mathcal{A}=\{A_i|i\in I\}$ be an open cover of $X$ (where $I$ is an arbitrary index set). Take any $A_0\in\mathcal{A}$, then $X\backslash A_0$ contains only a finite number of points $\{x_1,\ldots,x_n\}$. For any $1\leq k\leq n$, choose $A_k\in\mathcal{A}$ containing $x_k$ (such an element of $\mathcal{A}$ exists since $\mathcal{A}$ covers $X$). Then $\{A_0,\ldots,A_n\}$ is a finite subcover of $X$. Thus $X$ is compact.

• This obviously works any set endowed with the cofinite topology. I do have a question: in order for this proof to go through, don't we also need to show that the subspace topology on a subset is also cofinite? This isn't challenging to show, but this is, strictly speaking, needed in order for your proof to apply to all subspaces of a given space with the cofinite topology, right? May 25, 2017 at 20:49
• @user193319 If you want to prove that every subset of $\mathbb{R}$ with the cofinite topology is compact, then yes, you need to show that such a subset $X$ also has the cofinite topology before applying my argument above. May 25, 2017 at 23:14

Note that given any non-empty open set $U$, there are only finitely many points outside of $U$. So we only need a finite number of open sets to get a covering of $\Bbb R$.

The same goes for any non-empty subset of $\Bbb R$. If $\cal U$ is an open cover of some set $A$, then picking any open set $U$ in $\cal U$, the remainder of the entire space (and in particular $A\setminus U$) is finite. So we can find some finitely many open sets from $\cal U$ to cover this remainder.

Therefore every open cover has a finite subcover. So $A$ is compact, but $A$ is arbitrary, so every subset of $\Bbb R$ is compact.

• I'm trying to understand the axiom of choice. Is it correct that this argument uses it twice in the second paragraph: in the first sentence ("picking") and again in the second sentence ("find")? Sep 11, 2015 at 23:21
• No, the argument uses it exactly zero times. If you include choosing from finitely many sets using the axiom of choice, however, then you are correct. Sep 12, 2015 at 3:25
• Hm I'm confused. The open cover $\cal U$ isn't assumed to be finite so when you say "picking any open set $\cal U$ in U, aren't you picking an element out of a potentially infinite set? Also while $A\setminus U$ is finite, for any given point in $A\setminus U$ there may be infinitely many sets of the cover containing that point and you are saying you can choose one. It seems you are choosing a set out of a potentially infinite collection of sets in both these instances. Doesn't that require the axiom of choice? Sep 13, 2015 at 0:04
• The axiom of choice is needed when choosing from infinitely many sets simultaneously. Not when choosing from a single infinite set. Sep 13, 2015 at 4:37
• This answer should have received the most up votes in my opinion. Thanks Nov 25, 2021 at 3:53

A formal proof:

Let $$\mathcal{A}\subseteq\tau$$ and $$\mathbb{R}=\cup\mathcal{A}.$$ (We can suppose that $$\emptyset\notin \mathcal{A}.$$ Why?)

$$A\in \mathcal{A}\Rightarrow |\setminus A|<\aleph_0\Rightarrow (\exists x_1,x_2,\ldots,x_n\in \mathbb{R})(\setminus A=\{x_1,x_2,\ldots,x_n\})$$

$$\left.\begin{array}{rr}\Rightarrow (\exists x_1,x_2,\ldots,x_n\in \mathbb{R})(\mathbb{R}=A\cup (\setminus A)=A\cup\{x_1,x_2,\ldots,x_n\}) \\ \\ \mathbb{R}=\cup\mathcal{A}\end{array}\right\}\Rightarrow$$

$$\left.\begin{array}{rr}\Rightarrow (\exists B_1,B_2,\ldots, B_n\in\mathcal{A})(x_1\in B_1)(x_2\in B_2)\ldots (x_n\in B_n) \\ \\ \mathcal{A}^*:=\{A,B_1,B_2,\ldots,B_n\}\end{array}\right\}\Rightarrow$$

$$\Rightarrow (\mathcal{A}^*\subseteq\mathcal{A})(|\mathcal{A}^*|=n+1<\aleph_0)(\mathbb{R}=\cup\mathcal{A}).$$