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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?

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  • $\begingroup$ 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? $\endgroup$ Aug 16, 2013 at 9:43
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    $\begingroup$ @Dima: $[0,1]$ is not open. But it's still compact in this topology. $\endgroup$
    – Asaf Karagila
    Aug 16, 2013 at 9:53
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    $\begingroup$ 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. $\endgroup$ Aug 16, 2013 at 13:39

3 Answers 3

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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.

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    $\begingroup$ 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? $\endgroup$
    – user193319
    May 25, 2017 at 20:49
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    $\begingroup$ @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. $\endgroup$ May 25, 2017 at 23:14
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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.

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    $\begingroup$ 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")? $\endgroup$
    – Smithey
    Sep 11, 2015 at 23:21
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    $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Sep 12, 2015 at 3:25
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    $\begingroup$ 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? $\endgroup$
    – Smithey
    Sep 13, 2015 at 0:04
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    $\begingroup$ The axiom of choice is needed when choosing from infinitely many sets simultaneously. Not when choosing from a single infinite set. $\endgroup$
    – Asaf Karagila
    Sep 13, 2015 at 4:37
  • $\begingroup$ This answer should have received the most up votes in my opinion. Thanks $\endgroup$
    – sathishT
    Nov 25, 2021 at 3:53
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A formal proof:

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

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

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

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

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

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