Let $X$ a metric space, and $\mathcal M$ a family of closed subsets of $X$ with the finite intersection property, i.e. with the property that every finite intersection of elements of $\mathcal M$ is non-empty.

Show that $X$ is compact if and only if every family of closed sets with the finite intersection property have non-empty intersection.

Let define $\mathcal M:=\{A_\alpha\in X:A_\alpha=\overline A_\alpha\land \alpha\in I\}$ with the properties defined above, i.e.

$$\bigcap \mathcal M\neq\emptyset\iff\bigcap_{\alpha\in J} A_\alpha\neq\emptyset,\;\forall J\subseteq I:|J|\in\Bbb N$$

(The right implication is obvious.)

My strategy for this proof is to show that $X$ is complete and totally bounded, what is equivalent to be compact.


$X$ is complete: for every Cauchy sequence $(x_n)$ in $X$ start choosing $X$ as a closed neighborhood for $x_0$. Because $(x_n)$ is Cauchy for every $\epsilon=1/k$ exists some $N\in\Bbb N$ such that $d(x_n,x_m)<1/k$ whenever $n,m\ge N$.

Then for every Cauchy sequence on $X$ we can define the family of sets

$$\mathcal B:=\{B_k:B_k=\overline{\Bbb B}(x_N,1/k), k\in\Bbb N_{>0}\land B_0=X\}$$

what is a family of nested closed sets with the property of non-empty finite intersection. Then if $\bigcap\mathcal B\neq\emptyset$ then $X$ is complete, i.e every Cauchy sequence converges in $X$.

$X$ is totally bounded: if $X$ is not totally bounded then exists some $\epsilon>0$ such that doesnt exists a finite open cover of $X$ composed of open balls with centers in $X$. Then for some $\epsilon>0$ the family

$$C_\epsilon :=\{\Bbb B(x,\epsilon):x\in X\}$$

doesn't contain a finite subcover. Now define the family composed by the complements of $C_\epsilon$:

$$C_\epsilon^*:=\{(\Bbb B(x,\epsilon))^\complement:x\in X\}$$

then observe that the family $C_\epsilon^*$ have the finite intersection property because in other case $C_\epsilon$ would have a finite subcover.

But we have that

$$\bigcap C_\epsilon^*=\bigcap (\Bbb B(x,\epsilon))^\complement=\left(\bigcup \Bbb B(x,\epsilon)\right)^\complement=X^\complement=\emptyset$$

what is a contradiction about the assumptions on $X$, then $X$ is totally bounded.

Because $X$ is totally bounded and complete then is compact.$\Box$

  1. Can you check this proof please and comment any dubious or wrong step?

  2. If you know some alternative proof easier to this approach it will be very interesting to see it. Thank you.


This theorem holds also for topological spaces that are not metric, so using Cauchy sequences might not be the most suitable approach.

Let $X$ be a topological space and assume that the intersection of any family of closed subsets having the finite intersection property is non-empty. Let $\{O_i\}_{i\in I}$ be some open cover of $X$ and assume that it has no finite subcover. Hence for each finite $F\subseteq I$ there is some $x\in X$ such that $x\notin \bigcup_{i\in F}O_i$. Equivalently, for each finite $F\subseteq I$ there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$. Hence the family $\{X\setminus O_i\}_{i\in I}$ is a family of closed subsets with the finite intersection property. It follows that there is some $x\in\bigcap_{i\in I}(X\setminus O_i)$, i.e., $x\notin\bigcup_{i\in I}O_i$, contradicting that $\{O_i\}_{i\in I}$ is a cover of $X$. We conclude that $\{O_i\}_{i\in I}$ must have a finite subcover, so $X$ is compact.

Conversely, assume that $X$ is compact and let $\{C_i\}_{i\in I}$ be some family of closed subsets with the finite intersection property. Assume that $\bigcap_{i\in I}C_i$ is empty. Then $\{X\setminus C_i\}_{i\in I}$ is an open cover of $X$. By compactness, it has a finite subcover, so $\bigcup_{i\in F}X\setminus C_i=X$ for some finite $F\subseteq I$. But then $\bigcap_{i\in F}C_i=\emptyset$, contradicting that $\{C_i\}_{i\in I}$ has the finite intersection property. We conclude that $\bigcap_{i\in I}C_i\neq\emptyset$.

  • $\begingroup$ Oh, very interesting, thank you. The exercise is stated for $X$ being a metric space, this is the reason I choose Cauchy sequences. $\endgroup$ – Masacroso Nov 16 '16 at 23:31

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