Theorem 2.36 of Baby Rudin states:

If $\{K_\alpha\}$ is a collection of compact subsets of a metric space $X$ such that the intersection of every finite subcollection of $\{K_\alpha\}$ is nonempty then $\cap K_\alpha$ is nonempty.

Rudin proves this by contradiction:

Choose $K_1 \in K_\alpha$, assume no point of $K_1$ belongs to every $K_\alpha$. Then the sets $K_\alpha^c$ form an open cover of $K_1$. Since $K_1$ is compact there are finitely many indices $\alpha_1, \dots, \alpha_n$ such that $K_1 \subset K_{\alpha_1}^c \cup \dots \cup K_{\alpha_n}^c$ which means $K_1 \cap K_{\alpha_1} \cap \dots \cap K_{\alpha_n} = \emptyset$. Hence we have an empty intersection of a finite subcollection of $K_\alpha$ which contradicts our assumption.

My question is this: is there a way to prove this directly instead of by contradiction? So to derive the fact that $\cap K_\alpha$ is nonempty directly from an assumption that every finite subcollection is nonempty.

To be clear, I understand Rudin's proof, but I am looking for alternative proof that is done directly as opposed to by contradiction. Someone marked this as a duplicate, but I do not see any existing question that addresses this specific concern.


Well you can start by redefining the concept "compact" by stating that a space $X$ is compact if for every family $(F_{\alpha})_{\alpha\in A}$ of closed sets that has the so-called "finite intersection property" the intersection $\bigcap_{\alpha\in A}F_{\alpha}$ is not empty.

Here family $(F_{\alpha})_{\alpha\in A}$ has by definition this "finite intersection property" if for every finite $B\subseteq A$ the intersection $\bigcap_{\alpha\in B}F_{\alpha}$ is not empty.

Using this underlying definition of "compact" (equivalent with the definition that states that open covers must have finite subcovers) it is enough to prove here that in metric spaces every compact subspace $K$ is closed (so that the family of compact sets $(K_{\alpha})$ can be recognized as a family of closed sets).


The following statements are equivalent:

  • (1) Every collection of open sets that cover $X$ has a finite subcover.
  • (2) Every collection of closed sets that has the finite intersection propery has a non-empty intersection.


Let $(F_{\alpha})_{\alpha\in A}$ be a collection of closed sets that has the finite intersection property.

Now define $U_{\alpha}=F_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that for every finite $B\subseteq A$ we have $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}\neq X$$

This tells us that the collection $(U_{\alpha})_{\alpha\in A}$ has no finite subcover of $X$. It is a collection of open sets, so this allows us to conclude that $(U_{\alpha})_{\alpha\in A}$ does not cover $X$.

That means that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}\neq\varnothing$$as was to be shown.


Let $(U_{\alpha})_{\alpha\in A}$ be a collection of open sets that cover $X$.

Now define $F_{\alpha}=U_{\alpha}^{\complement}$ for $\alpha\in A$ and observe that: $$\bigcap_{\alpha\in A}F_{\alpha}=\bigcap_{\alpha\in A}U_{\alpha}^{\complement}=(\bigcup_{\alpha\in A}U_{\alpha})^{\complement}=\varnothing$$

That implies the existence of a finite $B\subseteq A$ such that $\bigcap_{\alpha\in B}F_{\alpha}=\varnothing$ (non-existence of such finite $B$ should contradict that closed family $(F_{\alpha})_{\alpha\in A}$ has the finite intersection property).

Then $$\bigcup_{\alpha\in B}U_{\alpha}=\bigcup_{\alpha\in B}F_{\alpha}^{\complement}=(\bigcap_{\alpha\in B}F_{\alpha})^{\complement}=\varnothing^{\complement}=X$$

Proved is now that cover $(U_{\alpha})_{\alpha\in A}$ has a finite subcover $(U_{\alpha})_{\alpha\in B}$ and we are ready.

  • $\begingroup$ But the question is why is this an equivalent definition. I would like a rigorous derivation of why these two definitions are equivalent in the form of a construction of your definition from Rudin's definition that every open cover contains a finite subcover. $\endgroup$ – Joe Feb 6 '18 at 16:51
  • $\begingroup$ I have added a proof of the equivalence. $\endgroup$ – drhab Feb 6 '18 at 18:29
  • $\begingroup$ But $X$ itself is not necessarily compact. We are talking about a collection of compact subsets of a metric space $X$. $\endgroup$ – Joe Feb 7 '18 at 0:20
  • 1
    $\begingroup$ Take some $K\in \{K_{\alpha}\}.$ For each $\alpha$ let $K'_{\alpha} =K\cap K_{\alpha}. $ Then $F=\{K'_{\alpha}\}$ is a family of closed subsets of the compact space $K,$ and $F$ has the FIP (Finite Intersection Property).... So $\phi \ne \cap F=$ $K\cap (\cap_{\alpha}K_{\alpha})=$ $\cap_{\alpha}K_{\alpha}.$ $\endgroup$ – DanielWainfleet Feb 7 '18 at 5:31
  • 1
    $\begingroup$ As @DanielWainfleet suggests. Pick out some $K$ from the collection and prove that family $K\cap K_{\alpha}$ has FIP. $\endgroup$ – drhab Feb 7 '18 at 8:05

Ordering the nonempty sets $K_\alpha$ and letting $L_i$ be the intersection of all the $K_\alpha$ for $\alpha\leq i$ we obtain a nested decreasing sequence of nonempty compact sets $L_i$ where $i$ runs through $\mathbb N$. Now we can apply Cantor's intersection theorem to conclude that the intersection is nonempty.

Cantor's intersection theorem does admit more direct proofs e.g, for $X=\mathbb R$. For instance, one can define $a_i$ to be the infimum of $L_i$, producing a monotone increasing sequence $(a_i)$. One can similarly define $b_i$ to be the supremum of $L_i$ to get a monotone decreasing sequence $(b_i)$. Both sequences are therefore bounded, and a point in the intersection of all the $L_i$ can be exhibited as the limit of $(a_i)$ or alternatively of $(b_i)$.

  • 2
    $\begingroup$ This only works for countable collections of compact subsets of $\mathbb R $. $\endgroup$ – Andrés E. Caicedo Feb 6 '18 at 13:56
  • $\begingroup$ @AndrésE.Caicedo, it is difficult to judge from the formulation of the question whether the OP is interested in more sophisticated versions of the result. To judge from his tone he is looking for the elementary version. I have the impression that the other answer is too sophisticated for him. $\endgroup$ – Mikhail Katz Feb 6 '18 at 14:00
  • 2
    $\begingroup$ It is not difficult, really. The theorem is about arbitrary collections of compact subsets of an arbitrary metric space. $\endgroup$ – Andrés E. Caicedo Feb 6 '18 at 14:02
  • $\begingroup$ Metrics have nothing to do with it. Compactness is a metric-independent property. At any rate I don't have a more constructive argument in the more general setting and hope that my remarks help the OP (probably freshman) understand the result. $\endgroup$ – Mikhail Katz Feb 6 '18 at 14:12
  • $\begingroup$ The fact that we are dealing with a metric space guarantees that compact sets are closed. I think that is essential here. Of course that will also work under a weaker condition like: $X$ is Hausdorff. $\endgroup$ – drhab Feb 6 '18 at 14:22

In metric spaces, compactness is equivalent to sequential compactness - i.e. every infinite sequence in a compact set has a convergent subsequence. Picking $ x_i \in K_i $ for each i, we get an infinite sequence of points in $ K_1$. By definition of sequential compactness this must converge to a point. By construction of our sequence, this must be in the intersection of all the K's.

  • $\begingroup$ But this is limiting us to a countable collection of compact sets. I'm looking for the general case. $\endgroup$ – Joe Feb 6 '18 at 17:38
  • $\begingroup$ Not to mention this wouldn't be a sequence in $K_1$ but a sequence in $\cup K_\alpha$ $\endgroup$ – Joe Feb 6 '18 at 17:39
  • $\begingroup$ Ohh my mistake! I was thinking of the nested sets problem, where the intersection of nested compact sets is non-empty. I'll edit to mention that. $\endgroup$ – Moon Bears-C- Feb 6 '18 at 17:42

Fix $K_1 \in \{ K_\alpha\}$. For any finite choice of indices $\alpha_1 , \dots, \alpha_n$,

$K_1 \cap \bigcap\limits_{i=1}^nK_{\alpha_i} \neq \emptyset \implies K_1 \not\subset \bigcup\limits_{I=1}^n K_{\alpha_i}^c$ which means that the open sets $\{K_\alpha^c\}$ do not have a finite subcover which covers $K_1$. Since $K_1$ is compact this means these sets do not cover $K_1$, hence there exists $p \in K_1$ such that $p \notin K_\alpha^c$ for all $\alpha \in X$. Thus $p \in K_\alpha$ for each $\alpha \in X \implies p \in \bigcap K_\alpha$ and specifically $\bigcap K_\alpha \neq \emptyset$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.