Why the Hausdorff Condition Matters
Compactness is not generally inherited by arbitrary subsets: even a subset of a compact space may fail to be compact. A useful way to obtain compact subsets is to find a set that is closed inside a compact space. For a finite intersection of compact sets, the Hausdorff condition supplies exactly the needed closedness: compact subsets of a Hausdorff space are closed.
Throughout, a subset \(K\) of a space \(X\) is called compact when it is compact with the subspace topology. Equivalently, every cover of \(K\) by sets open in \(X\) has a finite subcover. A space is Hausdorff if every two distinct points have disjoint open neighborhoods. The empty set is compact, since every open cover of it has the empty finite subcover.
The proof uses two facts. We give their arguments first so that the role of each hypothesis is explicit.
Two Tools: Closedness and Closed Subspaces
Proof. Let \(K\) be compact in a Hausdorff space \(X\). If \(K=\varnothing\), it is closed because its complement \(X\) is open. Otherwise, fix \(x\in X\setminus K\). For each \(y\in K\), the points \(x\) and \(y\) are distinct, so the Hausdorff property gives disjoint open sets \(V_y\) and \(U_y\) with \(x\in V_y\) and \(y\in U_y\). The sets \(U_y\), for \(y\in K\), cover \(K\). Compactness gives points \(y_1,\ldots,y_m\in K\) such that \(K\subseteq\bigcup_{j=1}^{m}U_{y_j}\). The set \(V=\bigcap_{j=1}^{m}V_{y_j}\) is open and contains \(x\), because it is a finite intersection of open neighborhoods of \(x\). It misses \(K\): if \(z\in K\), then \(z\in U_{y_j}\) for some \(j\), and \(U_{y_j}\cap V_{y_j}=\varnothing\), while \(V\subseteq V_{y_j}\). Thus every point of \(X\setminus K\) has an open neighborhood contained in \(X\setminus K\), so \(X\setminus K\) is open and \(K\) is closed. \(\square\)
This argument identifies where Hausdorffness is used: it provides disjoint neighborhoods of \(x\) and each point \(y\) of the compact set. Compactness then reduces the resulting family of neighborhoods to finitely many.
Proof. If \(F=\varnothing\), compactness follows from the empty finite subcover. Otherwise, let \(\{O_\alpha:\alpha\in A\}\) be an open cover of \(F\) in its subspace topology. For each \(\alpha\), there is an open set \(U_\alpha\) in \(Y\) such that \(O_\alpha=F\cap U_\alpha\). The collection consisting of all \(U_\alpha\) together with \(Y\setminus F\) covers \(Y\): points in \(F\) lie in some \(O_\alpha\subseteq U_\alpha\), and points outside \(F\) lie in \(Y\setminus F\). Since \(Y\) is compact, finitely many of these sets cover \(Y\). Removing \(Y\setminus F\), if it was selected, leaves finitely many \(U_\alpha\)'s that cover \(F\). The corresponding \(O_\alpha=F\cap U_\alpha\) therefore give a finite subcover of \(F\). Hence \(F\) is compact. \(\square\)
Proof of the Finite-Intersection Theorem
Proof. Let \(C=\bigcap_{i=1}^{n}K_i\), with \(n\geq1\). By Proposition 1, every \(K_i\) is closed in \(X\), because it is compact and \(X\) is Hausdorff. A finite intersection of closed sets is closed, so \(C\) is closed in \(X\). In particular, \(C=K_1\cap C\) is closed in the subspace \(K_1\): it is the intersection of \(K_1\) with a closed subset of \(X\). The space \(K_1\) is compact by assumption. Proposition 2, applied to the closed subset \(C\) of the compact space \(K_1\), now shows that \(C\) is compact. This also covers the case \(C=\varnothing\), which is compact. \(\square\)
The proof does not require \(X\) itself to be compact. It uses \(K_1\) as the compact space in which the intersection is closed. For \(n=1\), the conclusion simply says that \(K_1\) is compact, as already assumed. For any larger positive \(n\), the same argument applies without change.
Worked Example: Intersecting Compact Intervals
In \(\mathbb{R}\) with its standard topology, let \(K_1=[-2,2]\), \(K_2=[1,4]\), and \(K_3=[0,3]\). Find their intersection and verify its compactness.
A point belongs to all three intervals exactly when it is at least each left endpoint and at most each right endpoint. The largest left endpoint is \(\max(-2,1,0)=1\), and the smallest right endpoint is \(\min(2,4,3)=2\). Thus \(K_1\cap K_2\cap K_3=[1,2]\).
Each interval is compact by the Heine–Borel theorem: it is closed and bounded in \(\mathbb{R}\). The real line with its standard topology is Hausdorff, since distinct real numbers have disjoint open intervals around them (choose radii smaller than half their distance). The theorem therefore guarantees the intersection is compact. Directly, \([1,2]\) is closed and bounded, so Heine–Borel confirms the result.
Check: \(1\) belongs to all three intervals, and \(2\) belongs to all three; the endpoints are included because the intervals use square brackets. Values below \(1\) fail membership in \(K_2\), and values above \(2\) fail membership in \(K_1\). Hence the computed intersection is exactly \([1,2]\).
Worked Example: A Finite Hausdorff Space
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. Take \(K_1=\{a,b\}\), \(K_2=\{b,c\}\), and \(K_3=\{a,b,c\}\). Find their intersection and check the theorem's hypotheses.
The first two sets intersect in \(\{b\}\), and intersecting that with \(K_3=X\) still gives \(\{b\}\). Each \(K_i\) is compact: given any open cover of a finite set, choose one member of the cover containing each point; at most three selected sets cover \(K_i\). The space is Hausdorff because for any two distinct points \(x,y\), the open singletons \(\{x\}\) and \(\{y\}\) are disjoint neighborhoods. The theorem applies, and the computed intersection \(\{b\}\) is compact.
Check: The singleton \(\{b\}\) is compact directly: any cover of it contains at least one set that contains \(b\), and that one set is a finite subcover. This example also illustrates that the conclusion permits an intersection to be a proper subset of every set in the family.
Why the Hypothesis Cannot Be Dropped
Without Hausdorffness, a compact set need not be closed, so the proof's route from compact sets to a closed subset of \(K_1\) is unavailable. The failure can be genuine: two compact sets in a non-Hausdorff space can have a noncompact intersection.
Worked Example: Compact Sets with a Noncompact Intersection
Let \(N=\mathbb{N}\), and let \(X=N\cup\{p,q\}\), where \(p\) and \(q\) are distinct points not in \(N\). Give \(X\) the following topology. A subset \(U\subseteq X\) is open if either \(U\subseteq N\), or \(U\) contains \(p\) or \(q\) and \(N\setminus U\) is finite.
This is a topology. The empty set and \(X\) satisfy the rule. Any union of open sets that contains \(p\) or \(q\) contains a member of the union that already contains that point; that member omits only finitely many elements of \(N\), so the union does too. For a finite intersection that contains \(p\) or \(q\), every set being intersected contains that point and omits only finitely many elements of \(N\); the intersection omits at most the union of finitely many finite sets, which is finite. Intersections that contain neither special point are subsets of \(N\) and are open by definition.
Set \(K=N\cup\{p\}\) and \(L=N\cup\{q\}\). To check compactness of \(K\), take any open cover of \(K\) in its subspace topology. A member containing \(p\) contains all but finitely many points of \(N\), by the definition of the topology. Only finitely many points of \(K\) remain uncovered, and one cover member can be chosen for each such point. Together with the member containing \(p\), these choices give a finite subcover. Thus \(K\) is compact. For \(L\), a cover member containing \(q\) contains all but finitely many points of \(N\); choosing finitely many additional cover members handles the remaining points. Thus \(L\) is compact as well.
But \(K\cap L=N\), and the subspace topology on \(N\) is discrete: every subset of \(N\) is open in \(X\), hence open in \(N\). The cover of \(N\) by its singleton sets \(\{\{n\}:n\in N\}\) has no finite subcover, since a finite union of singleton sets is finite whereas \(N\) is infinite. Therefore \(K\cap L\) is not compact.
Finally, \(X\) is not Hausdorff. Every open neighborhood of \(p\) contains all but finitely many elements of \(N\), and the same is true of every open neighborhood of \(q\). Two such neighborhoods must intersect: the union of the two finite sets of omitted natural numbers is finite, and an infinite set \(N\) cannot be contained in that finite union. Hence no neighborhoods of \(p\) and \(q\) are disjoint.
A Common Mistake
Pitfall. It is incorrect to claim that the intersection of compact sets is always compact, without checking the ambient space's separation properties. In the example above, \(K\) and \(L\) are both compact, but concluding “therefore \(K\cap L=N\) is compact” gives the wrong answer. The singleton cover of \(N\) explicitly disproves compactness. In a Hausdorff space, the correction is to first prove each compact set closed, then observe that the intersection is closed in one compact member of the family. Without Hausdorffness, that closedness step is not justified.
The finite-family qualification is also worth keeping visible. For a nonempty finite family, one member such as \(K_1\) provides the compact space in which to apply the closed-subspace theorem. An empty family has intersection \(X\), and the conclusion would incorrectly assert that every Hausdorff space is compact. For example, \(\mathbb{R}\) is Hausdorff but not compact: the open cover \(\{(-m,m):m\in\mathbb{N}\}\) has no finite subcover.
Check Your Understanding
For each question, identify which compactness or separation hypothesis is being used.
- In a Hausdorff space, why is the intersection of two compact subsets closed in the first compact subset?
- Explain why the proof still works if the finite intersection is empty.
- Why does the theorem state \(n\geq1\), and give a Hausdorff space showing why the empty-family case cannot be included without another assumption.
- In the example with \(X=N\cup\{p,q\}\), give an open cover of \(N\) that has no finite subcover.
- Which step in the proof of Proposition 1 uses Hausdorffness, and which step uses compactness?