Turning Pointwise Separation into a Closed Complement
A closed set can be recognized by its complement: a subset \(K\) of a space \(X\) is closed exactly when \(X\setminus K\) is open. To show that \(K\) is closed, then, we need an open neighborhood around each point outside \(K\), with that neighborhood avoiding \(K\). In a Hausdorff space, any outside point can be separated from each individual point of \(K\). Compactness is what lets us combine those many separations into one neighborhood that avoids all of \(K\).
Throughout, \((X,\mathcal{T})\) is a topological space. A subset \(K\subseteq X\) has the subspace topology if its open sets are the sets \(K\cap U\) for \(U\in\mathcal{T}\). We call \(K\) compact when every cover of \(K\) by open sets in its subspace topology has a finite subcover. Equivalently, whenever \(K\subseteq\bigcup_{\alpha\in J}U_\alpha\) for a family of open sets \(U_\alpha\) in \(X\), some finite subfamily still covers \(K\). A space is Hausdorff if any two distinct points have disjoint open neighborhoods.
The theorem concerns the ambient topology on \(X\): its conclusion is that \(X\setminus K\) is open in \(X\). Its two hypotheses do different jobs. Hausdorffness supplies a pair of disjoint neighborhoods for each outside point and each point of \(K\); compactness reduces the resulting family of neighborhoods of points of \(K\) to finitely many.
Proof by Finite Subcover
Proof. We must show that \(X\setminus K\) is open. If \(K=\varnothing\), then \(X\setminus K=X\), which is open. Assume from now on that \(K\ne\varnothing\), and fix an arbitrary point \(x\in X\setminus K\). For every \(y\in K\), we have \(x\ne y\). Since \(X\) is Hausdorff, there are open sets \(U_y,V_y\in\mathcal{T}\) such that \(y\in U_y\), \(x\in V_y\), and \(U_y\cap V_y=\varnothing\). The sets \(U_y\), as \(y\) ranges over \(K\), cover \(K\), because each \(y\) belongs to its own set \(U_y\). Compactness gives points \(y_1,\ldots,y_n\in K\), with \(n\ge1\), such that \(K\subseteq U_{y_1}\cup\cdots\cup U_{y_n}\). Define \(V=V_{y_1}\cap\cdots\cap V_{y_n}\). A finite intersection of open sets is open, and every \(V_{y_i}\) contains \(x\), so \(V\) is an open neighborhood of \(x\). To see that \(V\) misses \(K\), suppose \(z\in V\cap K\). The finite subcover covers \(z\), so \(z\in U_{y_i}\) for some \(i\). But \(z\in V\) also implies \(z\in V_{y_i}\), contradicting \(U_{y_i}\cap V_{y_i}=\varnothing\). Hence \(V\cap K=\varnothing\), so \(V\subseteq X\setminus K\). We have found such an open neighborhood for every \(x\in X\setminus K\). Their union is \(X\setminus K\), which is therefore open. Thus \(K\) is closed. \(\square\)
The finite selection is the central step: it ensures that \(V\) is a finite intersection of open sets, and hence open. The separation chosen for each \(y\) may be different; the proof does not require a single neighborhood of \(x\) to work against all \(y\) before compactness is applied.
Worked Applications
Worked Example: The Closed Unit Interval
Let \(X=\mathbb{R}\) with its standard topology and \(K=[0,1]\). The closed interval \([0,1]\) is compact in this topology, by the open-cover compactness of closed intervals. The real line is Hausdorff: for distinct \(x,y\in\mathbb{R}\), choose \(r=|x-y|/3>0\); then the open intervals \((x-r,x+r)\) and \((y-r,y+r)\) are disjoint neighborhoods of \(x\) and \(y\). The theorem therefore says that \([0,1]\) is closed in \(\mathbb{R}\).
For a direct check of the conclusion, its complement is \((-\infty,0)\cup(1,\infty)\). Both rays are open in the standard topology, so their union is open. The theorem reaches this closedness conclusion using compactness and Hausdorffness, without requiring a separate description of the complement.
Worked Example: A Sequence Together with Its Limit
In \(X=\mathbb{R}\) with its standard topology, let \(K=\{0\}\cup\{1/n:n\in\mathbb{N}\}\), where \(\mathbb{N}=\{1,2,\ldots\}\). We first verify compactness from the open-cover definition. Let \(\mathcal{U}\) be any cover of \(K\) by sets open in the subspace topology on \(K\). Choose \(O\in\mathcal{U}\) containing \(0\). By the definition of the subspace topology, there is an open set \(G\subseteq\mathbb{R}\) such that \(O=K\cap G\). Since \(0\in G\) and \(G\) is open in \(\mathbb{R}\), there is \(\varepsilon>0\) such that \((-\varepsilon,\varepsilon)\subseteq G\).
Choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). For every \(n\ge N\), we have \(0<1/n\le 1/N<\varepsilon\), so \(1/n\in G\); also \(1/n\in K\), and therefore \(1/n\in O\). The point \(0\) is in \(O\) as well. Thus \(O\) covers \(0\) and every \(1/n\) with \(n\ge N\). There are only finitely many remaining points, namely \(1,1/2,\ldots,1/(N-1)\) when \(N>1\). For each remaining point choose one member of \(\mathcal{U}\) containing it. Together with \(O\), these choices form a finite subcover. If \(N=1\), there are no remaining points and \(O\) alone covers \(K\). This proves that \(K\) is compact. Since \(\mathbb{R}\) is Hausdorff, the theorem now gives that \(K\) is closed in \(\mathbb{R}\).
Check: the compactness argument used the strict inequality \(1/N<\varepsilon\); it ensures that every \(1/n\) with \(n\ge N\) lies inside the open interval \((-\varepsilon,\varepsilon)\). The finite number of uncovered points before the tail can be added individually to the subcover.
Worked Example: Finite Subsets of a Hausdorff Space
Let \((X,\mathcal{T})\) be Hausdorff, and take distinct points \(a,b\in X\). Set \(K=\{a,b\}\). To check compactness, consider any open cover of \(K\). At least one member of the cover contains \(a\), and at least one member contains \(b\). Choosing those one or two members gives a finite subcover, so \(K\) is compact. The theorem implies that \(\{a,b\}\) is closed in \(X\).
The same reasoning covers the degenerate finite cases: the empty set is closed because its complement is \(X\), and a singleton \(\{a\}\) is compact because any cover of it has a member containing \(a\). In a Hausdorff space the theorem then says that each singleton is closed. Thus the result also recovers the \(T_1\) property of Hausdorff spaces, where \(T_1\) means that every singleton is closed.
Why the Finite Step Cannot Be Skipped
A tempting but invalid argument is to choose, for every \(y\in K\), an open neighborhood \(V_y\) of \(x\) disjoint from a neighborhood of \(y\), and then claim that \(\bigcap_{y\in K}V_y\) is open. Arbitrary intersections of open sets need not be open. For a concrete calculation in \(\mathbb{R}\), each set \(V_n=(-1/n,1/n)\), for \(n\in\mathbb{N}\), is open and contains \(0\), but
Indeed, \(0\) lies in every interval. If \(t\ne0\), choose \(n\in\mathbb{N}\) with \(n\ge 1/|t|\); then \(1/n\le |t|\), so \(t\notin(-1/n,1/n)\). Hence the intersection contains no point other than \(0\), and \(\{0\}\) is not open in \(\mathbb{R}\). The correction is not to intersect neighborhoods indexed by every point of \(K\), but to use compactness to select finitely many of them. A finite intersection is open, which is exactly what the proof needs.
The theorem is useful whenever a compact set sits inside an ambient Hausdorff space: it immediately supplies closedness in that ambient space. Keep the roles of the hypotheses distinct. Separation gives the disjoint pairs of neighborhoods; compactness supplies a finite subcover; and the finite subcover makes the final neighborhood open while ensuring that it misses the entire compact set.
Check Your Understanding
For each question, identify where compactness or Hausdorffness is used.
- In the proof, why is \(V_{y_1}\cap\cdots\cap V_{y_n}\) an open neighborhood of \(x\)?
- Which part of the argument would fail if the cover \(\{U_y:y\in K\}\) had no finite subcover?
- Verify directly that \(\{0\}\cup\{1/n:n\in\mathbb{N}\}\) is compact using an arbitrary open cover.
- Explain why every finite subset of a Hausdorff space is closed, including the empty set.
- Give a family of open subsets of \(\mathbb{R}\) whose intersection is not open, and compute that intersection.