From Separating Points to Separating Closed Sets
In a Hausdorff space, any two distinct points have disjoint open neighborhoods. Normality asks for a stronger, set-wide separation: can two disjoint closed sets be placed in disjoint open sets? The challenge is that separating each pair of points separately does not immediately produce neighborhoods of the whole sets. Compactness supplies the finite selections that make the passage possible.
Throughout, compactness means that every open cover has a finite subcover. A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods. We use the convention that a space is normal if every pair of disjoint closed sets has disjoint open neighborhoods. The \(T_4\) condition means normal and \(T_1\); we will also verify the \(T_1\) condition for compact Hausdorff spaces.
The proof has two finite-subcover steps. First, fix a point of \(A\) and use compactness of \(B\) to find one neighborhood of that point disjoint from an open set containing all of \(B\). Then use compactness of \(A\) to combine finitely many such point-neighborhoods. Before carrying this out, we record why the closed sets in the theorem are compact.
Closed Subsets of a Compact Space
Proof. Let \(\{O_i:i\in J\}\) be an open cover of \(C\) in the subspace topology. For each \(i\in J\), there is an open set \(G_i\subseteq X\) with \(O_i=C\cap G_i\). The family consisting of \(X\setminus C\) and all the sets \(G_i\) is an open cover of \(X\): points outside \(C\) lie in \(X\setminus C\), and points in \(C\) lie in some \(O_i\subseteq G_i\). Compactness of \(X\) gives a finite subcover. Removing \(X\setminus C\), if it appears, leaves finitely many \(G_i\) that cover \(C\); the corresponding \(O_i\) cover \(C\). Hence \(C\) is compact. If \(C=\varnothing\), the empty subfamily covers it, so the conclusion also holds in that case. \(\square\)
For every \(a\in A\) and \(b\in B\), choose disjoint open neighborhoods \(U_{a,b}\) of \(a\) and \(V_{a,b}\) of \(b\), using the Hausdorff property.
The neighborhoods \(V_{a,b}\), as \(b\) ranges over \(B\), cover \(B\). Compactness of \(B\) gives a finite subcover. Intersect the matching \(U_{a,b}\) and unite the matching \(V_{a,b}\).
The resulting neighborhoods of its points cover \(A\). Compactness of \(A\) gives a finite subcover. Unite the selected neighborhoods of \(A\), but intersect the corresponding neighborhoods of \(B\).
We now verify that each step gives the claimed sets. Empty closed sets cause no difficulty: if \(A=\varnothing\), take \(U=\varnothing\), \(V=X\); if \(B=\varnothing\), take \(U=X\), \(V=\varnothing\). We can therefore suppose both are nonempty.
Proof of the theorem. Fix \(a\in A\). For each \(b\in B\), the Hausdorff property gives disjoint open sets \(U_{a,b}\) and \(V_{a,b}\) with \(a\in U_{a,b}\) and \(b\in V_{a,b}\). Since \(B\) is closed in compact \(X\), the lemma shows that \(B\) is compact. Thus finitely many of the sets \(V_{a,b}\), say \(V_{a,b_1},\ldots,V_{a,b_m}\), cover \(B\). Define \[ U_a=\bigcap_{j=1}^{m} U_{a,b_j}, \qquad V_a=\bigcup_{j=1}^{m} V_{a,b_j}. \] The set \(U_a\) is open, contains \(a\), and \(V_a\) is open and contains \(B\). Also \(U_a\cap V_a=\varnothing\): if a point belonged to this intersection, it would belong to \(U_{a,b_j}\cap V_{a,b_j}\) for some \(j\), contradicting the choice of those two disjoint neighborhoods.
As \(a\) ranges over \(A\), the sets \(U_a\) form an open cover of \(A\). Compactness of \(A\), again by the lemma, gives a finite subcover \(U_{a_1},\ldots,U_{a_n}\). Put \[ U=\bigcup_{i=1}^{n}U_{a_i}, \qquad V=\bigcap_{i=1}^{n}V_{a_i}. \] Both are open, \(U\) contains \(A\), and \(V\) contains \(B\). If a point belonged to \(U\cap V\), it would lie in some \(U_{a_i}\) and also in \(V_{a_i}\), contrary to \(U_{a_i}\cap V_{a_i}=\varnothing\). Hence \(U\cap V=\varnothing\), proving normality.
Finally, \(X\) is \(T_1\). For any \(x\in X\) and every \(y\ne x\), Hausdorffness supplies an open neighborhood of \(y\) that does not contain \(x\). The union of these neighborhoods, over all \(y\ne x\), is \(X\setminus\{x\}\), so \(X\setminus\{x\}\) is open and \(\{x\}\) is closed. Thus \(X\) is \(T_4\) under the stated convention. \(\square\)
Worked Separations
Worked Example: Two Closed Intervals
Let \(X=[0,1]\) with the subspace topology inherited from \(\mathbb{R}\), and let \(A=[0,1/3]\) and \(B=[2/3,1]\). These are disjoint closed subsets of \(X\). In the real line, \(X\) is compact by the Heine–Borel theorem; hence the theorem applies. We can also exhibit separating neighborhoods directly.
Set \(U=(-1,1/2)\cap X=[0,1/2)\) and \(V=(1/2,2)\cap X=(1/2,1]\). Each is open in the subspace topology because it is the intersection of \(X\) with an open real interval. Every point of \(A\) is at most \(1/3<1/2\), so \(A\subseteq U\). Every point of \(B\) is at least \(2/3>1/2\), so \(B\subseteq V\). A point cannot be both less than \(1/2\) and greater than \(1/2\), so \(U\cap V=\varnothing\).
Check: the cut point \(1/2\) is in neither neighborhood, and \(1/3\in U\) while \(2/3\in V\). Thus the neighborhoods cover the specified sets without needing to cover all of \(X\).
Worked Example: A Finite Discrete Space
Let \(X=\{1,2,3,4\}\) have the discrete topology, in which every subset is open. Every open cover of this finite set has a finite subcover: choose, for each of its four points, one cover member containing that point. Thus \(X\) is compact. It is Hausdorff because for distinct points \(x,y\), the open singletons \(\{x\}\) and \(\{y\}\) are disjoint neighborhoods.
Take \(A=\{1,3\}\) and \(B=\{2,4\}\). Both sets are closed, since every subset is closed in the discrete topology. Choose \(U=\{1,3\}\) and \(V=\{2,4\}\). These are open, contain \(A\) and \(B\), respectively, and satisfy \(U\cap V=\varnothing\).
Check: the four points are accounted for: \(1,3\in U\), \(2,4\in V\), and no point occurs in both sets.
Why the Intersection Matters
The two stages use different operations for a reason. When \(a\) is fixed, the finite union of the \(V_{a,b_j}\) must still cover all of \(B\), so we unite those sets. The corresponding \(U_{a,b_j}\) are intersected so the resulting neighborhood of \(a\) stays disjoint from every one of those \(V\)-sets. In the final stage, the union of finitely many \(U_{a_i}\) covers \(A\), while the intersection of their matching \(V_{a_i}\) remains disjoint from every selected \(U_{a_i}\) and still contains \(B\).
Common pitfall. In the last stage, replacing \(V=\bigcap_i V_{a_i}\) by \(V=\bigcup_i V_{a_i}\) does not in general preserve disjointness. For a concrete illustration in \(\mathbb{R}\), let \(A=\{-3/2,3/2\}\) and \(B=\{0\}\). For the point \(-3/2\), the open neighborhoods \(U_1=(-2,-1)\) and \(V_1=(-1/2,1/2)\) are disjoint and contain \(-3/2\) and \(0\). For \(3/2\), the open neighborhoods \(U_2=(1,2)\) and \(V_2=(-3,1/2)\) are disjoint and contain \(3/2\) and \(0\). Yet \(U_1\cup U_2=(-2,-1)\cup(1,2)\) intersects \(V_1\cup V_2=(-3,1/2)\) in \((-2,-1)\), which is nonempty. The union of the \(U_i\) is therefore not disjoint from the union of the \(V_i\). In the proof, the intersection \(V_1\cap V_2=(-1/2,1/2)\) avoids that problem.
Why Hausdorffness Is Needed
Compactness alone does not guarantee the conclusion. Consider \(X=\{a,b,c\}\) with topology \(\mathcal{T}=\{\varnothing,\{c\},\{a,c\},\{b,c\},X\}\). This is a topology: arbitrary unions of its members and finite intersections of its members are again on the list. Every finite topological space is compact, because from any open cover one can choose one member for each point, yielding at most finitely many selected sets.
The sets \(\{a\}\) and \(\{b\}\) are closed: their complements, \(\{b,c\}\) and \(\{a,c\}\), are open. But every open neighborhood of \(a\) is either \(\{a,c\}\) or \(X\), and every open neighborhood of \(b\) is either \(\{b,c\}\) or \(X\). Any such pair of neighborhoods contains \(c\) in both, so no disjoint open neighborhoods separate \(\{a\}\) and \(\{b\}\). In particular, \(X\) is not Hausdorff. This example shows why pointwise separation is essential to the theorem's argument; compactness supplies finite selections, but it does not create the initial disjoint neighborhoods.
The theorem is useful because it turns a local separation property into a global one when closed sets are compact. It also identifies precisely where the hypotheses enter: Hausdorffness separates each pair of points, while compactness makes it possible to replace infinitely many local choices by finitely many operations that preserve disjointness.
Check Your Understanding
For each question, identify where compactness or Hausdorffness is used.
- Why is a closed subset of a compact space compact? Describe how an open cover in the subspace topology is extended to a cover of the whole space.
- In the first finite-subcover step, why is \(U_a\) an intersection while \(V_a\) is a union?
- In the second step, explain why \(U=\bigcup_i U_{a_i}\) and \(V=\bigcap_i V_{a_i}\) are disjoint.
- For the three-point example, list the open neighborhoods of \(a\) and of \(b\), and explain why any such pair intersects.
- Why does the theorem imply \(T_4\), and not merely normality, under the convention used here?