Why Shrink an Open Neighborhood?
Separating two disjoint closed sets by open neighborhoods is a global condition: it concerns the whole space at once. A closely related way to express that separation is to start with one closed set inside an open set and find a smaller open neighborhood whose closure still fits inside the given open set. The closure condition matters: merely choosing a smaller open set is not enough, because its closure may reach the boundary of the original one.
Throughout, \((X,\mathcal{T})\) is a topological space. A set is closed if its complement is open. The closure \(\overline{A}\) of \(A\subseteq X\) is the intersection of all closed subsets of \(X\) containing \(A\). It is closed and contains \(A\): intersections of closed sets are closed, and every set in this intersection contains \(A\).
We first record the closure test used in the proof. A point \(x\) lies outside \(\overline{A}\) exactly when it has an open neighborhood disjoint from \(A\). Indeed, if \(x\notin\overline{A}\), then \(X\setminus\overline{A}\) is an open neighborhood of \(x\) disjoint from \(A\). Conversely, if an open neighborhood \(W\) of \(x\) misses \(A\), then \(X\setminus W\) is closed and contains \(A\). The definition of closure gives \(\overline{A}\subseteq X\setminus W\), so \(x\notin\overline{A}\).
The Closure-Shrinking Characterization
Proof. Suppose first that \(X\) is normal, and let \(F\) be closed and \(U\) open with \(F\subseteq U\). The set \(G=X\setminus U\) is closed and disjoint from \(F\). By normality there are disjoint open sets \(V,W\) with \(F\subseteq V\) and \(G\subseteq W\). For each \(x\in W\), the open neighborhood \(W\) of \(x\) misses \(V\). The closure test therefore gives \(x\notin\overline{V}\). Thus \(\overline{V}\subseteq X\setminus W\). Since \(G\subseteq W\), we have \(X\setminus W\subseteq X\setminus G=U\). Also \(V\subseteq\overline{V}\), so \(F\subseteq V\subseteq\overline{V}\subseteq U\). Conversely, suppose the stated shrinking condition holds. Let \(F\) and \(G\) be disjoint closed sets. Put \(U=X\setminus G\), which is open and contains \(F\). The condition gives an open \(V\) with \(F\subseteq V\subseteq\overline{V}\subseteq X\setminus G\). Set \(W=X\setminus\overline{V}\). Since \(\overline{V}\) is closed, \(W\) is open. The inclusion \(\overline{V}\subseteq X\setminus G\) implies \(G\subseteq W\); also \(V\cap W=\varnothing\) because \(V\subseteq\overline{V}\). Hence \(V,W\) are disjoint open neighborhoods of \(F,G\), respectively, and \(X\) is normal. These arguments also cover empty sets: the inclusions remain valid, and the shrinking condition can take \(V=\varnothing\) when \(F=\varnothing\). \(\square\)
The two inclusions involving \(V\) have distinct jobs. The inclusion \(F\subseteq V\) ensures that the original closed set is still covered. The inclusion \(\overline{V}\subseteq U\) keeps not just \(V\), but also all of its closure, away from the complement of \(U\). In the reverse direction, that last inclusion is what makes \(X\setminus\overline{V}\) an open neighborhood of the other closed set.
Using the Criterion
Worked Example: Shrinking in the Real Line
Give an open neighborhood of \(F=[-1,1]\) whose closure lies inside \(U=(-2,2)\), in \(\mathbb{R}\) with its standard topology.
Choose \(V=(-\frac{3}{2},\frac{3}{2})\). This set is open and contains every point of \([-1,1]\), since \(-\frac{3}{2}<-1\leq x\leq1<\frac{3}{2}\) for \(x\in F\). Its closure is \(\overline{V}=[-\frac{3}{2},\frac{3}{2}]\): points strictly between the endpoints belong to \(V\), each endpoint has every open neighborhood meeting \(V\), and a point outside the closed interval has an open interval around it disjoint from \(V\). Finally, $$ [-1,1]\subseteq(-\tfrac{3}{2},\tfrac{3}{2})\subseteq[-\tfrac{3}{2},\tfrac{3}{2}]\subseteq(-2,2). $$ The final inclusion holds because \(-2<-\frac{3}{2}\) and \(\frac{3}{2}<2\), so both endpoints of the closure lie strictly inside \(U\).
Check: the set chosen is open, contains \(F\), and its computed closure is contained in \(U\). Thus it satisfies every inclusion in the criterion.
Worked Example: Failure in the Finite-Complement Topology
Let \(X\) be an infinite set with the finite-complement topology: its open sets are \(\varnothing\) and the subsets whose complements in \(X\) are finite. Show that this space is not normal and that closure shrinking fails.
The closed sets are \(X\) and the finite subsets of \(X\). Choose distinct points \(a,b\in X\). Then \(F=\{a\}\) and \(G=\{b\}\) are disjoint closed sets. Any open neighborhood \(V\) of \(a\) is nonempty, so \(X\setminus V\) is finite. Likewise, any open neighborhood \(W\) of \(b\) has finite complement. Their intersection is $$ V\cap W=X\setminus\big((X\setminus V)\cup(X\setminus W)\big). $$ The set removed on the right is finite. Since \(X\) is infinite, its complement is nonempty. Therefore \(V\cap W\ne\varnothing\), so no such neighborhoods are disjoint and \(X\) is not normal.
There is also a direct failure of shrinking. Let \(U=X\setminus\{b\}\), an open set containing \(F=\{a\}\). Every open \(V\) containing \(a\) is cofinite. Any closed set containing this infinite \(V\) must be \(X\), because the other closed sets are finite. Hence \(\overline{V}=X\), which is not contained in \(U\) because \(b\notin U\). No eligible \(V\) exists.
Check: the argument rules out every open neighborhood \(V\) of \(a\), not just one attempted choice. The obstruction is that each such \(V\) has closure equal to all of \(X\).
Normality and the \(T_4\) Convention
The closure-shrinking condition characterizes normality as defined here. It does not by itself imply the \(T_1\) separation axiom. This distinction matters because some conventions define \(T_4\) to mean both \(T_1\) and normal. Under that convention, \(X\) is \(T_4\) exactly when it is \(T_1\) and has the closure-shrinking property.
Worked Example: Shrinking Without \(T_1\)
Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). Check normality, closure shrinking, and the \(T_1\) condition.
The only closed sets are \(\varnothing\) and \(X\). For disjoint closed sets, at least one must be empty. The open sets \(\varnothing\) and \(X\) are disjoint neighborhoods of \(\varnothing\) and \(X\), respectively, so the space is normal. For shrinking, if \(F=\varnothing\), take \(V=\varnothing\) for either possible open set \(U\). If \(F=X\), the only open \(U\) containing \(F\) is \(X\); take \(V=X\), whose closure is \(X\). Thus the shrinking condition holds in every case.
However, \(\{a\}\) is not closed: its complement \(\{b\}\) is not open. Therefore \(X\) is not \(T_1\), and it is not \(T_4\) under the convention requiring \(T_1\) as well as normality.
Check: this example does not contradict the theorem. It satisfies both normality and closure shrinking, while failing the additional \(T_1\) requirement.
A Common Pitfall
Common mistake: conclude that \(V\subseteq U\) is enough to ensure \(\overline{V}\subseteq U\). For example, in \(\mathbb{R}\) take \(V=U=(-1,1)\). Then \(V\) is open and \(V\subseteq U\), but \(\overline{V}=[-1,1]\) is not contained in \(U\): the points \(-1\) and \(1\) belong to the closure and not to \(U\). The correction is to choose a neighborhood with room to spare, such as \(V=(-\frac12,\frac12)\) when \(F=\{0\}\) and \(U=(-1,1)\). Its closure is \([-\frac12,\frac12]\), and \(-1<-\frac12\leq\frac12<1\), so \(\overline{V}\subseteq U\).
The criterion is useful because it converts separation of two closed sets into control of one closure. To separate \(F\) from a disjoint closed \(G\), take the open set \(U=X\setminus G\), shrink around \(F\), and use the open complement of the smaller neighborhood’s closure to cover \(G\). The closure condition is precisely what guarantees that the two resulting open neighborhoods do not meet.
Check Your Understanding
For each question, keep track of whether the required inclusion concerns the open set itself or its closure.
- In the normal-to-shrinking direction, why does an open set \(W\) disjoint from \(V\) also miss \(\overline{V}\)?
- In the shrinking-to-normal direction, identify the open set used for the closed set \(G\), and explain why it contains \(G\).
- In the finite-complement topology on an infinite set, why must two nonempty open sets intersect?
- Does the indiscrete two-point space satisfy the closure-shrinking condition? Is it \(T_1\)?
- Give a specific open \(V\subseteq U\) in \(\mathbb{R}\) for which \(\overline{V}\not\subseteq U\), and identify the points witnessing failure.