The Closed-Set Shrinking Lemma in Normal Spaces

Why Shrink an Open Neighborhood?

In separation arguments, it is often not enough to find an open set around a closed set: one needs room between that neighborhood and the edge of a larger open region. Normality provides exactly this room. The key is to control the closure of the smaller open set, not merely the set itself.

Throughout, $(X,\mathcal{T})$ is a topological space. A subset is closed if its complement is open. For $A\subseteq X$, its closure $\overline{A}$ is the intersection of all closed subsets of $X$ that contain $A$. In particular, $\overline{A}$ is closed, contains $A$, and is contained in every closed set containing $A$.

Definition: The space $X$ is normal if, whenever $C_1$ and $C_2$ are disjoint closed subsets of $X$, there are disjoint open sets $G_1,G_2$ such that $C_1\subseteq G_1$ and $C_2\subseteq G_2$. In this tutorial, this is the separation property meant by normality. If a convention includes the $T_1$ axiom in the definition of normality, the results below still apply.

The Closed-Set Shrinking Lemma

Theorem (Closed-set shrinking lemma): Let $X$ be normal. If $F$ is closed, $U$ is open, and $F\subseteq U$, then there is an open set $V$ such that $$ F\subseteq V\subseteq\overline{V}\subseteq U. $$

Proof. The sets $F$ and $X\setminus U$ are closed: $F$ is closed by hypothesis, and $X\setminus U$ is closed because $U$ is open. They are disjoint, since $F\subseteq U$. By normality, there are disjoint open sets $V,W$ with $F\subseteq V$ and $X\setminus U\subseteq W$. As $W$ is open, $X\setminus W$ is closed. Also $V\cap W=\varnothing$, so $V\subseteq X\setminus W$. The definition of closure now gives $\overline{V}\subseteq X\setminus W$. Finally, $X\setminus U\subseteq W$ implies $X\setminus W\subseteq U$. Combining these inclusions gives $F\subseteq V\subseteq\overline{V}\subseteq U$. $\square$

The argument also covers degenerate cases. If $F=\varnothing$, the choice $V=\varnothing$ works for any open $U$. If $U=X$, then $X\setminus U=\varnothing$, and the separation argument still applies; for example, $V=X$ works when $F=X$. The key step in every case is that the complement of the prescribed open set is a closed set disjoint from $F$.

The closure inclusion is the feature that makes the lemma stronger than merely finding an open neighborhood inside $U$. It says that even after adding all points forced in by taking the closure of $V$, the result remains inside $U$.

The Shrinking Property Characterizes Normality

The lemma is not just a consequence of normality. It also recovers normality, so it can be used as an equivalent formulation.

Theorem: A space $X$ is normal if and only if for every closed $F\subseteq X$ and open $U\subseteq X$ with $F\subseteq U$, there is an open $V$ satisfying $F\subseteq V\subseteq\overline{V}\subseteq U$.

Proof. The forward implication is the closed-set shrinking lemma. For the reverse implication, suppose the stated shrinking property holds, and let $C_1,C_2$ be disjoint closed subsets of $X$. Set $U=X\setminus C_2$. This is open and contains $C_1$, so the property gives an open $V$ with $C_1\subseteq V\subseteq\overline{V}\subseteq X\setminus C_2$. Define $W=X\setminus\overline{V}$. Because $\overline{V}$ is closed, $W$ is open. The inclusion $\overline{V}\subseteq X\setminus C_2$ implies $C_2\subseteq W$, and $V\cap W=\varnothing$ because $V\subseteq\overline{V}$. Thus $V,W$ are disjoint open neighborhoods of $C_1,C_2$, respectively. This is normality. $\square$

This converse checks the direction of the complement carefully: a neighborhood of $C_2$ is obtained by removing the closure of the neighborhood of $C_1$, not merely by removing $V$. The closure containment is what ensures that the resulting set is open and still contains $C_2$.

Worked Examples

Worked Example: Shrinking an Interval Neighborhood

Use the standard topology on $\mathbb{R}$. Let $F=[0,1]$ and $U=(-1,2)$. Find an open $V$ satisfying the lemma and verify every inclusion.

Choose $V=(-1/2,3/2)$. It is open in the standard topology. If $x\in F$, then $0\leq x\leq1$, and therefore $-1/2<x<3/2$; hence $F\subseteq V$.

The closure of an open interval $(a,b)$ in the standard topology on $\mathbb{R}$ is $[a,b]$: its endpoints are limits of points in the interval, and any point outside $[a,b]$ has an open interval neighborhood disjoint from $(a,b)$. Thus $\overline{V}=[-1/2,3/2]$. Since $-1<-1/2$ and $3/2<2$, every point of $\overline{V}$ lies in $(-1,2)=U$. We have verified $$ [0,1]\subseteq(-1/2,3/2)\subseteq[-1/2,3/2]\subseteq(-1,2). $$

Worked Example: Shrinking a Neighborhood on the Circle

Give $S^1=\{(x,y)\in\mathbb{R}^2:x^2+y^2=1\}$ the subspace topology inherited from the Euclidean topology on $\mathbb{R}^2$. Let $F=\{(1,0)\}$ and $U=\{(x,y)\in S^1:x>0\}$. Find a suitable $V$.

Consider $V=\{(x,y)\in S^1:x>1/2\}$. The coordinate function $(x,y)\mapsto x$ is continuous: the preimage of an open interval in $\mathbb{R}$ is the intersection of $S^1$ with an open vertical strip in $\mathbb{R}^2$. Therefore $U$, the preimage of $(0,\infty)$, and $V$, the preimage of $(1/2,\infty)$, are open in $S^1$. The point $(1,0)$ satisfies $1>1/2$, so $F\subseteq V$.

We claim that $\overline{V}=\{(x,y)\in S^1:x\geq1/2\}$. If $x<1/2$, a sufficiently small open interval about the first coordinate $x$ lies entirely below $1/2$; its coordinate preimage is a neighborhood in $S^1$ missing $V$. Such a point is not in $\overline{V}$. Points with $x>1/2$ are already in $V$. The only remaining points have first coordinate $1/2$, namely $(1/2,\sqrt{3}/2)$ and $(1/2,-\sqrt{3}/2)$. Each is in the closure: points on the circle with angles $\theta$ approaching $\pi/3$ from below have first coordinate $\cos\theta>1/2$, and points with angles approaching $-\pi/3$ from above also have first coordinate greater than $1/2$. Hence every neighborhood of either endpoint meets $V$.

For every point of $\overline{V}$, its first coordinate is at least $1/2$, and so is strictly positive. Thus $\overline{V}\subseteq U$, and $F\subseteq V\subseteq\overline{V}\subseteq U$, as required.

Worked Example: Equality at the Closure Boundary

Let $X=\{a,b,c,d\}$ have the discrete topology, in which every subset is open. Set $F=\{a,c\}$ and $U=\{a,b,c\}$. Find $V$ and compare its closure with $U$.

Every subset of a discrete space is also closed, because its complement is open. In particular, $F$ is closed and $U$ is open, with $F\subseteq U$. Take $V=F=\{a,c\}$. It is open. Since it is closed as well, its closure is exactly itself: $F\subseteq\overline{F}$ by definition, while the closed set $F$ contains $F$, so $\overline{F}\subseteq F$. Therefore $\overline{V}=\{a,c\}\subseteq\{a,b,c\}=U$.

Here the containment $\overline{V}\subseteq U$ is strict: $b\in U$ but $b\notin\overline{V}$. In contrast, $V\subseteq\overline{V}$ is equality. The lemma requires inclusions, not strict inclusions at every stage.

A Common Pitfall and Why the Lemma Matters

Common pitfall. One might try to use $V=U$ automatically. For instance, in the standard topology on $\mathbb{R}$, take $F=\{1/2\}$ and $U=(0,1)$. The choice $V=U$ contains $F$, but $\overline{V}=[0,1]$, which is not a subset of $U$: specifically, $0\in\overline{V}$ and $0\notin U$. The correction is to choose a smaller open set, such as $V=(1/4,3/4)$. Its closure is $[1/4,3/4]$, and $1/4>0$ and $3/4<1$, so $F\subseteq V\subseteq\overline{V}\subseteq U$.

That calculation illustrates why the lemma is useful: ordinary containment $V\subseteq U$ does not prevent the boundary of $V$ from reaching outside $U$. The lemma gives a neighborhood whose entire closure stays within the allowed region. In the construction of continuous functions that separate closed sets, one needs to choose open neighborhoods at successive stages while keeping their closures inside previously chosen open sets. The shrinking lemma supplies that local step; the later construction handles how to organize all the stages.

Check Your Understanding

For each question, distinguish the open set from its closure and state which hypothesis is being used.

In the standard topology on $\mathbb{R}$, let $F=[2,3]$ and $U=(1,5)$. Give one valid choice of $V$, and compute $\overline{V}$.
In the closed-set shrinking lemma, why must $X\setminus U$ be closed, and why is it disjoint from $F$?
Explain why $V\cap W=\varnothing$ and $W$ open imply $\overline{V}\subseteq X\setminus W$.
Show how the shrinking property produces disjoint open neighborhoods of two disjoint closed sets $C_1,C_2$.
In the standard topology on $\mathbb{R}$, why does $V=(0,1)$ fail the conclusion $\overline{V}\subseteq U$ when $U=(0,1)$?