Every Metric Space Is Normal

Separating Closed Sets Without a Positive Gap

A metric makes it possible to separate distinct points by small open balls. Normality asks for more: given two disjoint closed sets, can we find disjoint open sets containing them? A tempting approach is to take small neighborhoods of a fixed radius around each set. That approach requires a positive distance between the sets, which need not exist. Instead, we will let the neighborhood size vary from point to point, using distance-to-set functions.

Let $(X,d)$ be a metric space, with its metric topology: a set $U\subseteq X$ is open if every $x\in U$ has some $r>0$ such that $B_d(x,r)\subseteq U$, where $B_d(x,r)=\{y\in X:d(x,y)<r\}$. A set is closed if its complement is open.

Definition: A space is normal if any two disjoint closed subsets have disjoint open neighborhoods: for disjoint closed $A,B$, there are open sets $U,V$ such that $A\subseteq U$, $B\subseteq V$, and $U\cap V=\varnothing$. In the $T_4$ convention, normality also requires the $T_1$ property, meaning that every singleton is closed.

Every metric space is $T_1$. Indeed, if $x\ne p$, then $B_d(x,d(x,p)/2)$ does not contain $p$, by the triangle inequality. Thus every point of $X\setminus\{p\}$ has a ball contained in $X\setminus\{p\}$, so that complement is open and $\{p\}$ is closed. We will prove the stronger separation assertion in the definition of normality as well.

Distance to a Set

Definition: For a nonempty subset $C\subseteq X$, define the distance from $x\in X$ to $C$ by $$ d(x,C)=\inf\{d(x,c):c\in C\}. $$ This is a nonnegative real number. The definition in this form is for nonempty $C$.

The infimum need not be attained: there need not be a point $c\in C$ with $d(x,c)=d(x,C)$. The key fact is that the function $x\mapsto d(x,C)$ varies continuously, whether or not nearest points exist.

Lemma: If $C$ is nonempty, then for all $x,y\in X$, $$ |d(x,C)-d(y,C)|\le d(x,y). $$ Consequently, $x\mapsto d(x,C)$ is continuous.

Proof. Fix $c\in C$. The triangle inequality gives $d(x,c)\le d(x,y)+d(y,c)$. Since this holds for every $c\in C$, taking infima over $c$ gives $d(x,C)\le d(x,y)+d(y,C)$. Interchanging $x$ and $y$ gives $d(y,C)\le d(x,y)+d(x,C)$. Together these inequalities imply $|d(x,C)-d(y,C)|\le d(x,y)$. To verify continuity directly, let $\varepsilon>0$ and take $d(x,y)<\varepsilon$. The inequality gives $|d(x,C)-d(y,C)|<\varepsilon$, which is the definition of continuity at $x$. $\square$

Closedness gives another important fact. If $C$ is closed and $x\notin C$, then $X\setminus C$ is open and contains $x$. Therefore some $r>0$ satisfies $B_d(x,r)\subseteq X\setminus C$. For each $c\in C$, this implies $d(x,c)\ge r$, and hence $d(x,C)\ge r>0$. In particular, for a closed set, distance zero occurs exactly at its points: if $x\in C$, then $d(x,C)=0$; if $x\notin C$, then the argument just given shows $d(x,C)>0$.

The Separation Construction

Theorem: Every metric space is normal. More precisely, if $A$ and $B$ are disjoint closed subsets of a metric space $(X,d)$, there are disjoint open sets containing $A$ and $B$, respectively. No positive lower bound on the distances between points of $A$ and $B$ is required.

Proof. First suppose that $A$ or $B$ is empty. If $A=\varnothing$, take $U=\varnothing$ and $V=X$; these are disjoint open sets with $A\subseteq U$ and $B\subseteq V$. If $B=\varnothing$, take $U=X$ and $V=\varnothing$. Thus assume both are nonempty.

Define $$ U=\{x\in X:d(x,A)<d(x,B)\},\qquad V=\{x\in X:d(x,B)<d(x,A)\}. $$ We first show these sets are open. The distance functions are defined because $A$ and $B$ are nonempty. By the lemma, for any $x,y\in X$, $$ \big|(d(x,A)-d(x,B))-(d(y,A)-d(y,B))\big|\le 2d(x,y). $$ Let $x\in U$ and write $\delta=d(x,B)-d(x,A)>0$. If $d(x,y)<\delta/2$, the displayed inequality implies $$ d(y,A)-d(y,B)<d(x,A)-d(x,B)+\delta=0. $$ Thus $y\in U$, so $B_d(x,\delta/2)\subseteq U$. Every point of $U$ therefore has an open ball contained in $U$, proving that $U$ is open. For $x\in V$, use instead $\delta=d(x,A)-d(x,B)>0$; the same inequality gives $d(y,B)-d(y,A)<0$ whenever $d(x,y)<\delta/2$. Hence $V$ is open as well.

If $x\in U\cap V$, then both $d(x,A)<d(x,B)$ and $d(x,B)<d(x,A)$, which is impossible. Therefore $U\cap V=\varnothing$. Finally, if $a\in A$, then $d(a,A)=0$. Since $a\notin B$ and $B$ is closed, the closed-set fact above gives $d(a,B)>0$. Thus $a\in U$, proving $A\subseteq U$. For each $b\in B$, we have $d(b,B)=0$, and since $b\notin A$ with $A$ closed, $d(b,A)>0$; therefore $b\in V$ and $B\subseteq V$. The sets $U,V$ satisfy all the required conditions. As already shown, $X$ is $T_1$, so it is normal also under the $T_4$ convention. $\square$

The comparison is strict on purpose. A point where the two distances are equal belongs to neither of the constructed sets, but it does not need to belong to either one. The proof only requires that every point of $A$ lie in $U$, every point of $B$ lie in $V$, and $U,V$ be open and disjoint.

Worked Example: Closed Sets with Distance Zero in the Real Line

In $\mathbb{R}$ with its standard metric, let $$ A=\{n:n\in\mathbb{N}\},\qquad B=\left\{n+\frac{1}{2n}:n\in\mathbb{N}\right\}. $$ These sets are disjoint: for each $n\ge1$, $n<n+1/(2n)<n+1$, so each point displayed in $B$ lies strictly between consecutive integers and is not in $A$.

Both sets are closed. To check this directly, let $x\notin A$. The bounded interval $(x-1,x+1)$ contains only finitely many positive integers. If it contains any, the distances from $x$ to those finitely many integers have a positive minimum; choose a positive radius smaller than that minimum and smaller than $1$. If it contains none, choose any radius smaller than $1$. In either case a neighborhood of $x$ misses $A$, so $\mathbb{R}\setminus A$ is open. For $B$, every point $n+1/(2n)$ is greater than $n$; thus only finitely many such points can lie in $(x-1,x+1)$. The same finite-distance argument gives a neighborhood of each $x\notin B$ missing $B$. Hence $B$ is closed.

For every $n\in\mathbb{N}$, the points $n\in A$ and $n+1/(2n)\in B$ are separated by $$ d\left(n,n+\frac{1}{2n}\right)=\frac{1}{2n}. $$ These distances have infimum zero: given $\varepsilon>0$, choose $n>1/(2\varepsilon)$, so $1/(2n)<\varepsilon$. The distance between the sets is therefore zero. Nevertheless, the theorem supplies disjoint open neighborhoods: specifically, take $U=\{x:d(x,A)<d(x,B)\}$ and $V=\{x:d(x,B)<d(x,A)\}$. The proof shows they are open and disjoint, and that they contain $A$ and $B$, respectively.

Check: the construction does not require us to find a nearest point in $B$ for each $n$, or to calculate the full formulas for $d(x,A)$ and $d(x,B)$. It uses only the strict comparison, continuity, and positivity of distance from a point outside a closed set.

Worked Example: Two Closed Half-Lines

In $\mathbb{R}$, let $A=(-\infty,0]$ and $B=[2,\infty)$. Both are closed: their complements, $(0,\infty)$ and $(-\infty,2)$, are open. For $x\in\mathbb{R}$, direct minimization gives $$ d(x,A)= \begin{cases} 0,&x\le 0,\\ x,&x>0, \end{cases} \qquad d(x,B)= \begin{cases} 2-x,&x<2,\\ 0,&x\ge 2. \end{cases} $$ For $x\le0$, the first distance is $0$ and the second is $2-x>0$, so $x\in U$. For $0<x<2$, the comparison defining $U$ is $x<2-x$, which is equivalent to $x<1$. For $x\ge2$, $d(x,B)=0<d(x,A)$, so such points belong to $V$. It follows, including the equality case at $x=1$, that $$ U=(-\infty,1),\qquad V=(1,\infty). $$ These are open, disjoint, and contain $A$ and $B$, respectively. At $x=1$, both distances equal $1$, so $1$ belongs to neither neighborhood.

Check: $0\in A\subseteq U$, $2\in B\subseteq V$, and no real number is both less than $1$ and greater than $1$. Thus the computed neighborhoods have the required properties.

Worked Example: Separation in a Finite Metric Subspace

Let $X=\{0,2,5,9\}\subseteq\mathbb{R}$ with the metric inherited from the usual metric on $\mathbb{R}$. The topology on $X$ is discrete: for each point, a sufficiently small ball contains that point alone, so every subset of $X$ is open and closed. Take $A=\{0\}$ and $B=\{2\}$. The distances to these sets at the four points are $$ \begin{array}{c|cccc} x&0&2&5&9\\ \hline d(x,A)&0&2&5&9\\ d(x,B)&2&0&3&7 \end{array} $$ Comparing entries in the same column, $d(x,A)<d(x,B)$ only at $x=0$. Hence $U=\{0\}$. The strict reverse inequality holds at $x=2,5,9$, so $V=\{2,5,9\}$. These sets are disjoint and open in $X$, and they contain $A$ and $B$, respectively.

Check: at $x=5$, for example, the two distances are $5$ and $3$, so $5\in V$, not $U$. At $x=0$, they are $0$ and $2$, so $0\in U$. The computed sets agree with the strict inequalities in the construction.

Why a Fixed-Radius Argument Can Fail

A common mistake is to assume that disjoint closed sets have a positive distance from each other, then take small-radius neighborhoods around both. In the first example, for every $\varepsilon>0$, there are $a\in A$ and $b\in B$ with $d(a,b)<\varepsilon$. Thus no single positive radius can serve as a uniform separation scale for the two sets. If one writes $\delta=\inf\{d(a,b):a\in A,\ b\in B\}=0$ and tries to use balls of radius $\delta/3$, the proposed radius is $0$, which does not give open neighborhoods of the sets.

The correction is to compare the two distances separately at each point. A point nearer to $A$ than to $B$ belongs to $U$; a point nearer to $B$ than to $A$ belongs to $V$. The comparison changes continuously, which makes these sets open, while the strict inequalities make them disjoint. Closedness ensures that points of $A$ are strictly nearer to $A$ than to $B$, and vice versa. This is why the proof handles distance zero between the sets without any additional hypothesis.

Check Your Understanding

Use the distance-to-set construction and its proof to answer the following questions.

Why does a point $x\notin C$ have $d(x,C)>0$ when $C$ is closed? Identify where closedness is used.
For disjoint closed sets $A,B$, why does $d(x,A)<d(x,B)$ define an open condition?
In the first worked example, show that the infimum of the distances between $A$ and $B$ is zero, even though $A$ and $B$ are disjoint and closed.
What happens to the construction if $A=\varnothing$? Give disjoint open neighborhoods of $A$ and any closed set $B$.
In the half-line example, why does the point $1$ belong to neither constructed neighborhood?

What You'll Learn