Distance Gives a Quantitative Separation
In a metric space, a point outside a closed set is not merely excluded from the set: it has a positive distance from it. That distance supplies a numerical margin with which to build open neighborhoods that do not overlap. The key is to compare the radii of the neighborhoods with the distance between the point and the set.
Throughout, \((X,d)\) is a metric space. For a nonempty subset \(A\subseteq X\) and \(x\in X\), define the distance from \(x\) to \(A\) by
The set of distances is nonempty and bounded below by zero, so its infimum exists as a finite nonnegative real number: it is bounded above by \(d(x,a_0)\) for any chosen \(a_0\in A\). The infimum need not be the distance to a point of \(A\); for the separation argument, we need the infimum itself, not a point where it is attained.
Proof. Fix \(x,y\in X\). For every \(a\in A\), the triangle inequality gives \(d(x,a)\le d(x,y)+d(y,a)\). Taking the infimum over \(a\in A\) on both sides gives \(d(x,A)\le d(x,y)+d(y,A)\). Interchanging \(x\) and \(y\) gives \(d(y,A)\le d(x,y)+d(x,A)\). Together these inequalities imply \(\lvert d(x,A)-d(y,A)\rvert\le d(x,y)\). This bound also proves continuity: given \(\varepsilon>0\), if \(d(x,y)<\varepsilon\), then \(\lvert d(x,A)-d(y,A)\rvert<\varepsilon\).
For the closure claim, recall that in a metric space \(x\in\overline{A}\) means every open ball about \(x\) intersects \(A\). If \(d(x,A)=0\), then for any \(\varepsilon>0\) there must be some \(a\in A\) with \(d(x,a)<\varepsilon\); otherwise every such distance would be at least \(\varepsilon\), contradicting that the infimum is zero. Thus every ball \(B(x,\varepsilon)\) intersects \(A\), and \(x\in\overline{A}\). Conversely, if \(x\in\overline{A}\), every ball \(B(x,\varepsilon)\) intersects \(A\). Therefore \(d(x,A)<\varepsilon\) for every \(\varepsilon>0\), which forces \(d(x,A)=0\). \(\square\)
The continuity estimate is often called the \(1\)-Lipschitz bound. It says that the distance to a set cannot change by more than the point itself moves. In particular, a point outside a closed set has strictly positive distance from it.
The Separation Construction
Proof. Since \(F\) is closed, \(X\setminus F\) is open and contains \(x\). By the definition of the metric topology, there is an \(\varepsilon>0\) such that \(B(x,\varepsilon)\subseteq X\setminus F\). Thus \(d(x,y)\ge\varepsilon\) for every \(y\in F\), so \(\rho=d(x,F)\ge\varepsilon>0\).
The ball \(U\) is open and contains \(x\). By the distance theorem, the function \(z\mapsto d(z,F)\) is continuous. Hence \(V\), the preimage of the open interval \((-\infty,\rho/3)\), is open. For each \(y\in F\), \(d(y,F)=0<\rho/3\), so \(F\subseteq V\).
It remains to prove disjointness. Suppose a point \(z\) belonged to both \(U\) and \(V\). The distance estimate, applied to the points \(x,z\), gives \(d(x,F)\le d(x,z)+d(z,F)\). Membership in \(U\) and \(V\) says \(d(x,z)<\rho/3\) and \(d(z,F)<\rho/3\). Consequently \(\rho<2\rho/3\), which is impossible because \(\rho>0\). Thus \(U\cap V=\varnothing\). \(\square\)
There is also a direct ball description of \(V\): \(V=\bigcup_{y\in F}B(y,\rho/3)\). Indeed, if \(z\in V\), then \(d(z,F)<\rho/3\), so the definition of infimum gives some \(y\in F\) with \(d(z,y)<\rho/3\). Conversely, if \(d(z,y)<\rho/3\) for some \(y\in F\), then \(d(z,F)\le d(z,y)<\rho/3\). This description exhibits \(V\) as a union of open balls, each centered at a point of \(F\).
Worked Example: A Closed Interval in the Real Line
Use the standard metric on \(\mathbb{R}\). Separate \(x=0\) from the closed set \(F=[2,5]\).
For \(t\in[2,5]\), \(d(0,t)=t\), so \(d(0,F)=\inf\{t:2\le t\le5\}=2\). Thus \(\rho=2\), and the construction gives \(U=B(0,2/3)=(-2/3,2/3)\).
To compute \(V\), the distance from \(t\) to \([2,5]\) is \(2-t\) when \(t<2\), zero when \(2\le t\le5\), and \(t-5\) when \(t>5\). Therefore \(d(t,F)<2/3\) precisely when \(t>2-2/3=4/3\) and \(t<5+2/3=17/3\). Hence \(V=(4/3,17/3)\).
Check: \(F=[2,5]\subset(4/3,17/3)\), and every point of \(U\) is less than \(2/3\), while every point of \(V\) is greater than \(4/3\). Thus \(U\cap V=\varnothing\). The interval endpoints are excluded because the defining distance condition is strict.
Worked Example: A Half-Plane in the Euclidean Plane
Give \(\mathbb{R}^2\) the Euclidean metric \(d((u,v),(a,b))=\sqrt{(u-a)^2+(v-b)^2}\). Separate \(x=(0,0)\) from the closed half-plane \(F=\{(a,b):a\ge3\}\).
For any \((a,b)\in F\), \(a\ge3\), so \(a^2+b^2\ge9\) and \(d((0,0),(a,b))\ge3\). Equality occurs at \((3,0)\in F\). Thus \(d(x,F)=3\), and the point neighborhood is \(U=B((0,0),1)\).
For any \((u,v)\), the distance to \(F\) is \(\max(3-u,0)\). To verify this, if \(u\ge3\), the point \((u,v)\) itself belongs to \(F\), so the distance is zero. If \(u<3\), then for every \((a,b)\in F\), \(a-u\ge3-u>0\), and \((a-u)^2+(b-v)^2\ge(3-u)^2\). Equality is attained at \((a,b)=(3,v)\), proving the formula. It follows that \(V=\{(u,v):\max(3-u,0)<1\}=\{(u,v):u>2\}\).
Check: every point of \(F\) has first coordinate at least \(3\), so \(F\subseteq V\). If \((u,v)\in U\), then \(\sqrt{u^2+v^2}<1\), hence \(|u|<1\) and \(u<1\). Such a point cannot have \(u>2\), so \(U\cap V=\varnothing\).
Worked Example: The Discrete Metric
Let \(X=\{p,q,r,s\}\) with the discrete metric \(d(a,b)=0\) when \(a=b\) and \(d(a,b)=1\) when \(a\ne b\). Take \(x=p\) and \(F=\{q,s\}\).
Both distances from \(p\) to points of \(F\) equal \(1\), so \(d(p,F)=\inf\{1,1\}=1\). The point neighborhood is \(U=B(p,1/3)=\{p\}\), since the other points are distance \(1\) from \(p\). For any \(z\in F\), \(d(z,F)=0<1/3\). For \(z\notin F\), the distances from \(z\) to \(q\) and \(s\) are both \(1\), so \(d(z,F)=1\). Therefore \(V=\{z:d(z,F)<1/3\}=\{q,s\}\).
Check: \(F\subseteq V\) and \(U\cap V=\{p\}\cap\{q,s\}=\varnothing\). In this metric every singleton is the open ball of radius \(1/2\) about its point, so every subset is a union of open singletons and is open; its complement is also open, so every subset is closed.
Choosing Radii and Avoiding a Pitfall
The same calculation explains which radii are safe. If \(\rho=d(x,F)>0\), \(a>0\), and \(b>0\) satisfy \(a+b\le\rho\), then \(B(x,a)\) is disjoint from \(\bigcup_{y\in F}B(y,b)\). If a point \(z\) lay in both, there would be some \(y\in F\) with \(d(x,z)<a\) and \(d(z,y)<b\). The triangle inequality would give \(d(x,y)\le d(x,z)+d(z,y)<a+b\le\rho\), contradicting \(d(x,y)\ge d(x,F)=\rho\). The radii \(\rho/3\) used above are a convenient choice: their sum is \(2\rho/3<\rho\).
Common pitfall. A radius that reaches the distance \(\rho\) can be too large when used for both neighborhoods. In \(\mathbb{R}\), let \(x=0\) and \(F=\{2\}\). Here \(\rho=d(0,F)=2\). The incorrect choice \(U=B(0,2)=(-2,2)\) and \(V=B(2,2)=(0,4)\) gives \(1\in U\cap V\), since \(|1-0|=1<2\) and \(|1-2|=1<2\). The correction is to use radius \(2/3\) for each: \(U=(-2/3,2/3)\) and \(V=(4/3,8/3)\), which are disjoint and contain \(0\) and \(2\), respectively.
This construction separates one point from a disjoint nonempty closed set. Its hypotheses matter: if the set is not closed, a point outside it need not have positive distance from it. For example, in \(\mathbb{R}\), \(0\notin(0,1)\), but \(d(0,(0,1))=0\), because \((0,1)\) contains positive numbers arbitrarily close to zero. Closedness is what turns exclusion of the point into a positive separation radius.
Check Your Understanding
For each question, distinguish the distance infimum from any claim that a closest point exists.
- In \(\mathbb{R}\) with the standard metric, compute \(d(1,\{4,7\})\) and write the two neighborhoods produced by the radius-\(\rho/3\) construction.
- Prove that if \(A\ne\varnothing\) and \(d(x,A)>0\), then \(x\notin\overline{A}\).
- In the discrete metric on a set \(X\), find \(d(x,F)\) when \(F\ne\varnothing\) and \(x\notin F\).
- For \(x=0\) and \(F=[3,\infty)\) in \(\mathbb{R}\), compute \(\rho\), \(U\), and \(V\) from the theorem.
- Explain why the argument establishing positive distance from a closed set does not apply to \(F=(0,1)\) and \(x=0\) in \(\mathbb{R}\).