Let \(X\) be a metric space with metric \(d\); let \(A \subset X\) be nonempty.
Show that \(d(x, A) = 0\) if and only if \(x \in \overline{A}\).
Show that if \(A\) is compact, \(d(x, A) = d(x, a)\) for some \(a \in A\).
Define the \(\epsilon\)-neighborhood of \(A\) in \(X\) to be the set
\[ U(A, \epsilon) = \ts{x \mid d(x, A) < \epsilon}. \]
Show that \(U(A, \epsilon)\) equals the union of the open balls \(B_d(a, \epsilon)\) for \(a \in A\).
Assume that \(A\) is compact; let \(U\) be an open set containing \(A\). Show that some \(\epsilon\)-neighborhood of \(A\) is contained in \(U\).
Show the result in (d) need not hold if \(A\) is closed but not compact.
Solution
Let \[ d(x,A)=\inf\{d(x,a):a\in A\}. \]
If \(x\in\overline A\), every \(\varepsilon>0\) ball about \(x\) meets \(A\), so some \(a\in A\) satisfies \(d(x,a)<\varepsilon\). Hence \(d(x,A)=0\). Conversely, if \(d(x,A)=0\), then for every \(\varepsilon>0\) there is \(a\in A\) with \(d(x,a)<\varepsilon\), so every neighborhood of \(x\) meets \(A\). Thus \(x\in\overline A\).
The function \(a\mapsto d(x,a)\) is continuous on compact \(A\), so it attains a minimum at some \(a_0\in A\). Therefore \[ d(x,A)=d(x,a_0). \]
We have \[ x\in U(A,\varepsilon) \iff d(x,A)<\varepsilon \iff \exists a\in A:\ d(x,a)<\varepsilon \iff x\in\bigcup_{a\in A}B(a,\varepsilon). \] Hence \[ U(A,\varepsilon)=\bigcup_{a\in A}B(a,\varepsilon). \]
If \(U=X\), any \(\varepsilon>0\) works. Otherwise let \(F=X\setminus U\), which is a nonempty closed set disjoint from \(A\). Since \(A\) is compact, the continuous function \[ a\mapsto d(a,F) \] attains a positive minimum \(\delta>0\) on \(A\). Then \(U(A,\delta)\subset U\): if \(x\notin U\), then \(x\in F\), so \(d(x,A)\ge\delta\).
Take \(X=\mathbb R^2\), \[ A=\{(x,0):x\in\mathbb R\}, \] which is closed but not compact, and \[ U=\{(x,y):|y|<1/(1+|x|)\}. \] Then \(U\) is open and contains \(A\), but no \(\varepsilon\)-neighborhood of \(A\) lies in \(U\), since for sufficiently large \(|x|\) the vertical width \(1/(1+|x|)\) is smaller than \(\varepsilon\).