(a) Show $d(x,A)=0$ if and only if $x∈\overline{A}$.
(b) Show that if $A$ is compact, $d(x,A)=d(x,a)$ for some $a∈A$.
(c) Define the $ε$-neighborhood of $A$ by $$U(A,ε)=\{x~|~d(x,A) < ε\}$$ Show that $U(A,ε)=∪_{a∈A}B_d(a,ε)$.
(d) Assume $A$ is compact; let $U$ be an open set containing $A$. Show that some $ε$-neighborhood of $A$ is contained in $U$.
Proof: (a) ($⇒$) This implies that for every $ε > 0$ there is a point $a∈A$ such that $d(x,a) < ε$, implying every neighborhood of $x$ intersects $A$ and $x∈\overline{A}$. ($⇐$) To obtain a point $a∈A$ of distance $δ < ε$ from $x$ for given $ε > 0$, take any element from $A∩B_d(x,ε)$.
(b) Let $α=d(x,A)$. Then $d : x×A→ℝ$ is continuous, so consider the nonempty closed sets $d^{-1}([α,α+1/n])$; the family has the finite intersection property, and so their intersection contains a point $a∈A$ which is seen to satisfy $d(x,a)=α$.
(c) ($⊆$) Suppose $x∈U(A,ε)$, so that $d(x,A) < z < ε$ for some $z$; choose $a∈A$ such that $d(x,a) < z$ and we see $x∈B_d(a,ε)$. ($⊇$) This is clear since $d(x,A) ≤ d(x,a)$ for all $a∈A$ by definition.
(d) Observe the closed $K=X-U$. Letting $ε=\text{inf }d(k,A)$ for $k∈K$ we see that if $ε > 0$, then $A⊆U(A,ε)⊆U$. So we may assume a sequence $k_n∈K$ such that $d(k_n,A)→0$ is decreasing. By (b), we may obtain another sequence $k_n×a_n$ such that $d(k_n,a_n)→0$ is also decreasing. Since $A$ is compact, let $a∈A$ be a limit point of $a_n$. Reusing notation, if $ε > 0$, we shall find $k∈K$ such that $d(a,k) < ε$, so that $a∈A∩K$, a contradiction. To wit, let $n$ be such that $d(a_n,k_n) < ε/2$, and let $m > n$ be such that $d(a,a_m) < ε/2$. Then by the triangle inequality $d(a,k_m) < ε$, and we are done.$~\square$
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