How to show that if $A$ is compact, $d(x,A)= d(x,a)$ for some $a \in A$?

Question

I really think I have no talents in topology. This is a part of a problem from Topology by Munkres:

Show that if $A$ is compact, $d(x,A)= d(x,a)$ for some $a \in A$.

I always have the feeling that it is easy to understand the problem emotionally but hard to express it in math language. I am a student in Economics and I DO LOVE MATH. I really want to learn math well, could anyone give me some advice. Thanks so much!

I know that and I can feel that the conclusion holds just because of the closeness of $A$. But I have difficulty to express that feeling in math language. — Brain Zhang, Jun 10 '13 at 15:20
Warmup: Set $f(y) = d(x,y)$; can you show that $f$ is continuous? Next: do you know the extreme value theorem? — Nate Eldredge, Jun 10 '13 at 15:44
@NateEldredge thanks!!I got your point and prove it out!!can you give me some adviceds on learningmath well? thanks in advance! — Brain Zhang, Jun 10 '13 at 15:54

Luis Banegas Saybe · Answer 1 · 2017-04-10T03:58:56.243

Let $f$ : A $\longrightarrow$ $\mathbb{R}$ such that a $\mapsto$ d(x, a), where $\mathbb{R}$ is the topological space induced by the $<$ relation, the order topology.

For all open intervals (b, c) in $\mathbb{R}$, $f^{-1}((b, c))$ = {a $\in$ A $\vert$ d(x, a) $>$ b} $\cap$ {a $\in$ A $\vert$ d(x, a) $<$ c}, an open set. Therefore $f$ is continuous.

(Munkres) Theorem 27.4 Let $f$ : X $\longrightarrow$ Y be continuous, where Y is an ordered set in the order topology. If X is compact, then there exists points c and d in X such that $f$(c) $\leq$ $f$(x) $\leq$ $f$(d) for every x $\in$ X

By Theorem 27.4, $\exists$ r $\in$ A, d(x, r) = inf{ d(x, a) $\vert$ a $\in$ A}

Therefore d(x, A) = d(x, a) for some a $\in$ A

score 2 · Answer 2 · answered Jun 10 '13 at 15:09

2

Hint: Compact $\iff$ sequentially compact for metric spaces. Can you construct a sequence which must tend to the $a$ you want to find?

answered Jun 10 '13 at 15:09

not all wrong

16,568
2
37
58

yup, find open balls with radius 1/n and centered $a$, but I don't how to find the $a$ I want~~ – Brain Zhang Jun 10 '13 at 15:18

score 0 · Accepted Answer · answered Jun 10 '13 at 16:05

0

Hints:

(1) By definition of distance between points and set:

$$D:=d(x,A):=\inf_{a\in A}d(x,A)$$

(2) By definition of infimum:

$$\forall\,n\in\Bbb N\;\exists\,a_n\in A\;\;s.t.\;\;D\le d(x,a_n)\le D+\frac1n$$

(3) The sequence $\,\{a_n\}\,$ has a subsequence

$$\,\{a_{n_k}\}\;\;s.t.\;\;a_{n_k}\xrightarrow[k\to\infty]{}a\in A$$

(4) The function distance $\,d(x,.): X\to\Bbb R_+\;$ from a metric space $\,X\,$ is continuous

answered Jun 10 '13 at 16:05

DonAntonio

214,715

1

thanks so much! from your hints I know how to prove this problem!! I am really really grateful to your reply and moved by all people's enthusiasm on this site! – Brain Zhang Jun 10 '13 at 16:10

How to show that if $A$ is compact, $d(x,A)= d(x,a)$ for some $a \in A$?

3 Answers3

Linked