1. minimum distance

Let (X,d) be a metric space and let A be a compact subset of X. Let x be a point in X.

I want to show that there exists a point a* in A such that d(x,a*) is less than or equal to d(x,a) for all a in A.

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Attempt:

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

Then there is a sequence $(a_n)$ in A such that $d(x,a_n)\rightarrow D$ as $n \rightarrow \infty$

Because A is compact there is a subsequence $(a_{n_j})$ in A which converges to a point in A, call it a*

Then $d(x,a_{n_j}) \rightarrow d(x,a^*)$ as $n \rightarrow \infty$

Then... something... to give d(x,a*) = D which is equivalent to what I was trying to show

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I'm not sure... it's all a little cloudy to me... I feel like I'm on the right track but I may not be appreciating some of the subtleties...

Also, the question came with a hint to construct a sequence in A satisfying $d(x, a_n)$ less than or equal to D + 1/n, then use the compactnes of A... That's even cloudier to me :S

Thanks for any help

2. Originally Posted by RanDom
Let (X,d) be a metric space and let A be a compact subset of X. Let x be a point in X.

I want to show that there exists a point a* in A such that d(x,a*) is less than or equal to d(x,a) for all a in A.
Hi RanDom.

If $x\in A$ then we can take $a^*=x$ and the result is proved. So assume $x\notin A.$

Then for each $a\in A$ let $d(x,a)=\epsilon_a>0.$ Then $\{B_{\epsilon_a}(a):a\in A\}$ is an open cover for $A$ and by compactness there is a finite subcover $\{B_{\epsilon_{a_i}}(a_i):1\le i\le n\}.$ Hence there exists $1\le j\le n$ such that $d(x,a_j)$ is a minimum. Set $a^*=a_j.$

3. Thanks, that makes sense.

Can you show me how to do it following the hint that was given as well?