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Let $ (X,d) $ be a metric space and $ A \subseteq X $ be compact. Prove that for any $ y \in X $ there exists $ x \in A $ so that $d(y,x) = d(y,A)$ where $ d(x,y)=|x-y| $.

Since A is compact it is covered by finitely many balls of $X$.

So $ A \subseteq \bigcup_{j = 1}^N B(x,j)=B(x,N) .$

Let $y \in A$. Then $y \in B(x,N)$ and so $d(x,y) < N$.

And then I'm stuck on how to approach this problem.

Bach
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s_healy
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  • Is the metric really $|x-y|$? That seems oddly specific for the problem. – Randall May 20 '19 at 02:28
  • I think if you unpack the definition of $d(y, A)$ you will find it involves some kind of infimum (depending on construction) and you can then prove that such an infinum actually must be contained within A since it is compact and contains all its limit points. – Jack Crawford May 20 '19 at 02:36
  • @Randall Yeah, it was not given in the problem though so I do not know if we need that information. The metric was stated at the top of the problem set as the default metric. – s_healy May 20 '19 at 02:36

2 Answers2

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Hint: For every $x \in X$, the function $$y\in A\to d(x,y)\in \mathbb R$$ is continuous because $d(y,y')\le\delta$ implies $d(y,x)\le d(y,y')+d(y',x) \implies |d(y,x)-d(y,'x)|\le d(y,y')\le\delta$. Since $A$ is compact and $d(-,x)$ is a real valued continuous function, what can you say about the image?

Kat
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Your $d(y, A)$ is defined to be $d(y, A)=\inf\{ d(y,x):x\in A \}$. It follows that there is a sequence $\{x_n\}\subset A$, such that $$d(y,x_n)\xrightarrow{n\to\infty}d(y,A).$$ Since $A$ is compact, there exists a convergent subsequence $\{x_{n_k}\}$ of $\{x_n\}$ such that $$x_{n_k}\xrightarrow{n_k\to\infty}x_0\in A.$$ Now we claim that $d(y,A)=d(y,x_0).$ This follows from the triangle inequality: $$ d(y,x_0)\le d(y,x_{n_k})+d(x_{n_k},x_0) ,$$ if we let $n_k\to\infty$, then $d(y,x_0)\le d(y,A)$ hence $d(y,x_0)=d(y,A)$.

Bach
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