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Let $(X,d)$ be a metric space, $K \subset X$ a nonempty compact subset and $d(x,K) = \inf\{d(x,y):y \in K\}$. Show that there is a $y \in K$ such that $d(x,K) = d(x,y)$.

I have showed that the function $f_{K}(x) = d(x,K), \forall x \in X$ is continuous in $X$.

Now I need to prove that there is a $y \in K$ such that $d(x,K) = d(x,y)$. I found the following argument:

"Since $K$ is a compact set and $d_{K}(x) = d(x,K)$ is continuous, we have due to Weirstrass' theorem that $f$ admits maximum and minimum points. So we have a point $y \in K$ such that $f_{K}(x) = d(x,K) = d(x,y)$".

But to me, the above statement is wrong because $x$, although fixed, is in $X$.

Could someone explain it to me? Because all solutions that I found use this argument.

Aloizio Macedo
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Thiago Alexandre
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2 Answers2

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I think the best way to describe this argument is "confused". Instead, I would say fix $x\in X$ and define $g:X\to\mathbb{R}$ as $g(y)=d(x,y)$. Then $g$ is obviously continuous by a mere application of the triangular inequality. Now use Weierstrass's principle to say that since $K$ is compact, $g$ admits a minimum value, that is, there exists $y\in K$ such that $\inf\{d(x,z): z\in K\}=d(x,y)$; note that $\inf\{d(x,z): z\in K\}=d(x,K)$ by definition, so we are done.

Just dropped in
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Consider the function $f:K \to \mathbb R$ defined by $f(y)=d(x,y)$. This is a continuous function on $K$ so it attains its minimum. [ Use triangle inequality to verify that $|f(y_1)-f(y_2)| \leq d(y_1,y_2)$].

Kavi Rama Murthy
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  • So the argument is for any $x$ fixed, we have $d_{x}: {x} \times K \rightarrow \mathbb{R}$ defined by $d_x(x,y)$. Now, I know that I have a point $y \in K$ such that $d_x(x,y) \leq d_x(x,z) \forall z \in K$. Is it correct? – Thiago Alexandre Mar 31 '19 at 23:25
  • @ThiagoAlexandre Why are you defining a function on ${x}\times K$? $x$ is fixed throughout and you define $f$ exactly as I have defined in the answer. – Kavi Rama Murthy Mar 31 '19 at 23:31