2
$\begingroup$

$A$ : closed set in $\mathbb{R}^n$

$B$ : compact set in $\mathbb{R}^n$

$A \cap B=\phi.$

$d$ : distance function

Then, prove that \begin{equation} \text{For all } x \in B, \text{ there exists } a_x \in A \text{ such that } d(x, A)=d(x, a_x). \end{equation}

$\bigg(d(x,A)=\inf\{d(x,y) | y \in A\} \bigg)$

I think that I should use the continuity of distance function.

For all $x\in B$, define $f(x):=d(x,A) \ (x\in B).$

Because $B$ is compact and $f$ is continuous, $f(B)$ is also compact. Thus there exists maximum value and minimum value of $f(B)$.

But this idea didn't work.

I would like you to give me some ideas.

$\endgroup$
2
3
$\begingroup$

Compactness of $B$ is not needed!

$d(x,A)$ can be written as $\lim d(x,a_n)$ for some squence $(a_n) \subset A$. Now boundedness of $d(x,a_n)$ implies boundedness of $(a_n)$. Hence there is a subsequence converging to some $a_x$. Since $A$ is closed we see that $a_x \in A$. Can you see that $d(x,A)=d(x,a_x)$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.