# Given $x_{0}$, let $f(x) = \|x-x_{0}\|$. Show tha $f$ has a minumum on any closed, nonempty set $A \subset \mathbb{R^{n}}$

Given $$x_{0}$$, let $$f(x) = \|x-x_{0}\|$$. Show tha $$f$$ has a minumum on any closed, nonempty set $$A \subset \mathbb{R^{n}}$$.

I tried to do a test for reduction to the absurd, but it was a little difficult to get the result. Perhaps you find a simpler way to demonstrate this exercise using elements of real analysis.

• Any bounded closed subset of $\mathbb{R}^n$ is compact. Show that $f$ is continuous. – Dzoooks Oct 7 '19 at 23:09
• So $f$ is continuous and, as $f$ wanders off to $\infty$ in a very nice way as $\|x\|\to\infty$, you can always look for the minimum of $f$ on a bounded closed subset of $A$. Can you take it from here? – Maximilian Janisch Oct 7 '19 at 23:10
• @MaximilianJanisch Thanks, edited. – Dzoooks Oct 7 '19 at 23:12

Did you tried use the following fact?

All limited sequence in a closed set has a convergent subsequence to an element of this set.

Try construct a sequence in $$A$$ s.t. the distances to $$x_0$$ are decreasing (at least non-increasing).

Examine $$\ell=\inf_{x\in A} f(x)$$. We know $$\ell \ge 0$$, because $$f(x)\ge 0$$ always. We need to show that $$f$$ attains its infimum. Here's a sketch of how: let $$(x_n)$$ denote a sequence of elements in $$A$$ so that $$f(x_n)\to \ell$$ (why does such a sequence exist?). In particular, we see that $$\lVert x_n-x_0\rVert \le\ell+\varepsilon$$ for $$n$$ sufficiently large, so that the sequence is eventually contained in a compact neighborhood of $$x_0$$, and so up to passing to a subsequence, we may assume it converges.

So, $$(x_n)\to x_\infty$$, a limit point. Now, $$x_\infty\in A$$, because $$A$$ is closed. By continuity of $$f$$ (justify!), we see that $$f(x_\infty)=\ell$$ (justify!), and so $$f$$ attains its minimum on $$A$$.

Let $$C$$ be a closed subset and $$x\in C$$. We denote by $$r=\|x-x_0\|$$. Write $$L=B(x_0,2r)\cap C$$ is compact, since $$f$$ is continuous, the restriction of $$f$$ to $$L$$ attains its minimum at $$y$$. $$f(y)$$ is the minimum of the restriction of $$f$$ to $$C$$. Suppose that $$f(z) it is equivalent to saying that $$\|z-x_0\|<\|y-x_0\|. This implies that $$z\in B(x_0,2r)\cap C$$, contradiction.

Without loss of generality, $$x_0=0.$$ If $$0\in A$$, the minimum is $$0.$$ If not, let $$N$$ be the first integer such that $$A\cap \overline B_N(0)\neq \emptyset.$$ Now, $$A\cap \overline B_N(0)$$ is compact, and $$f$$ is continuous, so there is a $$y\in A\cap \overline B_N(0)$$ such that $$f(x)=\|x\|\ge \|y\|$$ for all $$x\in A\cap \overline B_N(0).$$ Now let $$x\in A.$$ There is a first integer $$n$$ such that $$x\in \overline B_n(0).$$ If $$n\le N$$, then $$f(x)=\|x\|\ge \|y\|$$. On the other hand, if $$n>N,$$ then again, $$\|x\|>\|y\|$$ because $$x\notin \overline B_N(0).$$ So, if fact, $$y$$ is a global minimum for $$f$$.

Hint Pick some $$a \in A$$. Let $$\|x_0-a\|=R$$.

Denote by $$K:= \{ x \in \mathbb R^n : \| x -x_0\| \leq R$$.

Show that $$A \cap K$$ is closed in $$K$$, and hence compact and non-empty.

Deduce that $$f$$ attains its minimum on $$A \cap K$$.