Is there a closed set which doesn't have a minimal distance to a point?

Question

I am currently studying functional analysis and in the lecture we had the following theorem:

Let $X$ be a reflexive normed space. Let $A \subset X$ be a convex, closed non-empty subset and let $x_0 \notin A$ be a point. Then there is an $x \in A$ such that $$||x_0 - x||_X=\text{dist}(x_0,A) = \text{inf}_{y \in A} \{||x_0 - y||\}$$

So there is actually a point in $A$ which has minimum distance to $x_0$. Now I wonder, what would be a counterexample of this statement, if we exclude the convexity? Or the reflexivity?

I already am pretty sure that there only can be such a counter example in an infinite dimensional vectorspace, since in a finite dimensional space, we can look at $\mathbb{R}^n$ (because all normes are equivalent) and can define a sequence $x_n$ in $X$ such that $||x_0 - x_n|| \rightarrow \text{dist}(x_0, A)$ for $n \rightarrow \infty$. The alomost every element in the sequence is in a ball of radius $\text{dist}(x_0, A) +1$ around $x_0$, so the sequence is bounded and therefore admits a convergent subsequence which converges in $A$, since it is closed.

If I didn't make a mistake, closedness should therefore be sufficiant in finite dimensions, but what about infinite? I am thinking for a while now and I just can't come up with a example..

Looking forward to your replies!

Hannes

Answer 1

The situation is "even worse" than illustrated in @MoisheKohan's apt comment.

That is, in non-Hilbert Banach spaces, closed affine subspaces need not have a point closest to $0$, or there may be infinitely-many points achieving the inf/max.

For example, in the Banach space $C^o[-1,1]$ (with sup norm), the affine subspace $\{f:\int_{-1}^0f-\int_0^1f=1\}$ has no norm-minimizing element.

On the other hand, in $L^1[0,1]$, the subspace $\{f:\int_0^1f=1\}$ has infinitely-many norm-minimizing elements. :)