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Which says that $\mathbb R^n$ the distance between a point $b$ and a set $X$ defined by$$ \inf \left\{ d(b,x) \mid x \in X \right\} $$

The proposition:

If $X$ is closed, this distance is reached at some point in the set $X$.

To prove it I assumed without loss of generality that $X$ is also bounded, because if not, it intersected with (e.g.) $$ B_b (n) = \left\{ x \in \mathbb R^n \mid d(b,x) \leqslant n \right\}, $$ where $n$ for example, can be chosen as the first natural number such that the intersection is not empty, and obviously the rest of the points of $X$, are at a greater distance, and will not be candidates.

Then define $$ f(x) = \left| x - b \right| \quad \text{for }x \in X $$ and since the domain is compact we conclude the result, that reaches its minimum at some point in the set. This demonstration clearly not true for any metric space, because being closed and bounded is not enough to be compact in general, but anyway maybe it can be shown in a more general way. That is my question, is this true?

Michael Hardy
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Daniel
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    If I am not mistaken, the following is a counterexample. Fix the metric space to be $\mathbb R \backslash {0}$. Take $b = -1$ and $X = (0,10]$. $X$ is closed and bounded in the metric space, but the infimum does not exist. – Srivatsan Sep 11 '11 at 18:32
  • Oops, sorry, there is a grave error in my previous comment. The infimum clearly exists, but it is not realized by any $x \in X$. – Srivatsan Sep 11 '11 at 18:41
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    Oh you are right, thanks!! I was expecting that the result was true )= – Daniel Sep 11 '11 at 18:49
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    @Srivatsan: One might think that completeness might save this. One example that shows that this doesn't help either would be the closed set $X$ given by the sequences whose only nonzero entry is the $n$-th entry and, say $1+\frac{1}{n}$, in any $\ell^p$ space and $b = 0$. – t.b. Sep 11 '11 at 18:53
  • So the only practically useful characterization. Are compact spaces is equivalent to closed and bounded. – Daniel Sep 11 '11 at 19:01
  • If $X$ would be a compact closed set in a metric space, then the infimum would be acheved, because if you would take a minimizing sequence $x_n$, this would have a convergent subsequence, and its limit would be the point where the minimum is acheved. By the fact that $X$ is closed, the minimizing point lies in $X$. – Beni Bogosel Sep 12 '11 at 10:59

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No, such a statement does not hold for a general metric space. Let $V$ be the metric space $[-1, 0) \cup (0, 1]$, and $X = (0,1]$ (verify that $X$ is closed and bounded in $V$). Then for $b = -1$, the set of distances $$ \{ |x-b| \,:\, x \in (0, 1] \} $$ has the infimum $1$, but no minimum value.

Theo's comment describes a stronger counterexample showing that this conclusion does not follow even assuming the completeness of the underlying metric space. Consider the space $\ell^p$ of sequences (say, $p=2$), and let $X$ be the set of sequences $a_n$, where $a_n$ is $1+\frac{1}{n}$ in the $n$th entry, and zero in every other entry. Boundedness of $X$ is clear; it is also closed since it has no limit points. However, for $b=0$, the distance $d(b,a_n)$ gets arbitrarily close to, but never, $1$.

t.b.
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Srivatsan
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  • basically, it would seem you need the sets to be compact, right? In a general metric space compact is equivalent to complete and totally bounded by a generalization of the proof of Heine-Borel. – Mark S. Sep 21 '17 at 18:10