Not reached distance to a closed non-empty set I have just begun studying topology besides my physics courses because I find I will need it later on for general relativity. In this book I'm reading, they define the distance of a point $x$ to a non-empty set $F$ of a metric space $E$ as $d(x,F) = \inf_{z \in F}d(x,z)$. So far, so good. 
However they state in a remark that, even if F is closed this limit may not be reached, meaning that there would be no point $z\in F$ such as $d(x,z)=d(x,F)$.
Now, my question is, can that only occur if we place ourselve in a metric space E that is not Complete ? Or can such a thing arise even if the space is not complete ? I know about this "completness" property only loosely because of a analysis course that I followed 2 years ago, so I am not sure about my reasoning.
In any case, could you provide an explicit example where this phenomenon occurs ? (I have a couple in head, but I am not so sure about them.
 A: Consider the space
$$\ell^2(\mathbb{N},\mathbb{C}) = \biggl\{ f \colon \mathbb{N}\to \mathbb{C} \;\Big\vert \sum_{k = 0}^\infty \lvert f(k)\rvert^2 < +\infty\biggr\}$$
with the norm
$$\lVert f\rVert = \sqrt{\sum_{k = 0}^\infty \lvert f(k)\rvert^2}.$$
This is a Hilbert space - it's $L^2(\zeta)$ where $\zeta$ is the counting measure on $\mathbb{N}$, so complete.
Let
$$e_n(k) = \begin{cases} 1 &, k = n \\ 0 &, k \neq n\end{cases}$$
and
$$F = \bigl\{ \bigl(1 + 2^{-n}\bigr)\cdot e_n : n \in \mathbb{N}\bigr\}.$$
Then $F$ is a closed subset of $\ell^2(\mathbb{N},\mathbb{C})$, and $\operatorname{dist}(0,F) = 1$, but $\lVert x - 0\rVert > 1$ for all $x\in F$.
It is immediate from the definitions that
$$\lVert (1+2^{-n})e_n\rVert = 1 + 2^{-n} > 1$$
for all $n\in \mathbb{N}$, and it's easy to see that
$$\operatorname{dist}(0,F) = \inf \{ \lVert x\rVert : x \in F\} = \lim_{n\to\infty} \lVert (1+2^{-n})e_n\rVert = 1.$$
It remains to see that $F$ is closed. Since
$$\lVert (1+2^{-n})e_n - (1+2^{-m})e_m\rVert = \sqrt{(1+2^{-n})^2 + (1+2^{-m})^2} \geqslant \sqrt{2}$$
for $m \neq n$, it follows that a sequence in $F$ is a Cauchy sequence if and only if it is eventually constant (pick $N\in \mathbb{N}$ so that $\lVert x_k - x_m\rVert < 1$ for $k,m \geqslant N$). If $x \in \overline{F}$, then there is a sequence $(x_k)$ in $F$ with $x = \lim_{k\to\infty} x_k$. So $(x_k)$ is a Cauchy sequence, hence eventually constant, but then $x_k = x$ for all large enough $k$, and therefore $x \in F$.
