is the span of compact and connected complete Let $A$ be a compact and connected subset of $\ell^2$
I would like to know if $span(A)$ is complete.
Thanks
 A: No, this is not true. Let $B := \{\frac{1}{2^n}e_n:~ n\in\mathbb N\} \cup\{0\}$, where $(e_n)_{n\in\mathbb N}$ is the standard base of $\ell^2$.
$B$ is not connected, but if we include all the lines between $\frac{1}{2^n}e_n$ and $\frac{1}{2^{n+1}}e_{n+1}$, i.e. the sets $\{\frac{t}{2^n}e_n+\frac{1-t}{2^{n+1}}e_{n+1}:~t\in[0,1]\}$, and the line between $0$ and $e_0$ and call this set $A$, then $A$ is path-connected, hence connected.
We check that $A$ is a counterexample:
$\operatorname{span}A$ is not complete: We have $\operatorname{span}A = \operatorname{span}B$ because we included only linear combinations of $B$ into $A$. But $\operatorname{span}B = c_{00}$ is the space of finite sequences, which is not closed (its closure is $\ell^2$). So $\operatorname{span}A$ is not complete.  
$A$ is compact: The line segment between two points $x,y$ is always compact as it is homeomorphic to $[0,1]$ via  the homeomorphism $t\mapsto tx+ (1-t)y$. We show sequential compactness: If $(x_j)_{j\in\mathbb N}$ is a sequence in $A$, then one out of two things can happen:


*

*We either only need finitely many of the $e_n$ to represent each $x_j$. Then $(x_j)_{j\in\mathbb N}$ is a sequence in the set of line-segments between those $e_n$, which is a finite union of compact sets, so $(x_j)_{j\in\mathbb N}$ must have a convergent subsequence.

*The other possibility is, that $(x_j)_j$ needs infinitely many of the $e_n$ to be represented, but then it must have a subsequence which converges to $0$: If $x_j\in\{\frac{t}{2^n}e_n+\frac{1-t}{2^{n+1}}e_{n+1}:~t\in[0,1]\}$, then $\|x_j\|\leq \frac{1}{2^n}$. 
A: Most of your hypotheses can be dropped.  We can show the following generalization, from which your question follows as a special case.

Let $X_0$ be a subspace (not necessarily closed) of a Banach space $X$.  Then $\overline X_0$ is complete.

Proof:  Let $\{x_n\}$ be a Cauchy sequence of $\overline X_0$.  Then it is Cauchy in $X$, hence it is convergent to some $x\in X$.  But since $\overline X_0$ is closed, we know $x\in\overline X_0$.  Therefore, $\overline X_0$ is complete.
