Question about the proof that a normed space is finite-dimensional if its closed unit ball is compact

I have a question about a minor point in the proof of the following theorem.

Let $B$ denote the unit closed ball in $(V,\|\cdot\|)$. If $B$ is compact then $V$ is finite-dimensional.

Proof. Suppose that $V$ is infinite-dimensional and take $\{x_n\}_{n\geq 1}\subset V$ linearly independent. In particular, any finite subset of $\{x_n\}_{n\geq 1}$ will be linearly independent. For $m\geq 1$, let $E_m=\mathrm{span}\{x_1,\dots,x_m\}$. This is a linear space for each $m\geq 1$ and $E_m\subsetneq E_{m+1}$. Since it is isomorphic to $\mathbb R^m$, it is closed and by Riesz's Lemma we take $y_{m+1}\in E_{m+1}\setminus E_m$ such that $\|y_{m+1}\|=1$ and $\|y_{m+1}-y\|>1/2$ for any $y\in E_m$. Hence, $\|y_{k+1}-y\|>1/2$ for any $y\in E_k$ for $k\leq m$. This way, we have constructed a sequence $\{y_m\}_{m\geq 1}\subset B$ such that $$m\neq k\implies \|y_m-y_k\|>\frac12.$$So this has no convergent subsequence and thus $B$ is not (sequentially) compact.

Now, I don't really see how we prove that all points in $(y_m)$ are more than $1/2$ apart. In particular, how does the equation in display mode follow from the previous stuff?

Also, why does the fact that $E_m$ is isomorphic to $\mathbb R^m$ imply that $E_m$ is closed. Do isomorphisms preserve closed sets?

Any ideas?

• You know that $\|y_{m+1}-y_{k}\| > 1/2$ for $k=1,2,3,\cdots m$. But you also know that $\|y_{m+1+j}-y_{m+1}\| > 1/2$ for $j =1,2,3,\cdots$. And, yes, isomorphisms preserve closed sets because they preserve open sets, and they preserve set complements. – DisintegratingByParts Apr 20 '16 at 20:11

$E_m$ is a closed subset of $E_{m+1}$, the Riesz lemma implies that there exists $y_{m+1}\in E_{m+1}$ such that for every $y\in E_m, \|y_{m+1}-y\|\geq 1/2$. In particular if $y=y_l, l\leq m$, $\mid y_{m+1}-y_l\|\geq 1/2$.
$E_m$ is closed, to see this, consider $e_1,..,e_m$ a base of $E_m$, a sequence $x_i=u^1_ie_1+..u^n_ie_n$ is a Cauchy sequence implies that it is the Cauchy sequence for the norm sup since every norm of a finite dimensional vector space are equivalent. This implies that $(u^l_i)$ is a Cauchy sequence of $R$ thus converges towards $u^i$, thus $(x_i)$ converges towards $u^1e_1+...+u^ne_n$. https://en.wikipedia.org/wiki/Riesz%27s_lemma#The_result