Show that a subspace contains a sequence of unit vectors that is norming Let $E$ be a normed space and let $E^*$ be its continuous dual space. A linear subspace $F\subset E^*$ is called norming for a subset $\,S\subset E\,$ if for all $x\in S$ we have $$\|x\|=\sup_{f\in F, \|f\|\leq 1} |f(x)|$$
Am trying to understand the proof of the following lemma in my notes:
Lemma. If $E_0$ is a separable subspace of $E$ and $F$ is a linear subspace of $E^*$ which is norming for $E_0$, then $F$ contains a sequence of unit vectors that is norming for $E_0$.
Proof. Choose a dense sequence $D=(x_n)\subset E_0\,$ and choose a sequence of unit vectors $(f_n)\subset F$ such that $|f_n(x_n)|\geq (1-\varepsilon_n)\|x_n\|$ for all $n\geq 1$, where the numbers $0<\varepsilon_n\leq1$ satisfy $\lim_{n\to\infty}\varepsilon_n= 0$.
The sequence $(f_n)$ is norming for $E_0$. To see this let $\delta>0$. Pick $n_0\geq 1$ such that $0<\varepsilon_{n_0}\leq \delta$ and $\|x-x_{n_0}\|\leq \delta$. Then $$(1-\delta)\|x\|\leq (1-\varepsilon_{n_0})\|x\|\leq (1-\varepsilon_{n_0})\|x_{n_0}\|+(1-\varepsilon_{n_0})\delta\leq |f_{n_0}(x_{n_0})|+\delta \leq |f_{n_0}(x)|+2\delta$$
Since $\delta>0$ is arbitrary it follows that $\|x\|\leq \sup_{n\geq 1} |f_n(x)|$.
Question. Don't we need to assume that $x$ is a limit point of $\{x_n\}$ to choose such $n_0$?
I think we need an extra step to approximate the norm of elements $x\in E_0\setminus D\,$ (I attempted a proof below).
Any feedback on this is very appreciated.
 A: I will try to write my own proof for clarity.
Let $D:=\{x_n : n\geq 1\}$ be a countable dense subset of $E_0$. We have the following
Proposition. Let $x\in E_0\setminus D$. Then each $\varepsilon$-ball  $B(x,\varepsilon)$ in $E_0$ contains $x_n$ for countably infinitely many $n$.
Proof. Suppose not. Then there exist $\varepsilon>0$ and $n_1,\dots, n_k$ such that the ball $B(x,\varepsilon)$ contains no $x_n$ other  than  $x_{n_1},\dots,x_{n_k}$. Let $\delta:=\min\{||x-x_{n_i}||:i=1,\dots,k$}. Since $x\notin D$, we have $\delta>0$. Then the ball $B(x,\delta)$ contains no $x_n$ whatsoever, which is a contradiction to the fact that $D$ is dense in $E_0$.
Note that the previous proposition fails if $x\in D$ and $x$ is an isolated point of $E_0$. Now we prove the lemma:
Let $x\in E_0$.
First assume $x=x_n$ for some $n$. Since $F$ is norming for $E_0$, for each $k\geq 1$, there exist $f^n_k\in F$ with $||f^n_k||=1$ such that $||x_n||-\frac{1}{k}\leq |f^n_k (x_n)|$. The sequence $(f^n_k)_{k\geq 1}$ then satisfies  $||x_n||-\frac{1}{k}\leq \sup_{k\geq 1}|f^n_k (x_n)|$ for each $k$ and so $||x_n||\leq \sup_{k\geq 1}|f^n_k (x_n)|$.
Next assume $x\in E_0\setminus D$. Since $F$ is norming for $E_0$, for each $n\geq 1$, there exist $f_n\in F$ with $||f_n||=1$ such that $||x_n||(1-\frac{1}{n})\leq |f_n (x_n)|$. Let $\delta\in]0,1]$. Since the ball $B(x,\delta)$ in $E_0$ contains $x_n$ for infinitely many $n$, we can pick $n_0$ large enough so that $0<\frac{1}{n_0}\leq \delta$ and $||x-x_{n_0}||\leq \delta$. Then we have the following inequalities:
$$(1-\delta)||x||\leq (1-\frac{1}{n_0})||x||\leq (1-\frac{1}{n_0})||x_{n_0}||+(1-\frac{1}{n_0})\delta$$
$$\leq |f_{n_0}(x_{n_0})|+\delta \leq |f_{n_0}(x)|+|f_{n_0}(x_{n_0}-x)|+\delta$$
$$\leq |f_{n_0}(x)|+ ||x-x_{n_0}||+\delta\leq |f_{n_0}(x)|+2\delta$$
$$\leq \sup_{n\geq 1} |f_n(x)|+2\delta$$
As $\delta\in]0,1]$, is arbitrary it follows that $||x||\leq \sup_{n\geq 1} |f_n(x)|$.
Now consider the countable set $C:=\cup_{n=1}^{\infty} \{f^n_k:k\geq 1\}\cup\{f_n:n\geq 1\} $. Clearly $C$ is a subset of unit vectors from $F$ and we have $||x||\leq \sup_{f\in C} |f(x)|$ for all $x\in E_0$. Moreover the reverse inequality follows from
$$||x||=\sup_{f\in F, ||f||\leq 1} |f(x)|\geq \sup_{f\in C} |f(x)|$$
Write $C=\{g_n:n\geq 1\}$. Then we finally have
$$||x||=\sup_{n\geq 1} |g_n(x)|$$
for all $x\in E_0$, with $g_n \in F$ and $||g_n||=1$ for each $n$.
A: Your question formulation hints at you are assuming that the limit point belongs to the sequence $D.\,$ Which is not the case in general.
'Limit point' is a topological notion, from scratch: Let $E$ be a topological space, and consider a subset $D\subset E$. Then $D\,$ is dense in a (usually larger) set $Z$, if $Z\subset\overline{D}$ with $\overline{D}$ being the closure of $D$. Any point in $\overline{D}$ is called a limit point of $D$.
This said written we have that $\,E_0=\overline{(x_n)}\,$. In the metric topology derived from the norm of $E$, this means that for every $x\in E_0$ and for every $\delta >0\,$ there is at least one sequence member $x_n$ with $\|x-x_n\|<\delta$.
Actually, more is true, and this in more generality:
Lemma$\;$ Consider a normed space $E$ (over $\mathbb C$ or $\,\mathbb R$, and 1-dimensional at least) and a dense subset $D\subset E$, i.e., $\overline D=E$.
For any open ball $\,B=B(x,\delta)=\big\{y\in E\:\vert\,\|x-y\|<\delta\big\}\,$ the set $\,B\cap D\,$ is infinite.
This holds true regardless of whether $\,x\in D\,$ or $\,x\in E\backslash D\,$.
Proof$\;$ by contradiction: Choose $x\in E$ and $\delta>0$. Suppose
$(B(x,\delta)\cap D)\backslash\{x\}$ is finite, and write $\{d_1,\dots,d_N\}$ for it. The latter cannot be empty as this would contradict the density of $D$.
We then have $\mu=\min\big\{\|x-d_k\|:1\leqslant k\leqslant N\big\} >0$. One kind of getting a  contradiction to the density of $D$ in $E$ is to look at points in the sphere $\big\{y\in E\,\vert\,\|x-y\| = \frac{\mu}{2}\big\}$ which all have a minimum gap of $\,\mu/2\,$ towards $D$.
$\quad\blacktriangle$
If $x\in E_0$ and $\delta>0$ are fixed, then there infinitely many $x_n\in D$ satisfying $\|x-x_n\|<\delta$. This allows to pick an index $n_0$ such that $\,0<\varepsilon_{n_0}\leq\delta\,$ holds as well.
Does this help to accept the proof of the lemma in (the) question to be complete?
