User:Caliburn/s/fa/Characterization of Separable Normed Vector Space

Theorem
Let $\mathbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.


 * $(1): \quad$ $X$ is separable
 * $(2): \quad$ $S_X = \set {x \in X : \norm x = 1}$ is separable
 * $(3): \quad$ there exists a countable set $\set {x_n : n \in \N} \subseteq X$ such that the closed linear span of $\set {x_n : n \in \N}$ is $X$.

$(1)$ implies $(2)$
This is immediate from Subspace of Separable Metric Space is Separable.

$(2)$ implies $(3)$
Let $\mathcal S = \set {x_n : n \in \N}$ be a everywhere dense subset of $S_X$.

Let:


 * $M = \paren {\map \span {\mathcal S} }^-$

be the closed linear span of $\mathcal S$.

We show that $M = X$.

Clearly $0 \in M$, from Closed Linear Span is Closed Linear Subspace.

Let $x \in X \setminus \set 0$ and $\epsilon > 0$.

Then:


 * $\ds \frac x {\norm x} \in S_X$

Then there exists $x_j \in \mathcal S$ such that:


 * $\ds \norm {\frac x {\norm x} - x_j} < \frac \epsilon {\norm x}$

since $\mathcal S$ is everywhere dense in $S_X$.

Then:


 * $\ds \norm {x - \norm x x_j} < \epsilon$

with:


 * $\norm x x_j \in \map \span {\mathcal S}$

Since $\epsilon > 0$ was arbitrary we have:


 * $x \in M$

from Condition for Point being in Closure.

So:


 * $M = X$

$(3)$ implies $(1)$
Let $\mathcal S = \set {x_n : n \in \N} \subseteq X$ be a countable set with closed linear span $X$.

First take $\Bbb F = \R$.

We show that:


 * $\ds {\span_\Q} \set {x_n : n \in \N} = \set {\sum_{i \mathop = 1}^n \alpha_i x_{n_i} : \alpha_i \in \Q \text { and } x_{n_i} \in \mathcal S \text { for each } i}$ is everywhere dense in $X$.

We will then show that:


 * ${\span_\Q} \set {x_n : n \in \N}$ is countable.

Clearly $0 \in \paren { {\span_\Q} \set {x_n : n \in \N} }^-$ from Closed Linear Span is Closed Linear Subspace.

Let $x \in X$ and $\epsilon > 0$.

Since the closed linear span of $\set {x_n : n \in \N}$ is $X$, there exists $x_{n_j} \in \set {x_n :n \in \N}$ and $\alpha_j \in \R$ such that:


 * $\ds \norm {x - \sum_{j = 1}^n \alpha_j x_{n_j} } < \frac \epsilon 2$

From Rationals are Everywhere Dense in Reals, for each $j$ there exists $\beta_j \in \R$ such that:


 * $\ds \size {\beta_j - \alpha_j} < \frac \epsilon {2 n \norm {x_{n_j} } }$

Then, we have:

So from Condition for Point being in Closure, we have:


 * $x \in \paren { {\span_\Q} \set {x_n : n \in \N} }^-$

Since $x \in X$ was arbitrary, we have that:


 * ${\span_\Q} \set {x_n : n \in \N}$ is everywhere dense.

Now take $\Bbb F = \C$.

We show that:


 * $\ds {\span_{\Q \sqbrk i} } \set {x_n : n \in \N} = \set {\sum_{i \mathop = 1}^n \alpha_i x_{n_i} : \alpha_i \in \Q \sqbrk i \text { and } x_{n_i} \in \mathcal S \text { for each } i}$

We will then show that:


 * ${\span_{\Q \sqbrk i} } \set {x_n : n \in \N}$ is countable.