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 $X$.
Proof
$(1)$ implies $(2)$
This is immediate from Subspace of Separable Metric Space is Separable.
$\Box$
$(2)$ 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:
- $M = X$
$\Box$
$(3)$ 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:
| \(\ds \norm {x - \sum_{j \mathop = 1}^n \beta_j x_{n_j} }\) | \(=\) | \(\ds \norm {x - \sum_{j \mathop = 1}^n \beta_j x_{n_j} + \sum_{j \mathop = 1}^n \alpha_j x_{n_j} - \sum_{j \mathop = 1}^n \alpha_j x_{n_j} }\) | ||||||||||||
| \(\ds \) | \(\le\) | \(\ds \norm {x - \sum_{j \mathop = 1}^n \beta_j x_{n_j} } + \norm {\sum_{j \mathop = 1}^n \paren {\beta_j - \alpha_j} x_{n_j} }\) | applying the triangle inequality | |||||||||||
| \(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \sum_{j \mathop = 1}^n \size {\beta_j - \alpha_j} \norm {x_{n_j} }\) | applying the triangle inequality $n$ times and using positive homogeneity | |||||||||||
| \(\ds \) | \(<\) | \(\ds \frac \epsilon 2 + \sum_{j \mathop = 1}^n \frac \epsilon {2 n \norm {x_{n_j} } } \norm {x_{n_j} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac \epsilon 2 + \frac \epsilon {2 n} \sum_{j \mathop = 1}^n 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
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.