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 $\Bbb F$.

The following statements are equivalent:

$(1): \quad$ $X$ 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 $\SS = \set {x_n : n \in \N}$ be a everywhere dense subset of $S_X$.

Let:

$M = \paren {\map \span {\SS} }^-$

be the closed linear span of $\SS$.

We show that $M = X$.

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

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

Then:

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

Then there exists $x_j \in \SS$ such that:

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

since $\SS$ is everywhere dense in $S_X$.

Then:

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

with:

$\norm x x_j \in \map \span {\SS}$

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 $\SS = \set {x_n : n \in \N} \subseteq X$ be a countable set with closed linear span $X$.

Let $\DD$ be a countable everywhere dense subset of $\GF$ such that $0 \in \DD$.

This exists from Real Number Line is Separable and Complex Plane is Separable.

We show that:

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

To do this we show that:

$\ds \map \span \SS \subseteq \map \cl { {\span_{\DD} } \set {x_n : n \in \N} }$

We will then have:

$\ds X = \map \cl {\map \span \SS} \subseteq \map \cl { {\span_{\DD} } \set {x_n : n \in \N} }$

from Set is Closed iff Equals Topological Closure and Topological Closure is Closed.

Then:

$\map \cl {\span_{\DD} \set {x_n : n \in \N} } = X$

Let:

$x \in \map \span \SS$

First note that if $x = {\mathbf 0}_X$, then $x \in {\span_{\DD} } \set {x_n : n \in \N}$ and hence:

$x \in \map \cl { {\span_{\DD} } \set {x_n : n \in \N} }$

Hence take $x \ne {\mathbf 0}_X$.

There exists $\alpha_1, \ldots, \alpha_n \in \GF$ such that:

$\ds x = \sum_{i \mathop = 1}^n \alpha_i x_{n_i}$ where $x_{n_i} \in \SS$ and $\alpha_i \in \GF$.

Without loss of generality suppose that $\alpha_i \ne 0$ and $x_{n_i} \ne 0$ for each $i$.

Let $\epsilon > 0$ and take $\beta_i \in \GF$ such that:

$\ds \cmod {\beta_i - \alpha_i} < \frac \epsilon {n \norm {x_{n_i} } }$ for each $1 \le i \le n$.

Then we have:

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

$x \in \map \cl { {\span_{\DD} } \set {x_n : n \in \N} }$

Hence:

$X = \map \cl { {\span_{\DD} } \set {x_n : n \in \N} }$

We just need to show that:

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

We exhibit a surjection from a countable set to ${\span_{\DD} } \set {x_n : n \in \N}$.

From Surjection from Natural Numbers iff Countable, it will follow that ${\span_{\DD} } \set {x_n : n \in \N}$ is countable.

Let:

$\ds S = \bigcup_{n \mathop = 1}^\infty \DD^n$

From Cartesian Product of Countable Sets is Countable, $\DD^n$ is countable.

From Countable Union of Countable Sets is Countable, $S$ is countable.

Define $\phi : S \to {\span_{\DD} } \set {x_n : n \in \N}$ by:

$\ds \map \phi {\tuple {\alpha_1, \ldots, \alpha_n} } = \sum_{i \mathop = 1}^n \alpha_i x_i$

where $\tuple {\alpha_1, \ldots, \alpha_n} \in \DD^n$.

Since $0 \in S$, this is surjective.

Hence ${\span_{\DD} } \set {x_n : n \in \N}$ is countable, finishing the proof.

$\blacksquare$