Separability of Normed Vector Space preserved under Isometric Isomorphism

Theorem

Let $\struct {X, \norm \cdot_X}$ be a separable normed vector space.

Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space that is isometrically isomorphic to $\struct {X, \norm \cdot_X}$.

Let $T : X \to Y$ be an isometric isomorphism between $\struct {X, \norm \cdot_X}$ and $\struct {Y, \norm \cdot_Y}$.

Then $\struct {Y, \norm \cdot_Y}$ is separable.

Let $\mathcal S = \set {x_n : n \in \N}$ be a countable everywhere dense subset of $X$.

We show that $\map T {\mathcal S} = \set {T x_n : n \in \N}$ is a countable everywhere dense subset of $Y$.

Let $y \in Y$ and $\epsilon > 0$.

Since $T$ is a bijection, there exists $x \in X$ such that $y = T x$.

Since $\mathcal S$ is everywhere dense in $X$, there exists $j \in \N$ such that:

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

Since $T$ is a linear isometry, we have:

$\norm {\map T {x - x_j} }_Y = \norm {x - x_j}_X$

and so:

$\norm {\map T {x - x_j} }_Y < \epsilon$

That is:

$\norm {T x - T x_j}_Y < \epsilon$

so:

$\norm {y - T x_j}_Y < \epsilon$

Since $\epsilon > 0$ and $y \in Y$ were arbitrary, and $T x_j \in \map T {\mathcal S}$, we have that:

$\map T {\mathcal S}$ is everywhere dense in $Y$.

$\map T {\mathcal S}$ is countable.

So:

$\struct {Y, \norm \cdot_Y}$ is separable.

$\blacksquare$