# Is this a correct proof that an isomorphic isometry preserved separability?

Let $(X,\|\cdot\|_X)$ be a separable normed space and let $(Y,\|\cdot\|_Y)$ be a normed space. Assume they are both infinite dimensional. Let $$T:(X,\|\cdot\|_X) \longrightarrow (Y,\|\cdot\|_Y)$$ be a isomorphic isometry. What I'm interested to know is if $(Y,\|\cdot\|_Y)$ also will be separable? What I am really interested to know is wheter or not there can exist such a map between $c$ and $l^\infty$, but it doesn't hurt to be a bit more general.

My attempt:

Take $y\in Y$. Then there exist $x \in X$ such that $T(x) = y$. Also there exist a countable dense subset $X'$ of $X$ such that for every $\epsilon > 0$ there exist $x' \in X'$ such that $\| x' - x \|_X < \epsilon$. But since $T$ is an isometry we get

$$\| x' - x \|_X = \| T(x' - x) \|_Y = \| T(x') - y \|_Y < \epsilon$$

But $T(x') \in T(X')$ and $T(X')$ is countable (T is bijective) so therefore $(Y,\|\cdot\|_Y)$ must separable.

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Your proof is correct. Here, it is actually sufficient to assume that there is a continuous surjective operator $T$ from $X$ to $Y$. Your proof goes through for such $T$. – Yury Nov 22 '12 at 23:42

Yes, your proof is correct. But I would write it slightly differently: first I would introduce $X'$, and take its image in $Y$ as a countable subset. Afterwards I would prove that $T(X')$ is dense in $Y$.

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Good point, I agree. Thank you for your quick answer! – DoubleTrouble Nov 22 '12 at 23:40