Is every separable metrizable space a closed subspace of a convex set in a normed space?

Question

This is from Infinite-Dimensional Topology, Prerequisites and Introduction. Here a space means a metrizable and separable space.

Lemma 1.2.3. For every compact space $X$ there exists an isometric $i:X\to C(X)$ such that for every $Y\subseteq X$ the set $i(Y)$ is closed in the convex hull of $i(Y)$.

Remark. Actually every space is a closed subspace of a convex set in a normed space, because every space can be thought of as a subspace of the compact Hilbert cube $Q$ (theorem 1.4.18).

But theorem 1.4.18 only says every space is homeomorphic to a subspace of $Q$. How can we obtain then that every space $X$ is a closed subspace of a convex set in a normed space? Wouldn't we need actually that $X$ can be thought of as a closed subspace of $Q$? But wait, then $X$ need be compact which is the case we want to avoid.

What do you think?

Thank you.

Answer 1

Lemma 1.2.3. Indeed says what you state and the proof is clear (IMHO), in fact it says a little more, namely that $i(X)$ lies in the closed ball of radius $\operatorname{diam}(X)$, by part (2).

Now if $X$ is a space we embed $X$ into $Q$ using some embedding $e$, such that $e(X) \subseteq Q$ and $e(X) \simeq X$. We apply the lemma to $Q$ and get $i: Q \to C(Q)$ such that $i(e[X))$ is a closed subset of $\operatorname{conv}(i(e(X)))$. So $X$ embeds (via $e\circ i$) into $\operatorname{conv}(i(e(X)))$ as a closed subset. The codomain is indeed a convex subset of a normed linear space (namely $C(Q)$, which is even a Banach space).

So the closedness is from subcondition (1) of lemma 1.2.3 in the book, that for every $Y \subseteq X$ the image $i(Y)$ is closed in $\operatorname{conv}(i(Y)$.