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

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.

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 $\operatorname{conv}(i(Y)$.
• Oh, I see, it was $i(e(X))$ the one who is a closed subspace of a convex. And $X$ is homeomorphic to $i(e(X))$. Thank you. Commented Oct 31, 2017 at 1:38