# Every linear topological space embeds into the Tychonoff product of linear metric spaces

I need a reference to the following (known?)

Fact. Every topological vector space $$X$$ over the field of real numbers is topologically isomorphic to a linear subspace of the Tychonoff product of linear metric spaces.

This seems to be a basic fact in the theory of topological vector spaces, but I did not find it in the classical textbooks: Bourbaki, Schaefer, Rolewicz, Robertson&Robertson. If you know a suitable reference, please help!

• In other words, you claim that for every fixed neighborhood of $0$ in $X$ there is a continuous semi-metric on $X$ whose open ball is contained in that neighborhood. If I haven't made a stupid mistake somewhere, this is certainly true: just take a sequence of symmetric neighborhoods $U_k$ where $U_0$ is your original neighborhood and $(k+1)(U_{k+1}+U_{k+1})\subset U_k$ and declare the distance between $x$ and $y$ to be $\inf\{\sum_i 2^{-k_i}:x-y=\sum z_i, z_i\in U_{k_i}\}$. However (if I haven't made a mistake) it is rather a qualifier exam problem than a basic fact. Am I missing anything? – fedja May 11 at 10:59
• This fact should be true with a proof written by @fedja but with an extra-condition: $[-1,1]\cdot U_k\subset U_{k-1}$. Writing about "basic" fact, I did not have in mind that it is simple, but that it should be known and had to be included into basic textbooks dedicated to topological vectors spces. – Taras Banakh May 11 at 11:10
• It may be there in some form in the exercise sections, but who knows? Do you have any interesting corollaries you can derive from it? And yes, you are right: I was a bit sloppy when writing the conditions on $U_k$ :-) – fedja May 11 at 11:18

I believe that a proof can be given in the following way: For each $$0$$-neighbourhood $$U$$ choose a decreasing sequence of $$0$$-neighbourhoods $$U_n$$ with $$U_0\subseteq U$$, $$U_n+U_n\subseteq U_{n-1}$$ and $$\lambda U_n \subseteq U_{n}$$ for all $$|\lambda|\le 1$$. Then the topology $$\tau_U$$ on $$X$$ having $$(U_n)_{n\in\mathbb N}$$ as a basis of the $$0$$-neighbourhood filter is semi-metrizable by the Birkhoff-Kakutani theorem (at least Wikipedia tells it like that). Then $$X$$ is isomorphic to the diagonal in the product of all ($$X,\tau_U)$$. If $$X$$ is Hausdorff you can pass to the associated Hausdorff spaces $$X_U$$ of the $$(X,\tau_U)$$ by factoring out $$\overline{\{0\}}^{\tau_U}$$. If $$q_U:X\to X_U$$ is the quptient map you still get an embedding $$X\to \prod_U X_U$$, $$x\mapsto(q_U(x))_{U}$$. You can also take the completions of $$X_U$$ to make $$X$$ isomorphic to a subspace of the product of complete linear metric spaces.
EDIT. The result is contained in $3 (4) of the Springer Lecture Notes in Mathematics 639, Topological Vector Spaces by Adasch, Ernst, and Keim. • This results seems to be absent in Kothe, too (if I have not missed anything). Of course the proof is more-or-less standard. – Taras Banakh May 11 at 14:22 This is wildly off topic but I couldn‘t resist giving the complete picture. Every tvs (lcs, uniform space) is naturally isomorphic to a subspace of a product of $$F$$ spaces (complete metrisable ones) (Banach spaces, complete metric spaces). Complete ones are isomorphic to closed subspaces in all three cases (an if and only if condition, of course). A tvs (lcs, uniform space) is emeddable as a closed subspace of a product of metrisable tvs‘s (normed spaces, metrisable spaces) if and only it satisfies a weaker completeness condition which I will make explicit in the case of uniform spaces——-a net $$(x_\alpha)$$ which satisfies the following condition is convergent: for each continuous semi-metric $$d$$ there is a $$\gamma$$ so that $$d(x_\alpha,x_\beta)=0$$ whenever $$\alpha,\beta \geq \gamma$$. • I think you mean$d(x_\alpha,x_\beta)<\varepsilon$(for a given$\varepsilon\$). – Sergei Akbarov May 11 at 12:04