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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!

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  • $\begingroup$ 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? $\endgroup$
    – fedja
    Commented May 11, 2019 at 10:59
  • $\begingroup$ 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. $\endgroup$
    – Taras Banakh
    Commented May 11, 2019 at 11:10
  • $\begingroup$ 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 $U_k$ :-) $\endgroup$
    – fedja
    Commented May 11, 2019 at 11:18

3 Answers 3

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In Jarchow's Locally Convex Spaces Theorems 2.7.3 and 2.9.2 on pages 39 and 43 together say that a Hausdorff topological vector space is linearly homeomorphic to a dense subspace of a projective limit of a projective system of metrizable topological vector spaces. Noting that in section 2.6 (p. 37) Jarchow defines the projective limit as a certain subspace of the product, this contains the required assertion provided that the space is Hausdorff.

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I don't have it at hand but I would look at Köthe's book Topological Linear Spaces.

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.

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  • $\begingroup$ This results seems to be absent in Kothe, too (if I have not missed anything). Of course the proof is more-or-less standard. $\endgroup$
    – Taras Banakh
    Commented May 11, 2019 at 14:22
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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 $\alpha,\beta \geq \gamma$.

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  • $\begingroup$ I think you mean $d(x_\alpha,x_\beta)<\varepsilon$ (for a given $\varepsilon$). $\endgroup$
    – Sergei Akbarov
    Commented May 11, 2019 at 12:04
  • $\begingroup$ Of course I don‘t. That would make the third claim absurd $\endgroup$
    – user131781
    Commented May 11, 2019 at 13:14
  • $\begingroup$ Ah, I see. I think you should give a reference. $\endgroup$
    – Sergei Akbarov
    Commented May 11, 2019 at 14:56
  • $\begingroup$ user131781, I have doubts that I was right when deleting my question about compact sets in nuclear spaces. Would you like this question to be "undeleted" so that you could write an answer? $\endgroup$
    – Sergei Akbarov
    Commented May 12, 2019 at 7:03
  • $\begingroup$ In the locally convex case, the condition can be reformulated as the completeness of the space considered as a topological group where a basis of the $0$-neighbourhoods is given by the kernels of continuous seminorms. This has been called seminorm-kernel-completeness in my Derived Functors in Functional Analysis. The characterization projective limit of normed spaces $\Leftrightarrow$ sk-completeness is simple but nevertheless nice. I would be very interested in getting a reference for this little result. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 7, 2022 at 13:56

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