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The following statement is well-known: for a Fréchet space $V$ and a closed subspace $W \subseteq V$ the quotient $V / W$ is again complete and hence a Fréchet space. For the particular case of a Banach space the same statement holds true.

Beyond the metrizable case this is no longer correct. So my first question is about a good counter-example, i.e. a complete locally convex space $V$ with a closed subspace $W$ such that $V / W$ is no longer complete.

My second question is whether counter-examples necessarily arise beyond the metrizable case, i.e. does every complete lcs have a closed subspace with a non-complete quotient? In other words, does the above quotient property characterize Fréchet spaces?

My third question is how the situation looks like for sequentially complete lcs with sequentially closed subspace. Are there any positive results/situations where the quotient is at least sequentially complete again?

Thanks a lot.

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4 Answers 4

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A counterexample for both the first and third questions can be found in Counterexamples in Topological Vector Spaces by Khaleelulla (p. 108).

Let $W$ denote the space of all $\mathbb C$-valued sequences $(x_n)$ and $\Phi$ the space of finite sequences. Let $E=E_1\oplus E_2$ where $E_1$ is the topological countable direct sum of copies of $W$ and $E_2$ is the topological countable product of copies of $\Phi$. $E$ is a complete locally convex space but the quotient $E/M$ where $M=\{(u,u):\ u\in E_1\cap E_2 \}$ is not even sequentially complete.

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  • $\begingroup$ thanks, I will try to get this book in the library. Any ideas for the second question? $\endgroup$ Mar 7, 2011 at 13:36
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I would like to add the following important case (which generalizes Bill Johnson's answer): An LS-spaces is defined as a countable inductive limit of Banach spaces $X_n\hookrightarrow X_{n+1}$ with compact inclusions (sometimes also called Silva spaces or DFS-spaces because they are precisely the strong duals or Frechet-Schwartz spaces). It follow from results of J. SEBASTIAO E SILVA from 1955 that LS-spaces are complete and quotients by closed subspaces are again LS (and henceforth complete).

A nice and quite typical example of an LS-space is the space of germs of analytic functions on a compact set. Another important example is the space $\mathscr E'(\Omega)$ of distributions with compact support.

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Yet another answer: EVERY incomplete Hausdorff (locally convex) topological vector space is a counterexample! It is due to the late Susanne Dierolf who proved (Manuscripta Math., 17(1):73–77, 1975) that every topological vector space is a quotient of a complete one.

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  • $\begingroup$ Wow. I had no inkling! $\endgroup$ Apr 19, 2020 at 22:35
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For the second question, consider the direct sum $Z$ of infinitely many copies of the scalar field. Given any subspace $X$, any algebraic complement $Y$ to the subspace is also a topological complement to $X$, so the quotient $Z/X$ is linearly homeomorphic to $Y$, which is again a direct sum of copies of the scalar field.

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  • $\begingroup$ Bill, there seems to be some confusion over terminology, see the recent answer. Could you add some clarification? $\endgroup$
    – Yemon Choi
    Jan 5, 2015 at 12:01
  • $\begingroup$ @Yemon. The direct sum is what you call $c_{00}$. Your erased objection to the answer below is correct. All subspaces are closed and all linear operators are continuous. $\endgroup$ Jan 5, 2015 at 14:35
  • $\begingroup$ Thanks, Bill; I must confess I didn't realize this was complete because I'm so accustomed to the norm topology $\endgroup$
    – Yemon Choi
    Jan 5, 2015 at 15:37
  • $\begingroup$ Im very sorry, I just realized that you wrote direct SUM, I was thinking about the direct PRODUCT. Like this, of course, your answer is correct and everything I wrote is nonsense. I think I will delete my "answer" now... $\endgroup$
    – Tom
    Jan 6, 2015 at 16:13

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