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Let $X$ be a topological vector space.

Let us say that $X$ enjoys sequential separablity if there exists a sequence $\{x_n\}$ in $X$ such that for every $x\in X$ there exists a subsequence of $\{x_n\}$ converging to $x$

Example. Suppose that $X$ is a TVS with $X=\bigcup_1^{\infty} X_n$ such that all $X_n$'s are relatively separable metrizable. Then $X$ is sequentially separable. Thus, every seprable normed space is sequential separable. Morover the weak-star topology on the dual of separable normed spaces is also sequntialy separable.

Q. Any example of a separable TVS which is not sequentially separable?

Indeed, for a given separable TVS, what are the necessary or sufficient conditions for sequentially separable property?

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    $\begingroup$ The approximation property is a well established important notion for Banach (and topological vector) spaces. You should look for a different name. $\endgroup$
    – Jochen Wengenroth
    Jan 15, 2022 at 13:02
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    $\begingroup$ Maybe sequentially separable? $\endgroup$
    – Jochen Wengenroth
    Jan 15, 2022 at 13:05
  • $\begingroup$ @ Jochen Wengenroth, Good suggestion. $\endgroup$
    – ABB
    Jan 15, 2022 at 14:18
  • $\begingroup$ But say is is sequentially separable, or else enjoys sequential separability. $\endgroup$
    – Gerald Edgar
    Jan 16, 2022 at 19:18
  • $\begingroup$ Another way of stating the property: There exists a sequence $(x_n)_n$ in $X$ such that for any $x\in X$ there exists a subsequence of $(x_n)$ converging to $x$. $\endgroup$
    – Pietro Majer
    Jan 17, 2022 at 10:14

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The product of at most continuumly many separable spaces is separable. Therefore, if $|I|=2^{\aleph_{0}}$, then $\mathbb{R}^{I}$ is a separable locally convex topological vector space.

On the other hand, $|\mathbb{R}^{I}|>2^{\aleph_{0}}$. If $X_{n}\subseteq\mathbb{R}^{I}$ for all $n$, and each $X_{n}$ is finite, then there are at most $2^{\aleph_{0}}$ many sequences of the form $(x_{n})_{n\in\omega}$ where $x_{n}\in X_{n}$ for each $n\in\omega$. Therefore, there are at most $2^{\aleph_{0}}$ many elements $x\in\mathbb{R}^{I}$ where $x_{n}\rightarrow x$ for some sequence $(x_{n})_{n\in\omega}$ with $x_{n}\in X_{n}$ for each $n\in\omega$.

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  • $\begingroup$ By the same idea in your argument, one may say that the cardinal number of any separable topological space is at most of continuum. So, it seems something is wrong in your proofs! $\endgroup$
    – ABB
    Jan 17, 2022 at 6:47
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    $\begingroup$ @AliBagheri. Can you elaborate? It is true that if $X$ is a Hausdorff space, and $A$ is a countable subset of $X$, then there are continuumly many elements $x\in X$ where $a_{n}\rightarrow x$ for some $(a_{n})_{n\in\omega}\in A^{\omega}$, but there are plenty of ways to have $x\in\overline{A}$ without there being a sequence in $A$ that converges to $x$. For example, $\mathbb{N}$ is dense in the Stone-Cech compactification $\beta\mathbb{N}$, but no sequence in $\mathbb{N}$ converges to an element in $(\beta\mathbb{N})\setminus\mathbb{N}$. $\endgroup$
    – Joseph Van Name
    Jan 17, 2022 at 14:29
  • $\begingroup$ Thanks for your pay attention. As you said, $\beta\mathbb{N}$ is a separable topological space which is not sequentially separable (that is a very nice example). But, what I am looking for is a separable topological vector space which is not sequentially separable. $\endgroup$
    – ABB
    Jan 18, 2022 at 8:04

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