# An approximation property in a separable topological vector space

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?

• The approximation property is a well established important notion for Banach (and topological vector) spaces. You should look for a different name. Jan 15 at 13:02
• Maybe sequentially separable? Jan 15 at 13:05
• @ Jochen Wengenroth, Good suggestion.
– ABB
Jan 15 at 14:18
• But say is is sequentially separable, or else enjoys sequential separability. Jan 16 at 19:18
• 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 $x$. Jan 17 at 10:14

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 $$n\in\omega$$.
• @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 $(\beta\mathbb{N})\setminus\mathbb{N}$. Jan 17 at 14:29
• 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.