# Sequential topological vector spaces

Since I'm dealing with the distinction between sequential continuous and continuous maps at the moment I came to ask myself once again what can be said about spaces where these two notions agree (sequential spaces). Of course we all know that metric spaces and more generally first-countable spaces are sequential and in the literatur it seems that often metrizability or first-countability is only assumed in order to not need to distinguish between sequential continuity and continuity.

I'm mostly interested in spaces that arise naturally in functional analysis, i.e. subspaces of topological vector spaces. A well known theorem says that a hausdorff topological vector space is metrizable iff it is first-countable. I tried to find out what could be said about sequential t.v.s. Are sequential t.v.s. metrizable too? Are there any reasonable t.v.s. that are sequential but not metrizable?

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The space of tempered distributions is sequential (for its usual strong topology). See, e.g., Dudley, and the references therein.

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That's interesting to know. – Johannes Hahn Aug 21 '10 at 19:56
This is because it is a countable inductive limit of a sequence of Banach spaces with compact interconnecting mappings (they are even nuclear). These spaces were studied in detail in the fifties by the Portuguese mathematician J. Sebastião e Silva, motivated by his work on spaces of distributions and holomorphic functions. They are consequently known as Silva spaces and have many nice properties in addition to the one at issue here---see, for instance, Köthe's monumental treatise. Many of the standard spaces of distributions are nuclear Silva spaces and this is very useful in their theory – jbc Oct 16 '12 at 13:51

The space of all functions $\mathbb R\to\mathbb R$ with the topology of pointwise convergence is obviously sequential but not first countable.

Indeed, for every $x\in\mathbb R$, the set $U_x:=\{f:|f(x)|<1\}$ is open. If the space had a countable base of neighborhoods of 0, every set of the form $U_x$ would contain an element of the base. So some element $U$ of the base would be contained in infinitely many sets $U_{x_1}, U_{x_2}, \dots$ and hence in the intersection $V=\bigcap U_{x_i}$. But 0 is not in the interior of $V$ because there is a sequence of functions outside $V$ that pointwise converges to 0. Namely the $n$th member of the sequence is the characteristic function of the set $\{x_i:i\ge n\}$.

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Oh damn it. I should have seen that... Thank you. – Johannes Hahn Aug 21 '10 at 19:50
Wait a minute... having a closer look, I think this space isn't sequential. Define $A:=\lbrace f:\mathbb{R}\to\lbrace 0,1\rbrace | |\lbrace x | f(x)=0\rbrace| < |\mathbb{R}| \rbrace$. Choose a wellordering of $\mathbb{R}$ such that any initial segment has less than $|\mathbb{R}|$ elements. Then the net $f_i$ of characteristic functions of $\lbrace x | x\geq i\rbrace$ converges to $0$ so that $0\in\overline{A}$... – Johannes Hahn Aug 22 '10 at 16:50
... But there is no sequence in $A$ converging to $0$: If $f_n\in A$, then $B:=\bigcup_n \lbrace x | f_n(x)=0\rbrace$ has less than $|\mathbb{R}|$ elements. Since $f_n|\mathbb{R}\setminus B = 1$, the sequence cannot converge to $0$. – Johannes Hahn Aug 22 '10 at 16:50
The topology is of pointwise convergence of sequences. By definition, a set is closed iff it contains limits of all its converging sequences. Your $A$ is closed in this topology. – Sergei Ivanov Aug 23 '10 at 3:29
Okay, this convergence does work. Thanks. – Johannes Hahn Aug 23 '10 at 9:55

Some of the common spaces of distributions are sequential but not metrizable.

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