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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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3 Answers 3

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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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  • $\begingroup$ That's interesting to know. $\endgroup$
    – Johannes Hahn
    Commented Aug 21, 2010 at 19:56
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    $\begingroup$ 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 $\endgroup$
    – jbc
    Commented Oct 16, 2012 at 13:51
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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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    $\begingroup$ 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 $0\in\overline{A}$... $\endgroup$
    – Johannes Hahn
    Commented Aug 22, 2010 at 16:50
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    $\begingroup$ ... 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 $0$. $\endgroup$
    – Johannes Hahn
    Commented Aug 22, 2010 at 16:50
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    $\begingroup$ 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. $\endgroup$
    – Sergei Ivanov
    Commented Aug 23, 2010 at 3:29
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    $\begingroup$ But it is not evident (probably it is not true) that the sequential topology guggested by S.Ivanov turns the space $\mathbb R^{\mathbb R}$ into a topological vector space. In fact, sequential topological groups are rare: under some conditions (like the countablity of a loca cs-network) they are either metrizable or $k_\omega$-spaces, see arxiv.org/abs/0911.4075 $\endgroup$
    – Taras Banakh
    Commented Mar 25, 2016 at 9:08
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    $\begingroup$ @David-Halmos In my first comment I made a mistake stating that $R^R$ is not TVS. It is TVS but is not sequential. But if you take the sequential modification of this topology (cositing of sequentially open sets), then $R^R$ with this new sequential topology is not a TVS. This is what I had in mind. Feel free to ask any questions. $\endgroup$
    – Taras Banakh
    Commented Jun 17, 2017 at 18:39
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Some of the common spaces of distributions are sequential but not metrizable.

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