TLDR: What are examples of (function-)spaces that are not sequential? When does this matter?

As a simple analyst, I am most happy if I can just work with sequences all the time. In most situations this is totally fine, as many many spaces one encounters in one's daily life are actually sequential (or even 1st countable, or even better metrisable). Now recently I was a bit shocked to find out that the seemingly familiar space of test-functions $\mathscr{D}(\mathbb{R}^d)$ (with its usual LF-topology) actually fails to be sequential. But hadn't I learned that one can verify whether a linear functional on $\mathscr{D}(\mathbb{R}^d)$ is a distribution by checking continuity with sequences? Well in that case it is true (Proposition 21.1 in Trèves TVS book), but only because we looked at linear functionals.

This got me thinking that there might actually be a bunch of spaces around, not pathological counterexamples, but real spaces one encounters in the wild, that fail to be sequential. In some cases, like above, this might not be problematic, but potentially for non-trivial reasons. In order to become more aware of these subtleties I would like to collect some more examples of this.

An answer should ideally contain the following:

  • A concrete example or a class of examples of non-sequential spaces, which are widely used or naturally show up in analysis. My main interested lies in topological vector spaces that appear as function spaces in some context. The example should not be some 'patholocial counterexample' (this is of course a bit vague).
  • An instance of where it matters that the space is non-sequential. Or a warning, of when one needs to be more careful with it and use filters or nets.
  • Loop holes or special situations where it suffices to focus on sequences nevertheless.

I'll make a start:

  • Test-functions: $\mathscr{D}(\mathbb{R}^d)$ (with standard LF-topology) is not sequential. In particular a function $f:\mathscr{D}(\mathbb{R}^d)\rightarrow \mathbb{C}$ might be sequentually continuous, but not continuous (example by PhoemueX). However, if $f$ is linear then sequential continuity implies continuity (Corollary after Proposition 13.1 in Trèves' TVS book). The same is true for other LF-spaces.
  • Distributions: $\mathscr{D}'(\mathbb{R}^d)$ (with the strong topology) is not sequential. A sequence of distributions converges strongly if and only if it converges weakly, but this is not true when sequences are replaced by nets/filters. The same result (for sequences) holds in strong duals of Montel spaces (Corollary 1 to Proposition 34.6 in Trèves)
  • An infinite dimensional Banach space, equipped with the weak topology is not sequential. However, despite this we have that compactness = sequential compactness (Eberlein-Smulian theorem).

Finally, here are some spaces that are sequential: $\mathscr{S}(\mathbb{R}^d)$ (Schwartz-space), $\mathscr{D}(M)$ (distributions on compact manifold $M$), the dual of a separable locally convex space with the weak$^*$-topology, ...

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    $\begingroup$ I don't know if this is the kind of example you're looking for, but the Stone-Cech compactification $\beta \mathbb N$ of the countable discrete space $\mathbb N$ contains no converging sequences at all (except those that are eventually constant). en.wikipedia.org/wiki/… $\endgroup$
    – Will Brian
    Sep 4, 2020 at 14:22
  • $\begingroup$ Thanks for that example, Will. This is not quite what I was after - I have tried to make the question more precise. $\endgroup$
    – Jan Bohr
    Sep 5, 2020 at 10:29
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    $\begingroup$ I would like to add a remark to the failure of $\mathscr D(\mathbb R^3)$ being sequential: It is of course true that a linear map is continuous if it is sequentially continuous, but this fails for linear maps defined on closed subspaces! $\endgroup$ Sep 5, 2020 at 11:31
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    $\begingroup$ The space $C_p([0,1])$ of all continuous real-valued functions on the unit interval with the topology pointwise convergence is a pretty natural example of a non-sequential topological vector space which is a function space. More generally, for a compact space $X$, $C_p(X)$ is sequential if and only if $X$ is scattered (see Arhangel'skii's "Topological Function Spaces" book for more information). $\endgroup$ Sep 5, 2020 at 11:51

2 Answers 2


The unit ball of the dual of a separable normed space is weak$^*$ sequentially compact, but this fail dramatically for non-separable spaces: The sequence of evaluations $\delta_n: \ell_\infty\to\mathbb R$ is probably the first sequence in $\ell^*_\infty$ that comes to mind and it has no weak$^*$-convergent subsequence: This would be a sequence of integers $n_1<n_2<\cdots$ such that for every bounded sequence $(x_n)_{n\in\mathbb N}$ of scalars one had a limit $\lim\limits_{k\to\infty} x_{n_k}$.

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    $\begingroup$ That's a great example. I'll add a (probably trivial) observation: The example proves at the same time that $(\ell_\infty,w^*)$ is not sequential and that in general one does not have compactness$ \Rightarrow $sequential compactness. $\endgroup$
    – Jan Bohr
    Sep 7, 2020 at 9:23

The sequentiality does not match well with an algebraic structure. For example, the following result of Banakh and Zdomskyy characterizes sequential topological groups with countable $cs^*$-character:

Theorem. A topological group $G$ with countable $cs^*$-character is sequential if and only if $G$ is either metrizable or contains an open $\mathcal M\mathcal K_\omega$-subgroup.

Let us recall that a topological space $X$ has countable $cs^*$-character if for every point $x\in X$ there exists a countable family $\mathcal F_x$ of subsets of $X$ such that for every neighborhood $O_x\subseteq X$ of $x$ and every sequence $\{x_n\}_{n\in\omega}\subseteq X$ that converges to $x$, there exist a set $F\in\mathcal F_x$ such that $F\subseteq O_x$ and $F$ contains infinitely many points of the sequence $(x_n)$.

A topological space $X$ is $\mathcal{MK}_\omega$ if there exists a countable cover $\mathcal C$ of $X$ by compact metrizable subspaces such that a subset $F\subseteq X$ is closed if and only if for every compact set $C\in\mathcal C$ the intersection $C\cap F$ is closed in $C$.

In this paper of Banakh and Repovs the above result of Banakh--Zdomskyy was extended to rectifiable spaces and topological left-loops.

In fact, the above theorem, is a corollary of the following result of Banakh:

Theorem. If a perfectly normal topological group $G$ contains a topological copy of the Frechet-Urysohn fan $S_\omega$ and a closed topological copy of the metric fan $M$, then $G$ is not sequential.

The metric fan is the subspace $$M=\{0\}\cup\{\tfrac1{n}+\tfrac{i}{nm}:n,m\in\mathbb N\}$$ of the complex plane. The Fr'echet-Urysohn fan is the set $M$ endowed with the strongest topology that coincides with the Euclidean topology on each subspace $K_m=\{0\}\cup\{\frac1{n}+\tfrac{i}{nm}:n\in\mathbb N\}$, $m\in\mathbb N$. It is easy to see that the Fr'echet-Urysohn fan is an $\mathcal{MK}_\omega$-space.

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    $\begingroup$ @JanBohr All sequential non-metrizable topological vector spaces you mentioned are $\mathcal{MK}_\omega$. $\endgroup$ Sep 7, 2020 at 16:22
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    $\begingroup$ @JanBohr Concerning topological groups with countable $cs^*$-character, then natural examples are LF-space or more generally, inductive limits of sequences of metrizable spaces. $\endgroup$ Sep 7, 2020 at 16:24
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    $\begingroup$ Concerning possible applications, we can consider the inductive limit $\mathbb R^\infty=\lim \mathbb R^n$ of finite-dimensional Euclidean spaces. It is an $\mathcal MK_\omega$-group and hence a sequential topological vector space. On the other hand, for any infinite-dimensional metrizable space $X$ the product $X\times\mathbb R^\infty$ is not sequential since it is not metrizable and not $\mathcal{MK}_\omega$ but has countable $cs^*$-character. $\endgroup$ Sep 7, 2020 at 16:27
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    $\begingroup$ @JanBohr Since topological vector spaces are connected, they contain an open $\mathcal{MK}_\omega$-subgroup if and only if they are $\mathcal{MK}_\omega$-spaces. Since $\mathcal D(\mathbb R^d)$ is not $\sigma$-compact, it is not $\mathcal{MK}_\omega$ neither. $\endgroup$ Sep 8, 2020 at 12:28
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    $\begingroup$ @JanBohr Another way to prove that $\mathcal D(\mathbb R^d)$ is not sequential is to observe that this space has countable network and hence is perfectly normal and it contains closed topological copies of the metric fan and Frechet-Urysohn fan. $\endgroup$ Sep 8, 2020 at 12:48

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