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Let $X$ be a Hausdorff locally convex vector space. Recall (my reference is the book of H. Jarchow, Locally Convex Spaces. B.G. Teubner, 1981) that we say that $X$ is a semi-Montel space if every bounded subset of $X$ is relatively compact (equivalently, every closed and bounded subset of $X$ is compact), and a Montel space if it is semi-Montel and satisfy one (hence all) of the following conditions (equivalent under the semi-Montel hypothesis, see Proposition 11.5.1, pp. 230 of Jarchow's book):

  • $X$ is reflexive;
  • $X$ is barrelled;
  • $X$ is quasi-barrelled.

It is known that the strong dual of a Montel space is also Montel (Jarchow, Proposition 11.5.4, pp. 230-231). In the proof of Theorem 4.11 (5), pp. 39-40 of the book of A. Kriegl and P.W. Michor, The Convenient Setting of Global Analysis (AMS, 1997), it is shown that if $X$ is the strong dual of a Fréchet-Montel space (hence $X$ is a Montel space), then $X$ is a compactly generated topological space (also called a k-space or a Kelley space), i.e. the topology of $X$ is the final topology with respect to the inclusions of compact subsets of $X$. However, it seems to me that the proof of this assertion uses only the fact that $X$ is Montel.

Question: Are Montel spaces compactly generated, or is there a counter-example to this claim?

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    $\begingroup$ Since we are looking at vector spaces here, each question that about a topology has slight variations asking about vector space topologies (i.e. topologies that turn the vector space into a topological vector space), LCTVS topologies etc. So let me ask: What can we say about the TVC or LCTVS topology generated by inclusions of compact sets? $\endgroup$
    – Johannes Hahn
    Commented Oct 15, 2014 at 0:26
  • $\begingroup$ @JohannesHahn Well, it seems to me that these inclusions must be extended to the inclusions of the vector subspaces generated by each compact subset, otherwise we cannot guarantee that the final topology will be linear. Since we also want a locally convex topology, it suffices to consider absolutely convex compact subsets (for $X$ semi-Montel, these are the bipolars of bounded subsets of $X$). The picture that seems to emerge is that, for $X$ semi-Montel, the LCTVS topology generated by the inclusions of compact subsets is the bornologification of $X$. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 15, 2014 at 2:04
  • $\begingroup$ In view of that, if $X$ is bornological (hence quasi-barrelled) and semi-Montel (hence Montel), I'm willing to bet that $X$ is compactly generated. That would actually be enough for the purposes I have in mind... $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 15, 2014 at 2:08
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    $\begingroup$ Kriegl-Michor really mean by $kX$ the finest topology (not necessarily locally convex) making all inclusions of compact subsets continuous. The proof uses the the Banach-Dieudonne theorem for which metrizability is quite essential. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 15, 2014 at 11:16
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    $\begingroup$ A counterexample to the Banach-Dieudonne theorem for non-metrizable spaces was first given by Komura [link.springer.com/article/10.1007%2FBF01361183] $\endgroup$
    – Jochen Wengenroth
    Commented Oct 16, 2014 at 10:12

2 Answers 2

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Komura's example mentioned in the comment is just a big product $\mathbb R^{\mathbb R}$ which is a Montel space and its (strong) dual $X$ is thus also Montel. As Komura showed the finest topology $\tau^f$ which agrees on all compact (=equi-continuous) sets with the weak* (and hence with the strong) topology is not a vector space topology. In particular, it is different from the strong topology. This should show that $X$ is not compactly generated.

The story is quite different if you consider the finest locall convex topology such that all inclusions $K\hookrightarrow X$ ($K$ compact) are continuous. This is indeed the associated bornological locally convex topology.

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  • $\begingroup$ That is an interesting counter-example. In this case, $X$ has the Mackey topology and its strong dual $\mathbb{R}^{\mathbb{R}}$ coincides as a vector space with its algebraic dual, hence $X$ is even bornological. Moreover, if one assumes that the continuum hypothesis is true, even $\mathbb{R}^{\mathbb{R}}$ itself is (ultra)bornological. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 16, 2014 at 21:17
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    $\begingroup$ I don't have any references at hand, but I think that products of "moderate cardinality" are bornological, in particular, I believe that $\mathbb R^{\mathbb R}$ is bornological even without assuming the continuum hypothesis. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 17, 2014 at 10:10
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    $\begingroup$ The Mackey-Ulam theorem is discussed in Bonet's and Perez-Carreras' book "barrelled locally convex spaces". They write that if a product $\mathbb R^I$ fails to be bornological then the cardinality $d$ of $I$ is "strongly inaccessible". In particular, $a<d$ $\Rightarrow$ $2^a < d$ which is certainly wrong for $c=card(\mathbb R) = 2^{\aleph_0}$. $\endgroup$
    – Jochen Wengenroth
    Commented Oct 18, 2014 at 10:29
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    $\begingroup$ @PedroLauridsenRibeiro Ulam did not require any hypothesis on the size of $\mathfrak{c}$ to show it has no non-principal $2 = \{0,1\}$-valued measures, the measures that matter for the Mackey-Ulam theorem. The argument: If $\mu : \mathcal{P}(2^{\mathbb{N}}) \rightarrow 2$ is a countably-additive measure, let $U_{n,b} = \{ f \in 2^{\mathbb{N}} \mid f(n) = b \}$, for $n \in \mathbb{N}$ and $b \in 2$. Define $g \in 2^{\mathbb{N}}$ by taking $g(n)$ to be the unique $b$ such that $\mu(U_{n,b}) = 1$. Then $\{g\} = \bigcap_{n \in \mathbb{N}}U_{n,g(n)}$ and $\mu(\{g\}) = 1$ by countable additivity. $\endgroup$
    – Robert Furber
    Commented Dec 16, 2021 at 19:18
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    $\begingroup$ @RobertFurber ah sorry, that was a byproduct of the gaps on my knowledge of axiomatic set theory... I've mistaken the result by Banach and Kuratowski that the cardinality of $\mathbb{R}$ isn't real-valued measurable for Ulam's result. The former does need the continuum hypothesis. Thanks for the explanation! $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Dec 16, 2021 at 23:47
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An(other) example of a Montel space which is not compactly generated is $\kern.4mm\mathscr D\kern.4mm(\kern.4mm\mathbb R\kern.4mm)$ . This follows from Theorem 6.1.4(iii) and Proposition 6.2.8(ii) on pages 190 and 195 in

A. Frölicher and A. Kriegl: Linear Spaces and Differentiation Theory, Wiley, Chichester 1988.

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  • $\begingroup$ Hmm... Come to think of it, this actually could be inferred indirectly from the discussion in the book of Kriegl-Michor (cited in the question) as well: Theorem 4.11 (3), pp. 39-40 states that the Kelley topology of a strict inductive limit of a sequence of Fréchet spaces coincides with its $c^\infty$ topology (i.e. the final topology induced by all smooth curves). If these Fréchet spaces are finite-dimensional and the sequence is strictly increasing, then Proposition 4.26 (iii), pp. 45 entails that the $c^\infty$ (and hence the Kelley) topology is not a vector space topology. $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 20, 2014 at 1:53
  • $\begingroup$ I do not have the book of Frölicher and Kriegl at hand, but I guess these results are the ones you quoted, right? $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 20, 2014 at 1:54
  • $\begingroup$ @Pedro Lauridsen Ribeiro: Otherwise YES, but instead Proposition 4.26 (ii), and the spaces should be infinite-dimensional. N.B. Many results in Kriegl and Michor's book are taken almost verbatim from the book of Frölicher anf Kriegl. $\endgroup$
    – TaQ
    Commented Oct 20, 2014 at 19:54
  • $\begingroup$ Oops, that was a typo (which unfortunately I can no longer edit out), sorry... Thanks! $\endgroup$
    – Pedro Lauridsen Ribeiro
    Commented Oct 21, 2014 at 1:45

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