# When is a sequentially closed cone, closed?

The following question I also posed here, but still got no answer. Let $X$ be a locally convex, Hausdorff topological vector space and $C\subseteq X$ a convex cone, which is sequentially closed. What are criteria, that would imply that $C$ is closed (in the topology of X)?Are there also "testifyable" criteria?

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A side note: currently you seem to have three unregistered accounts. I suggest registering one of them and then merging the others into it. –  Yemon Choi Feb 9 '13 at 20:26

Since this is too long for a comment, I post it as an answer:

Sorry, but I do not agree with Peter Michor's answer. There are certainly better examples, but this the first I can remember: There are countable inductive limits $X=\lim\limits_{\to} X_n$ of Frechet spaces which are not Hausdorff (take a decreasing sequence of open connected sets $U_n$ in the complex plane with empty intersection and $X_n=H(U_n)$ the Frechet space of holomorphic functions on $U_n$ together with the injective restriction maps). $X$ is the a quotient of the direct sum $\bigoplus X_n$ which is certainly bornological and the kernel of the quotient map is sequentially closed (because convergent sequences in the direct sum are located and convergent in some finite sum) but it is not closed because the quotient is not Hausdorff.

The situation is better for metrizable spaces (of course, this is trivial) as well as for so-called Silva spaces (also called LS or DFS-spaces, countable inductive limits of Banach spaces with compact inclusions): In these cases, sequentially closed subspaces are closed.

By 8.5.28 in the book of Bonet and Perez-Carreras, Barrelled Locally Convex Spaces, even sequentially closed subsets of Silva spaces are closed.

Edit. A simpler example (but possibly less relevant for analytical applications) is the space $X=\mathbb R^I$ endowed with the product topology (point-wise convergence of functions $f:I \to \mathbb R$) if $I$ is uncountable and of moderate cardinality (e.g. $I=\mathbb R$). Then $X$ is bornological (due to the cardinality restriction) and $L=\lbrace f\in X: \lbrace i\in I: f(i)\neq 0\rbrace \text{countable}\rbrace$ is sequentially closed and dense in $X$.

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So what does barreldness imply on the seminorms generating the topology of the space? –  andy teich Feb 14 '13 at 12:46
I do not understand this question. Barreledness is rather close to bornologicity, for instance, every (locally) complete bornological space is ultrabornological and hence barrelled. This means that barrelledness will not help you very much to conclude closed from sequentially closed. –  Jochen Wengenroth Feb 14 '13 at 13:02
...sorry, I meanz bornologicity! –  andy teich Feb 14 '13 at 14:35
If you can write down the seminorms "explicitely", what condotions do you have to impose in order that thes space is bornologic? Maybe my question is not well defined... –  andy teich Feb 14 '13 at 14:57
As for almost all locally convex properties, bornologicity does not reflect properties of single seminorms. The essential point is always the relation between them or how many of them you need. A trivial example: If only countably many seminorms describe the locally convex topology the space is (semi-) metrizable and hence bornological. –  Jochen Wengenroth Feb 15 '13 at 7:26

In ordered vector spaces the question is restated as follows: When does the Archimidean property imply that the positive cone of an ordered vector space is closed? So that should help in your search.

Here is a simple example of a condition involving only the cone.

Suppose that $X$ is a topological vector space with convex cone (wedge) $X_+$ . Suppose that there is $e\in X_+$ that is an order unit. That is, for each $x\in X$ there is $\lambda >0$ satisfying $$x\in \lambda e -X_+$$ In this case $X_+$ is sequentially closed if and only if it is closed. This is because $e$ must be an internal point of $X_+$. Supposing that $X_+$ is sequentially closed; if $x\in \overline{X_+}$, then $\alpha e + (1-\alpha) x$ is an internal point of $X_+$ for all $\alpha\in (0,1)$ because $X_+$ is convex. In particular $$\frac{1}{n+1} e + (1- \frac{1}{n+1}) x$$ is an internal point of $X_+$. Thus, $x$ is in $X_+$.

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If $X$ is bornological (carries the finest locally convex topology compatible with the given family of bounded sets, or the the given dual space), then sequentially closed implies closed.

Edit: As Jochen pointed out, this is wrong. Sorry.

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Lieber Herr Michor, do you agree with the counterexample to your claim that I posted as an answer? –  Jochen Wengenroth Feb 13 '13 at 15:29