Talk:Topological vector space

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[edit] dual always Hausdorff?

If V is a Hausdorff topological vector space, is V ' also a Hausdorff topological vector space? Probably some Hahn-Banach consequence? AxelBoldt, Thursday, July 4, 2002

The weak* topology on V* is always Hausdorff — even if V isn't. This is because V separates the points of V*. So if v* and w* are distinct elements of V* — say because they behave differently on v ∈ V — then they are separated by the preimages (under v: V* → K) of any disjoint neighbourhoods of 〈v,v*〉 and 〈v,w*〉. As you can see, no Hahn-Banach Theorem (or any form of choice whatsoever) is needed. — Toby Bartels, Saturday, July 6, 2002

PS: Why'd you change "V^*" to "V'"? So ugly! (and IME far less common).

I don't agree on the last thing. D' is much more common than D* for the (topological) dual of D, and it's the same for all other function spaces. The star often denotes the group of units or the nonzero elements. At least in functional analysis, the topological dual is almost always denoted by a prime, AFAIK. — MFH:Talk 21:05, 28 February 2006 (UTC)

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I create Locally convex topological vector space as a meso-stub yesterday (and I am about to use this link Locally convex space to create a redirect). The problem is that I didn't get very far, since I didn't want to create too much overlap here - this page is really about locally convex spaces! One way around this is to shift much of the content from here to there, and create specific content on this page. In particular, a non-locally-convex example. --AndrewKepert 21:43, 11 Mar 2004 (UTC)

[edit] not necessarily Hausdorff

The very important notion of "associated Hausdorff space" would not make sense if all TVS were Hausdorff. I was already about to delete "Hausdorff" from the intro, but now hesitate until I'm sure that it's not used elsewhere (please, those who know well this area of wiki, help me...)

I think your justification,

"added common Hausdorff condition just to make it easier to write Locally_compact article"

is almost hilarious! We cannot change an extremely well-established definition just to make something else easier!

Please, if somebody has serious objections, tell me quickly, else I will ASAP delete "Hausdoff" before more and more pages risk referring to this definition.

(Already, I dislike a bit

A topological vector space is a real or complex vector space endowed with...

I would much prefer

A topological vector space is a vector space over a topological field endowed with...

or at least

A real or complex vector space is a topological vector space if it is endowed with...

this way, at least, you don't "block" people working in a slightly more general setting than you. But well, here I would agree that the latter are a minority (and of course minorities should not be treated on the same footing than the "main stream people", and not be able to use other's work... sorry for the sarcasm). --— MFH: Talk 18:12, 8 Apr 2005 (UTC)

I would guess the Hausdorff property is copied from [[1]]. I think it would be better to remove the property but I am no expert on the topic and do not know what is common in the literature. As for the second point I prefer A topological vector space is a vector space over a field endowed with. No need to limit oneself to real or complex vector spaces. Considering this is not a basic topic we can assume the reader is comfortable working with vector spaces over arbitrary fields. MathMartin 21:08, 9 Apr 2005 (UTC)

I agree with the fact that a topological vector space is not necessarily real or complex. A part of the theory works in a general contest (even if -I admit- I do not know anyone working on general TVS). Moreover, the requirement about the scalar multiplication (beeing continuous from K x X to X) is not completely correct. In the general definition of a TVS one requires that, for any k in K, the application that maps x to k x is continuous. Gala.martin 18:35, 27 February 2006 (UTC)

This would be important. Can you give a reference? I think that the current version is correct: it has to be continuous in both variables. I happened to work on things that are not TVS just becase scalar multiplication is not continuous (in the first variable !) - in fact, this seems to me always the case in ultrametric normed vector spaces (just like |x+x|=|x|, |k x|=|x| goes not to zero when k goes to zero.) — MFH:Talk 20:58, 28 February 2006 (UTC)

If your field has not a topological structure, what does it mean to be continuous in both the variables? But I am a little confused, now that I write. I can remember that Schaefer required continuity in both the variables... unfortunately I am abroad now, and cannot look for reference. Gala.martin 17:23, 1 March 2006 (UTC)

You are perfectly right. Shame on me. You can equip the field with its natural uniformity, and then, of course, continuity in both the variables is natural. I mean, you can give the definition you want, but you do not go that far without this assumption. You were really unlucky not to have continuity in the scalar variable!! Gala.martin 17:28, 1 March 2006 (UTC)

How about "... vector space over a field (often taken to be the real or complex numbers) ...": so as to give the novice a gentle hint. linas 01:34, 28 February 2006 (UTC)

I agree with you fine fellows, and have performed the generalization. I also removed the comment about product topologies being used: what other topology can be meant on the product? It struck me as an awkward parenthetical remark so I removed it. If you object, feel free to complain or restore. -lethe ^{talk +} 19:16, 1 March 2006 (UTC)