Talk:Topological vector space

message: The article current reads: "A vector space is an abelian group with respect to the operation of addition, and in a topological vector space the inverse operation is always continuous (since it is the same as multiplication by −1). Hence, every topological vector space is an abelian topological group.". The reasoning here sounds wrong. The addition is a function +: V^2 to V, where V^2 has the product topology. It is this function which must be continuous.
--Kaba3 (talk) 21:17, 22 February 2012 (UTC)
thread_title:  Abelian topological group 

message: 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)
thread_title: dual always Hausdorff?

message: 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)
replies: text: I would guess the Hausdorff property is copied from []. 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)

text: 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)
replies: text:  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)
replies: text:  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)
replies: text:  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)

text:  If the current definition is correct (scalar multiplication has to be continuous in both variables), then the sentence "The multiplication operation is continuous at 0 if and only if for any neighborhood U of 0 and any scalar λ there exists another neighborhood V of 0 such that λV is contained in U." under the illustration is incorrect. Jaan Vajakas (talk) 12:57, 20 February 2008 (UTC)

text:  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)
thread_title:  not necessarily Hausdorff 

message: Text says:
:More generally, a linear transformation from X to a Hausdorff vector space Y is continuous if and only if its kernel is closed.
I doubt it!!! --Bdmy (talk) 19:55, 1 January 2009 (UTC)
replies: text: Ouch.  Indeed, I'm not sure why I accepted the edit at face value. siℓℓy rabbit (talk) 20:31, 1 January 2009 (UTC)
thread_title:  Closed kernel 

message: The article currently says that weakly bounded sets are relatively compact.  But this is not true.  Weakly bounded sets are not even originally bounded in some cases (for instance, there are topological vector spaces which lack any functionals, and so are weakly bounded in themselves).  But it appears to be false also for the test case of a Hilbert space (the closed unit ball is weakly bounded, but not compact).  I'm not sure what was intended here. siℓℓy rabbit (talk) 00:09, 5 January 2009 (UTC)
replies: text: It should read "weakly bounded sets are relatively weakly compact."; otherwise, of course, that's not true. But you're right I forgot about the case when the dual space is trivial. So, some further assumption is needed; maybe locally convex-ness. Thanks for spotting the error. Since this is one of the most frequently used fact, it should be mentioned somehow. But it has to be stated correctly :) -- Taku (talk) 00:16, 5 January 2009 (UTC)
replies: text: Dear Taku, in any non-reflexive Banach space the unit ball is weakly bounded but not relatively weakly compact. Also, I believe that an extensive discussion of weak topologies is more appropriate in the article locally convex topological vector space. --Bdmy (talk) 10:58, 5 January 2009 (UTC)
thread_title:  Weakly bounded is relatively compact? 

message: The definition of Cauchy sequence should read: for every neighborhood V of 0, x_n - x_m \in V when n and m are large.
ChriS (talk) 20:42, 19 July 2009 (UTC)
thread_title:  Cauchy definition 

message: In the Definition section X × X → X is referred to as addition.  Is that correct?
Confluente (talk) 23:48, 22 December 2009 (UTC)
replies: text: Yes it is; sum or addition: it is the map + : X × X → X taking (x,y)\in\ X\times X into x+y\in X. But notice that if not specified  "X × X → X" denotes any map from the Cartesian product  X × X  to X. --pma (talk)  00:09, 23 December 2009 (UTC)
thread_title:  X × X → X  

message: I don't get it. Why is it so that Cauchy-sequences are always bounded but Cauchy-nets are not? If there is an unbounded Cauchy-net, one can pick an unbounded Cauchy-sequence out of it. The same thing puzzles me with the "sequential completeness". This is from the article: 
:Every Cauchy sequence is bounded, although Cauchy nets or Cauchy filters may not be bounded. A topological vector space where every Cauchy sequence converges is sequentially complete but may not be complete (in the sense Cauchy filters converge). Every compact set is bounded.
If there is a Cauchy-net which does not converge, one can extract a sequence out of it which does not converge, and this sequence is Cauchy as a subset of a Cauchy-net. Therefore "sequential completeness" implies completeness. I commented on this also in Talk:Locally convex topological vector space. Am I wrong? Lapasotka (talk) 08:08, 28 July 2011 (UTC)
thread_title:  Cauchy-nets vs Cauchy-sequences 

message: Why is that formulation more precise? Topological vector spaces get their uniform structure for free, since they are topological groups. --Chricho ∀ (talk) 00:45, 30 March 2012 (UTC)
thread_title:  “a uniform structure to be precise” 

message: From the article: Local convexity is the minimum requirement for "geometrical" arguments like the Hahn–Banach theorem.
This is probably true in some sense, but the Hahn–Banach theorem (general version) itself needs only a vector space structure for its statement and proof. YohanN7 (talk) 17:50, 30 December 2012 (UTC)
thread_title:  Hahn-Banach theorem 

message: Can you guys include the simplest example of a vector space that isn't a TVS?  That would help me understand what the point of this concept even is.  Would it have to be non-metrizable?  Etc.  — Preceding unsigned comment added by 2601:681:8002:9270:3105:85E1:FB5D:4472 (talk) 16:01, 27 July 2016 (UTC)
thread_title:  non-example, please 

message: I saw that you fixed the precompact link. I looked at the Schaefer–Wolff book used as one of the main sources for this article and there the definition of precompact is given as having compact completion. Poking around on WP, I found this definition (and its equivalence to total boundedness) at Totally_bounded_space#In_topological_groups, so I'm going to redirect the precompact link to that page. Lester Mobley (talk) 15:19, 24 December 2020 (UTC)
thread_title:  Precompact