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the first paragraphs at topological vector space seem odd to me.
I’d think it should start out saying that a topological vector space is a vector space over a topological field $k$, such that etc.. Then the following remark presently in the entry, about the internalization using the discrete topology is moot.
I edited it to clarify that any topological field may be used. But it remains true that (for a fixed field $k$) the category $k Vect$ is an algebraic category that may be internalized straightforwardly into $Top$ and this this internalization is not the correct category. It has to be a sort of topological internalization, and that is more than just internalization.
What we can do is take the category of pairs $(k, V)$, where $k$ is a field and $V$ is a vector space over $k$ and internalize that into $Top$; this is the category of pairs $(k, V)$ such that $k$ is a topological field and $V$ is a topological vector space over $k$. (Actually, this is better behaved if we generalize from fields to something purely algebraic such as commutative rings.)
Thanks for the edits!
I am still not a fan of that paragraph with the “but” in there, since it comes across as highlighting a problem or at least an awkwardness where there really is none. I find it misleading. What you say about pairs is the natural and obvious thing which should be emphasized. The other thing is more an oddity which I find wouldn’t need to be highlighted much in the Idea section. But I won’t inisist.
I really think the problem is that the whole thought was not expressed. Evidently a TVS according to the nLab is just an internal vector space in $Top$ if we use the two-sorted notion of vector space, where one of the sorts is the field. Beware that it is not the same thing as an internal vector space if we use the one-sorted notion (where the ground field is given in advance, but subsumed under the unary operations of the theory).
Other parts read awkwardly to me, for example the blatant lie that TVSes are path-connected. The trouble perhaps is that the article starts off in great generality, but then apparently forgets that generality and thinks just of real or complex TVS a paragraph or two later, leading one to say something that is obviously not generally true. As far as I know, it is quite rare to consider TVS over anything but a local field – usually $\mathbb{R}$ or $\mathbb{C}$ but occasionally also something like $\mathbb{Q}_p$. But I’ve never seen mention of TVS over say $\mathbb{Q}$; it’s just not in the spirit of the subject AFAIK.
I went ahead and made some edits to reflect my last comment. I also added something to vector space on the two-sorted notion.
But I’ve never seen mention of TVS over say $\mathbb{Q}$; it’s just not in the spirit of the subject AFAIK.
My first impression is that this violates the spirit because the subject is really about complete TVSes and this requires the field to be complete as well.
Actually, $\mathbb{Q}$ is complete if we use the discrete topology, and people do consider TVSes over this, only they call them topological abelian groups instead (with a certain property).
Toby: I agree with both of those comments (I had in mind $\mathbb{Q}$ with a topology which renders it incomplete).
This issue (#1 and following) comes up maybe more pronouncedly also in the first line at Banach space.
Either the first line there should explicitly name $\mathbb{R}$ as the ground field (which is the standard thing to do) or it should be more clear on what the general statement is. Or both.
I added a paragraph to Banach space to say that we are basically talking over $\mathbb{R}$ on that page.
Sorry to be nitpicky but: the very first sentence
A Banach space $\mathcal{B}$ is both a vector space and a normed space, such that the norm induced metric turns $\mathcal{B}$ into a complete metric space, and the induced topology turns $\mathcal{B}$ into a topological vector space.
might that better be turned into
A Banach space $\mathcal{B}$ is both a vector space (over a topological field, usually taken to be $\mathbb{R}$) and a normed space, such that the norm induced metric turns $\mathcal{B}$ into a complete metric space. The induced topology then turns $\mathcal{B}$ into a topological vector space.
?
Re #10: if we want to go this way, then the topological field should be a normed topological field, and there should obviously be a compatibility between the norm on the vector space and a norm on the ground field.
Anyway, I put in “over $\mathbb{R}$”.
In the Idea section, everything is actually fully general (at least for now), so I would not say ‘over $\mathbb{R}$’ there. (Instead, I said that at the end of that section in my last edit.) So I made it over a normed field.
Also, ‘both a vector space and a normed space’ is meaningless, since a normed space is already a vector space; this was probably supposed to be a metric space instead.
I would like to extend the range of full generality form the Idea section into later sections; the Definition section seems ripe for this. But I want to be careful, because we don't want to accidentally state something that's false over some fields.
Thanks for editing the first sentence at Banach space. Now I like it.
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