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I'm given the exercise to show that a finite dimensional linear subspace of an infinite topological vector space $X$ is nowhere dense (which I can do), and then to show that if $X$ is the union of countably many finite dimensional linear subspaces, then $X$ is of the first category. However, the premise of the second statement seems impossible to me if the underlying field is the real or complex numbers. Could someone help me to understand intuitively how this could be possible? More specifically, how is it that in passing from, say $\mathbb{R}^1, \mathbb{R}^2,\mathbb{R}^3....$ to $\mathbb{R}^\omega$ that second countability can be lost?

user140776
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  • Well for one thing, $\mathbb{Q}^\omega$ is already uncountable. So the usual construction for separability breaks down. I also think you might be confusing "union" and "span". A union of countably many finite dimensional subspaces is basically a bunch of extremely thin sets (thinner even than planes in 3D, in some sense) considered together. In infinite dimensions these can all be "perpendicular" to each other and consequently not have a limit point. (I realize not every TVS is Hilbert, but I'm trying to give some visualization.) – Ian May 18 '16 at 13:14
  • The exercise, as stated, says "union". Do you think this was possibly a typo? – user140776 May 18 '16 at 13:20
  • No, it was supposed to be "union". But I think you may be thinking that a union of finite dimensional linear subspaces is "big" by confusing it with the span of those subspaces instead. – Ian May 18 '16 at 13:21
  • Oh I see. That wasn't what I was thinking. I can see that any countable union of finite dimensional subspaces is necessarily of the fist category. But what I cant see is how any such union could be equal to a vector space over R. I see that its a subset, of course, but I cant picture it being a subspace. I just see a union of a bunch of lines through the origin which isnt a subspace if the lines are all different. – user140776 May 18 '16 at 13:23
  • I don't think the union will itself be a vector space but I don't see how the question said that it was. – Ian May 18 '16 at 13:24
  • It begins with the statement "let X be an infinite dimensional tvs" then part b of the problem says "Suppose that X is a union of countably many finite dimensional linear subspaces, Prove X is of the first category" So this is what I meant when I said that the premise of part b seems false. – user140776 May 18 '16 at 13:25
  • I think the wording might just be bad; I think it might be intended to be "let $X$ be an infinite dimensional TVS and let $Y \subset X$ be a union of countably many finite dimensional linear subspaces, prove that $Y$ is first category in $X$". I agree, I don't think a TVS over $\mathbb{R}$ or $\mathbb{C}$ can be first category in itself. – Ian May 18 '16 at 13:26
  • hmm....this is definitely possible. – user140776 May 18 '16 at 13:30

2 Answers2

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Essentially the only vector space which is the union of finite dimensional ones is $X=\lbrace x\in \mathbb K^N: \exists\, n\in\mathbb N\; \forall\, k\ge n\quad x_k=0\rbrace$. Anyway, if $X=\bigcup\limits_{n\in\mathbb N} L_n$ with finite dimensional subspaces $L_n$ then -- by definition of second category -- there would be $n\in\mathbb N$ such that $\overline{X_n}$ has interior points. Now you have to know two things: In (Hausdorff) TVS closed subspaces are always closed and the only subspace with interior points is the whole space $X$.

Jochen
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The key point is that the dimensions of the finite-dimensional subspaces can be getting larger and larger, and they can be nested in each other. So instead of envisioning a union of a bunch of lines, you should be imagining the union of a line, a plane containing the line, a 3-space containing the plane, and so on. The union of such an ascending sequence of finite-dimensional subspaces will again be a subspace, since any two points are contained in some common step of the union (whichever of the two points appeared later) and so any linear combination of them is as well.

More precisely, let $X$ be any infinite-dimensional vector space and let $\{x_1,x_2,\dots\}$ be an infinite linearly independent set. Let $A_1$ be the span of $\{x_1\}$, let $A_2$ be the span of $\{x_1,x_2\}$, let $A_3$ be the span of $\{x_1,x_2,x_3\}$, and so on. Then each $A_n$ is finite dimensional; let $A=\bigcup A_n$. For any $x,y\in A$, let $n$ be such that $x\in A_n$ and $m$ be such that $y\in A_m$; without loss of generality $n\geq m$. Then $x$ and $y$ are both in $A_n$, and so $ax+by\in A_n$ as well for any scalars $x$ and $y$. Thus $A$ is a subspace. It might happen to be the case that $A$ is all of $X$.

Eric Wofsey
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