Can a topological vetor space over $\mathbb{R}$ or $\mathbb{C}$ ever be considered "first category"? 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?
 A: 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 $X$.
A: 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 $X$.
