Finitely spaned subspaces of topological vector spaces Let $F$ be a proper finite dimensional subspace of a TVS (with infinite dimension). I have already shown that $F$ will be closed. It looks like $F$ must have an empty interior (possibly subject to some constraints on the underlying field).
Could anyone give me a tip?
 A: As said in the comments: Suppose that $X$ is a topological vector space and that $Y\subset X$ is a proper subspace, regardless of dimension. Then $Y$ has empty interior.
EDIT: All this post is valid under the assumption that our underlying field is $\mathbb{R}$ or $\mathbb{C}$ endowed with their usual topology.
Proof: Suppose on the contrary that $Y$ has an interior point, say $y\in Y$. Then we may find an open neighborhood $V$ of $y$ such that $V\subset Y$. Since $Y$ is proper, we have an element $x\in X\setminus Y$.
Consider the directed set $(0,1)$ directed under $\alpha\prec \alpha'$ if and only if $\alpha'<\alpha$ (so when im taking the limit as $\alpha\in(0,1)$ i am actually letting $\alpha$ approach $0$). Now consider the net $(x_\alpha)_{\alpha\in(0,1)}$ given by $x_\alpha=(1-\alpha)y+\alpha x$. Note that $$\lim_{\alpha\in(0,1)}x_\alpha-y=\lim_{\alpha\in(0,1)}\alpha(x-y)=0$$
because $\alpha$ becomes very small and $X$ is a topological vector space, so scaling is continuous. Thus $x_\alpha\to y$, so we may find $\alpha_o\in(0,1)$ such that $x_{\alpha_o}\in V$, i.e. $(1-\alpha_o)y+\alpha_ox\in V$. But $V\subset Y$, so $(1-\alpha_o)y+\alpha_ox\in Y$. Now $Y$ is a subspace and $y\in Y$, so $(1-\alpha_o)y\in Y$, so $(1-\alpha_o)y+\alpha_ox-(1-\alpha_o)y\in Y$, i.e. $\alpha_ox\in Y$, i.e. $x\in Y$, a contradiction.
A comment: don't feel dumb.
How to come up with this proof yourself:

*

*Think about how to prove this in a normed space.

*draw a picture.

*Now throw the norm in the trashcan and translate everything in topological language.

*Say out loud "that's all folks!"

A: If the underlying field $k$ is endowed with the discrete topology, the TVS $X$ may also be endowed with the discrete topology, and all subsets are open, including finite  dimensional subspaces.
Conversely, if $0$ is not an isolated point of $k$, and $U$ is a non-empty open set contained in a proper subspace $Y\subsetneq X$, then pick $u\in U$ and we obtain an open set $U-u$ by translating $U$.
Pick any $v\notin Y$ so we have a continuous map $f\colon k\to X$ given by: $$f\colon \lambda \mapsto \lambda v.$$
Then $f^{-1}(U-u)=\{0\}$ is open, contradicting $0$ not being isolated.
