Can every Hausdorff topological space be homeomorphically embedded in a topological vector space?

Question

It's true that for any metric space, we can isometrically embed it in a Banach space, so that the image of the metric space by that embedding is a linearly independent set.

Is the analogous theorem for topological spaces true? Often topological vector spaces are required to be Hausdorff, so with that definition the theorem could only potentially be true for topological spaces which are also Hausdorff.

So, to make it clear what I mean, rephrasing the title. Given a topological space $ (X, \tau) $ is it possible to find a topological vector space $ V $ and an embedding (homeomorphic onto the image) $ f:X \rightarrow V $ such that $ f(X) $ is a linearly independent subset of $ V $? Or can we at least somehow weaken this statement to make it true?

Answer 1

Any completely regular (Hausdorff) topological space can be imbedded in $\mathbb{R}^J$ for some index set $J$, which is a topological vector space. Indeed, this is true if and only if a topological space is completely regular. Thus any Hausdorff topological space that is not completely regular cannot be imbedded in a topological vector space over the reals.

Answer 2

Not an answer, but too long for a comment:

Matt Samuel's comment isn't quite true: here's an embedding (=map which is a homeomorphism onto its image) of $[0, 1]$ into a topological vector space over $\mathbb{R}$, whose image is linearly independent:

The space is $V=\prod_\mathbb{R}\mathbb{R}$ - elements of $V$ are maps $\mathbb{R}\rightarrow\mathbb{R}$, with addition and scalar multiplication defined in the obvious way.

The embedding is given by the following. For a real $r$, let $[\cdot]_r: \mathbb{R}\rightarrow\mathbb{R}$ be the function such that if $s\le r$, then $[s]_r=r-s$, and if $s>r$ then $[s]_r=0$. Our embedding, then, is the following:

$$x\mapsto "y\mapsto [y]_x."$$

It's easily checked that this is a topological embedding, and that its image is a linearly independent subset of $V$.