Open neighbourhoods in topological vector spaces

Question

It is well known that each open ball in a Banach space is homeomorphic to the whole space. Can we extend this to topological vector spaces? In other words, does every non-void open set in a non-discrete, Hausdorff topological vector space contain a set that is homeomorphic to the whole space? What if we assume local convexity?

Answer 1

Suppose that the field used is $\mathbb R$.

Hausdorff topological vector spaces are Tychonoff. Thus for any closed set $C$ and any point $p\notin C$ there is a continuous function which is zero on $C$ and $1$ on $p$.

Let $U$ be an open subset of $V$. By translating we can suppose that $0\in U$. Let $f:V\to \mathbb R$ be a function so that $f(0)=1$ and $f=0$ on $U^c$ (which is closed by definition). Let $A=f^{-1}((0,\infty])$. $A$ is open because $f$ is continuous and $A\subset U$ because $f=0$ on $U^c$. Moreover $0\in A$. For any $x\in V$ define $\varphi(x)=\max\{f(x),0\}$ and for any $v\in A$ define $F(v)=\int_0^v 1+ \frac{1}{\varphi(x)}dx$ where the integral is made along the segment between $0$ and $v$. $F$ is continuous with values in $[0,\infty]$. Define $B=F^{-1}([0,\infty[)$. $B$ is an open subset of $A$ and contains $0$.

Define $g(x)=xF(x)$. We show that this is an homeo between $B$ and $V$. First note that $g(0)=0$ and $g$ is continuous because $F$ is. Moreover, $F$ is monotone on half-lines because on $B$ we have $\varphi=f$ and $1/f(x)>0$.

This immediately implies that $g$ is injective.

Finally, let's see that $g$ is onto. For any $v\in V$ either $\mathbb R_+v\in B$ or there is $\lambda>0$ such that $\lambda v\in B$ (because $B$ is open). In both cases $F(B\cap \mathbb R_+v)=[0,\infty)$ so there is $v'\in\mathbb R_+\cap B$ such that $v'F(v')=v$