Show that $M^{-1}(O)=\{(r,x) \in \mathbb{R} \times V \mid rx \in O\}$ is open if $O$ is open and $(V,\mathcal{T})$ is a topological vector space.

Question

I have been spending a bit too much time on this problem so any help is appreciated. $(V,\mathcal{T})$ is a topological vector space and $O\in \mathcal{T}$ is an open set in $V$. Define $M^{-1}(O)=\{(r,x) \in \mathbb{R} \times V \mid rx \in O\}$. Show that $M^{-1}(O)$ is open in $\mathbb{R} \times V$ (in the standard product topology). As I understand it, for any given $(r,x) \in M^{-1}(O)$, I need to find an open set $W$ in $\mathbb{R} \times V$ such that $(r,x) \in W$ and $W \subset M^{-1}(O)$ but I am unable to come up with such a set.

Answer 1

$(V,\tau)$ is a topological vector space (over $\Bbb R$) iff $V$ is a vector space (with $+,-,\cdot$ as operations, say), $\tau$ is a topology (often assumed to be at least Hausdorff) and such that

1. $p: V \times V \to V$ defined by $p(x,y)=x+y$, is continuous where all products have the product topogy (here and later).
2. $m: V \to V$ defined by $m(x)=-x$, is continuous.
3. $M: \Bbb R \times V \to V$ defined by $M(r,x)=r\cdot v$, is continuous.

Then your $M^{-1}[O]$ is just open because of the third axiom. So nothing to prove.

Answer 2

I think you are overcomplicating this.

In a topological vector space $V$, by definition the scalar multiplication map

$$f: \mathbb{R} \times V \to V: (r, v) \mapsto rv$$

is continuous.

Thus if $O$ is open in $V$, then the inverse image $f^{-1}(O) = \{(r,v) \in \mathbb{R} \times V: rv \in O\}$ is open in $\mathbb{R} \times V$, as desired.