U open neighbourhood of the origin $\Rightarrow$ $\exists N$ open neighbourhood of the origin s.t $\alpha N \subset U \ \forall |\alpha | \leq 1$

Question

$(V,T)$ topological vector space. U open neighbourhood of the origin $\Rightarrow$ $\exists N$ open neighbourhood of the origin s.t $\alpha N \subset U \ \forall |\alpha | \leq 1$

How could one prove that the above is true? I was thinking that maybe one can use that $U$ is absorbing. So for $\forall x \in N, \ \exists |\alpha| \leq 1 \in \mathbb{F}$ s.t $x \in \alpha U$ and then find some $\alpha$ which works $\forall x \in N$.

Answer 1

If you take any vector space $V$ with the trivial topology $\{\varnothing, V\}$, you will find no such $N$. The only open neighborhood of any point is $V$ itself and you cannot scale it down to be a proper subset of $V$. If you dont talk about proper subsets, then $N=U$ is sufficient.

Note: I believe that this definition gives a topological vector space, because any map into such a topological space is continuous. So addition and scalar multiplication are continuous.

Note: Of course, this is no counter proof if we assume $T_1$ for such spaces. (thanks Igor)

Answer 2

My suggestion would be to take a set $N$ of the form

$$N(U):=\frac12 \{x\in U\mid \alpha x\in U \text{ for all } |\alpha|\leq 1\}.$$

If $U=V$, then take $N(U')$ for some other open neighborhood $U'\subset V$ of the origin, which exists because of $T_1$. I think it is easy to see, that such an $N$ satisfies $\alpha N\subset U$ for all $|\alpha|\leq 1$. It remains to show that $N$ is open (it is non-empty, because $0\in N$).