Absorbing at Each point means Open?

Question

In a topological vector space, it seems that open sets have the property that they are absorbing at each of their points, since $(\alpha,x)\to \alpha x$ is continuous.

I am wondering if the converse is true. Is it correct that a set that is absorbing at each of its points is open?

• Absorbing: a set $S$ is absorbing in a vector space $X$ if for every $x \in X$ there exist a $r \geq 0$, s.t. $\forall \;|\alpha|>r,\; x \in \alpha S $.

• Absorbing at $x_0$: $S$ is absorbing at $x_0$ if $S-x_0$ is absorbing.

Answer 1

For example, in an infinite dimensional Banach space, open balls are not open in the weak topology.