Every LCTVS Has a Base of Cl. Abs.-Con. Abs. Nbds. of the Origin

Every LCTVS Has a Base of Closed Absolutely Convex Absorbent Neighbourhoods of the Origin

Recall from the Every TVS Has a Base of Closed and Balanced Neighbourhoods of the Origin page that if $E$ is a topological vector space then $E$ has a base of closed and balanced neighbourhoods of the origin.

This is proven by taking an arbitrary base of neighbourhoods $\mathcal U$ of the origin, then taking a balanced neighbourhood of the origin $V_U$ with $V_U + V_U \subseteq U$, and then showing that $\mathcal V = \{ \overline{V_U} : U \in \mathcal U \}$ is a base of closed and balanced neighbourhoods of the origin.

When $E$ is furthermore a locally convex topological vector space, then more can be said.

Proposition 1: Let $E$ be a locally convex topological vector space. Then $E$ has a base of closed absolutely convex absorbent neighbourhoods of the origin.

• Proof: If $\mathcal U$ is a base of neighbourhoods of the origin then by the proposition on the page above, $E$ has a base of closed and balanced neighbourhoods of the origin, call in $\mathcal V$. But since $E$ is a locally convex topological vector space and since $\mathcal V$ is itself a base of neighbourhoods of the origin, by the proposition on Locally Convex Topological Vector Spaces over the Field of Real or Complex Numbers page, each $V \in \mathcal V$ is absolutely convex and absorbent.

• Thus $\mathcal V$ is a base of closed absolutely convex absorbent neighbourhoods of the origin. $\blacksquare$