Topological supremum of family of linear topologies

Question

In Hans Jarchow - Locally Convex Spaces 2.4.4 (c) it says:

The topological supremum of any family of linear topologies on a fixed vector space is linear.

I couldn't find a proof in the book and neither succeeded to proof it myself, any ideas?

Answer 1

OK, I didn't thought about it the way Tom Cruise mentioned, the supremum as the topology generated by the union. I guess it's pretty much straight forward now, let's say we have the linear topologies $\tau_\alpha$ on the vector space $X$ and $\tau$ is the supremum of the $\tau_\alpha$, i.e. the topology generated by sets of the form $\cap_{i<n} A_{\alpha(i)}$ with $A_{\alpha(i)}\in\tau_{\alpha(i)}$.

To see that $\tau$ is linear we have to check that it is translation invariant and contains a neighbourhood base ${\cal U}$ around $0$ of balanced, absorbing sets such that $$\forall U\in{\cal U}.\exists V\in{\cal U}.V+V\subseteq U.$$

The basic open sets are translation invariant since $\cap_{i<n}A_{\alpha(i)}+x=\cap_{i<n}(A_{\alpha(i)}+x)$ and one can show similarly that the collection of arbitrary unions of basic open sets, i.e. $\tau$, is translation invariant as well.

Then choose neighbourhood bases ${\cal U}_\alpha$ of $0$ in $\tau_\alpha$ with the properties mentioned above and set ${\cal U}$ as the collection of sets $\cap_{i<n}U_{\alpha(i)}$ with $U_{\alpha(i)}\in{\cal U}_{\alpha(i)}$. Obviously this is a neighbourhood base in $\tau$ around $0$ which is also balanced and absorbing and if we choose $V_{\alpha(i)}\in{\cal U}_{\alpha(i)}$ with $V_{\alpha(i)}+V_{\alpha(i)}\subseteq U_{\alpha(i)}$ then we get $$\cap_{i<n} V_{\alpha(i)}+\cap_{i<n} V_{\alpha(i)}\subseteq \cap_{i<n} U_{\alpha(i)}$$ which completes the proof.