In a topological vector space, show if $A$ and $B$ are bounded, then $A + B$ is bounded?

Question

I get as far as this before I am stuck:

Pick any neighbourhood of $0$ and call it $U$. Then there exists $a, b$ such that $A \subseteq aU$ and $B \subseteq bU$. So hence $ A + B \subseteq aU+bU$. This last part should probably be easy, but I'm having trouble showing that the latter is contained in another scaling of U. Help finishing this proof?

Answer 1

By continuity of multiplication by a scalar, we can take $V\subset U$ balanced, that is, $\lambda x\in V$ if $x\in V$ and $|\lambda|\leqslant 1$ and such that $V+V\subset U$ (continuity of addition). Then take $a,b\gt 0$ such that $A\subset aV$ and $B\subset bV$: then $$A+B\subset aV+bV=(a+b)\left(\frac a{a+b}V+\frac b{a+b}V\right)\subset (a+b)(V+V)\subset (a+b)U.$$