Let $X$ be a topological vector space and let $W$ be an open set which contains $0$ (a neighborhood of $0$). How do you prove that $0$ has neighborhoods $V_1$ and $V_2$ such that $V_1 + V_2 \subseteq W$?
(This was part of a proof of theorem 1.10 in Rudin's functional analysis which I don't get).
By the definition of a topological vector space, the addition function is continuous. Thus the preimage of $W$ is an open subset of $X\times X$, which contains $\langle 0,0\rangle$ since $0+0=0\in W$. By the definition of a product topology, there is an open basis set $V_1\times V_2$ containing $\langle 0,0\rangle$ and contained in $+^{-1}(W)$. Thus for any $\langle x,y\rangle \in V_1\times V_2$, $\langle x,y\rangle \in +^{-1}(W)$, which is to say $x+y\in W$ whenever $x\in V_1, y\in V_2$.