# Topological Vector Space

Show:

Let $\{U_{\lambda}: \lambda \in L\}$ be a base of neighbourhoods at $0$ in a topological vector space $\mathcal{X}$. Then $\{U_{\lambda}+ U_{\lambda}: \lambda \in L\}$ is also a base of neighbourhoods.

I have an intuition that this is to proved using the continuity of the "+" operator, but am not able to proceed.

-

Your intuition is correct! Let $U$ be a neighbourhood at $0$, since $+: X\times X\to X$ is a continuous map then $+^{-1}(U)$ is open and $(0,0)\in +^{-1}(U)$. By definition of the product topology there are $\lambda_1, \lambda_2$ so that $U_{\lambda_1}\times U_{\lambda_2}\subseteq +^{-1}(U)$. Using the fact that $U_{\lambda_1}\cap U_{\lambda_2}$ is a neighbourhood at $0$ we can find a $\lambda$ such that $U_{\lambda}\subseteq U_{\lambda_1}\cap U_{\lambda_2}$. It follows that $U_\lambda\times U_\lambda\subseteq +^{-1}(U)$, in other words $U_\lambda+ U_\lambda=+(U_\lambda\times U_\lambda)\subseteq U.$