Why local basis of $0$ in topological vector space generates the topology by translation?

On Rudin there's the following definition:

A local base of a topological vector space $X$ is thus a collection $B$ of neighborhoods of $0$ such that every neighborhood of $0$ contains a member of $B$.The open sets of $X$ are then precisely those that are unions of translates of members of $B$.

But in his book it is only shown that a set $E$ is open if and only if each of its translates is open.

Let $U$ be an open set in the topology how do we prove that $U$ is generated by union of translates of $B$?

Solution 1:

Let $U$ open in $X$. For $x \in U$, $U - x$ is an open neighbourhood of $0$. Therefore, there is an $U_x \in B$ such that $U_x \subseteq U - x$, hence $x + U_x \subseteq U$. Taking the union, $$ \bigcup_{x \in U} (x + U_x) \subseteq U $$ As $x \in x + U_x$ foreach $x \in U$, we have equality, that is $$ \bigcup_{x \in U} (x + U_x) = U $$ Therefore $U$ is a union of translates of elements of $B$.