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In Rudin's functional analysis , let $X$ denote a topological vector space . He states that the open sets of $X$ are precisely those that are unions of translates of members of the local base of $0$ .

Since for any nonempt subset $E$ , we can find some $x\in E$ , then we have $$E=(-x+E)+x$$ and $-x+E$ is a neighborhood of $0$ , so it suffice to show every neighborhood of $0$ can be write as a union of the local base of $0$ . However , for each neighborhood of $0$ , by definition of local base , we can only find an element of the local base which is a subset of the neighborhood , so how can we get the desired conclusion ?

J.Guo
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1 Answers1

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It's just a matter of using the definitions: Take an open set $U$ and let $x\in U$. Then, there is a basis element $B$ for the topology of $X$ such that $x\in B\subseteq U$. By the continuity of the vector space operations, $B-x$ is an open set containing the origin, so that there is an open basis element $N_x$ such that $0\in N_x\subseteq B-x$. But then, $x+N_x\subseteq B\subseteq U$. Then $U=\bigcup_{x\in U}x+N_x$ is a union of translates of basis elements of $X$, as desired.

Matematleta
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