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Proposition. Let $G$ be an abelian (probably not Hausdorff) topological group, let $A$ be a subset of $G$, and let $\mathfrak{U} = \{U\subseteq G \mbox{ : neighbourhood of }0\}$. Then, the closure $\overline{A}$ is written as $$ \overline{A} = \bigcap_{U\in\mathfrak{U}}(A + U) = \bigcap_{U\in\mathfrak{U}}\overline{A + U}. $$ Question. I want to prove $\bigcap\overline{A + U}\subseteq \bigcap(A + U)$. Let $x\in \bigcap\overline{A + U}$, and let $U\in\mathfrak{U}$. There exists a $U'\in \mathfrak{U}$ such that $U' + U' \subseteq U$. A PDF I'm reading says that $\overline{A + U'}\subseteq A + U' + U'$, why?

K. Y.
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    Every neighborhood $U$ of $0$ contained a symmetric ball $V$ such that $V+V\subset U$: By continuity, there are $W,W'$ such that $W+W'\subset U$. Let $V=(W\cap W')\cap(-W)\cap(-W')$. – Mittens May 10 '21 at 01:00
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    As for the proposition you are trying to proof: Notice that $x\in\overline{A}$ iff $(x+V)\cap A\neq\emptyset$ for any open neighborhood $V$ of $0$, which is equivalent to $x\in A-V$ for every such neighborhood. Since $V$ is an open neighborhood of $0$ iff so is $-V$ – Mittens May 10 '21 at 01:05
  • Thanks for your comment. – K. Y. May 10 '21 at 07:23

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Let $x\in \overline{A + U'}$, then $x - U'$ is a neighbourhood of $x$. Therefore $(x - U')\cap (A + U')\neq\varnothing$, and there exists $u$, $v\in U'$, and $a\in A$ such that $x - u = a + v$. Hence $x = a + u + v\in A + U' + U'$.

K. Y.
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