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Let $G$ be a topological group with identity element $e$. If $A,B$ are subsets of $G$ we define

$A^{-1} := \{ a^{-1} \mid a \in A\} \quad \quad AB:=\{ab \mid a \in A , b \in B\}$

Let $A \subseteq G$, then $\overline{A} = \bigcap \{AU | U \text{ neighborhood of } e\} = \bigcap \{AU^{-1} \mid U \text{ neighborhood of } e\} $

Proof:

  1. If $U$ is a neighborhood of $e$ also $U^{-1}$ is a neighborhood of $e$:

If $U$ is a neighborhood of $e$ then there is $A \subseteq U$ open s.t. $e \in A$. Then $e= e^{-1} \in A^{-1} \subseteq U^{-1}$ and $A^{-1}$ is open because the inverse is an homeomorphism. Then $U^{-1}$ is a neighborhood of $e$.

  1. Let $x \in G$, then every neighborhood of $x$ is in the form $xU$ where $U$ is some neighborhood of $e$:

Let $V$ be a neighborhood of $x$, then $U:=x^{-1}V$ is a neighborhood of $e$ and $xU = V$.

  1. Thesis:

Let $x \in \bigcap \{AU \mid U \text{ neighborhood of } e\}$. By contradiction suppose there exists a neighborhood of $x$, call it $V$, s.t. $V \cap A = \emptyset$. By 2. there exists $W$ neighborhood of $e$ s.t. $V=xW$. Then $e \notin A(xW)^{-1} = AW^{-1}x^{-1}$ and then $x \notin AW^{-1}$ which is a contradiction. Then $V \cap A \ne \emptyset$ for every $V$ neighborhood of $x$, i.e. $x \in \overline{A}$ and then $\bigcap \{AU \mid U \text{ neighborhood of } e\} \subseteq \overline{A}$.

Let $x \in \overline{A}$, i.e. $\forall \, V$ neighborhood of $x$ we have $V \cap A \ne \emptyset$, then $\forall \, U$ neighborhood of $e$ we have $(xU) \cap A \ne \emptyset$, then $\forall \, U$ neighborhood of $e$ we have $e \in AU^{-1}x^{-1}$ i.e. $\forall \, U$ neighborhood of $e$ we have $x \in AU^{-1}$. Then $x \in \bigcap \{AU \mid U \text{ neighborhood of } e\}$ and then $\overline{A} \subseteq \bigcap \{AU \mid U \text{ neighborhood of } e\}$.

Then the first equality is proved. The second one follows from 1.

Is it correct?

Elías Guisado Villalgordo
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Bremen000
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    Looks good to me... BTW it is customary to assume (wlog) that structures like groups, rings, fields, and many other things are anti-transitive, that is, its members are not subsets. This makes it possible to write $xU$ for ${x} U$ without ambiguity, as it is almost always done in print..... I would prefer you wrote $({x}U)\cap A$ (5th-last line), with extra brackets, to distinguish it from ${x}(U\cap A).$...... I recommend as supplemental reading the first 30 or so pages of Harmonic Analysis, Volume I by Hewitt & Ross, as a clear concise introduction to topological groups – DanielWainfleet Feb 16 '18 at 00:38
  • Thank you very much (for answering and for your advices!) – Bremen000 Feb 16 '18 at 07:33
  • Another recommendation from my side: In the proof for $\bigcap{AU \vert U ;\text{neigborhood of}; e} \subseteq \bar{A}$ I would state that you want to prove by contradiction and then assume the existence of a $x$-neighborhood $V$ such that $V\cap A \neq \emptyset$ – Peter Wildemann Feb 16 '18 at 09:33
  • One more (non-math) hint: when you write $x \in \bigcap {AU | U \text{ neighborhood of } e}$, you can make it look better by replacing the vertical bar with "\mid", as in "x \in \bigcap {AU \mid U \text{ neighborhood of } e}" which displays as $x \in \bigcap {AU \mid U \text{ neighborhood of } e}$ (You can see that @egreg did this in the answer below.) – John Hughes Feb 16 '18 at 14:25
  • Fixed everything! Thank you all! – Bremen000 Feb 16 '18 at 15:26

1 Answers1

2

Looks good to me. You can shorten the proof of 1: taking the inverse is a homeomorphism, so $U^{-1}$ is a neighborhood of $e^{-1}=e$ because it's the inverse image under “taking the inverse” of the neighborhood $U$ of $e$ (continuity suffices).

Part 2 is good.

For the sake of brevity, let me denote by $\mathscr{U}_e$ the set of neighborhoods of $e$.

In part 3 you can avoid most notation and also contradiction.

Suppose $x\in\bigcap\{AU\mid U\in\mathscr{U}_e\}$ and $x\notin\overline{A}$ and consider an arbitrary neighborhood $xW$ of $x$ (for some $W\in\mathscr{U}_e$). By assumption and 2, $x\in AW^{-1}$, which means that, for some $a\in A$ and $w\in W$, $x=aw^{-1}$. Hence $xw=a\in A$ and $xW\cap A\ne\emptyset$. Therefore $\bigcap\{AU\mid U\in\mathscr{U}_e\}\subseteq\overline{A}$.

Conversely, suppose $x\in\overline{A}$ and let $W\in\mathscr{U}_e$. Then $xW^{-1}\cap A\ne\emptyset$ so we can write $xw^{-1}=a$ for some $w\in W$ and $a\in A$. Hence $x=aw\in AW$. As $W$ is arbitrary, we conclude $\overline{A}\subseteq\bigcap\{AU\mid U\in\mathscr{U}_e\}$

egreg
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