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Let $A, B$ be two subsets of a topological Haudorff Group. Need to show that

  • $\forall B: A$ open $\Rightarrow$ $AB$ open
  • $A,B $ compact $\Rightarrow$ $AB$ compact
  • $A$ closed and $B$ compact $\Rightarrow AB$ closed
  • $A,B $ closed than $AB$ does not necesarry need to be closed

So far i know that every $T1$ Group is a Hausdorff (right?). I dont know how to do this since I have not subgroups here but subsets... Anyone an idea?

Thanks

José Carlos Santos
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Averroes2
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1 Answers1

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Let $G$ be your group.

  • $AB=\bigcup_{b\in B}Ab$, which is an union of open sets.
  • $AB$ is the image of $\pi\colon A\times B\longrightarrow G$ defined by $\pi(x,y)=xy$ and $A\times B$ is compact.
  • There is an answer here.
  • Take, in $(\mathbb{R},+)$ (endowed with its usual topology) $A=\mathbb{N}$ and $B=\left\{-n+\frac1n\,\middle|\,n\in\mathbb{N}\right\}$. Then $A$ and $B$ are closed, but $A+B$ is not (it contains the sequence $\left(\frac1n\right)_{n\in\mathbb N}$, but not its limit).
José Carlos Santos
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  • Can you explain the "Sketch of the proof" part of the answere you have linked for the third point? I think i know what his idea is but I can't really follow his steps. – thehardyreader Nov 07 '17 at 22:25
  • I can't explain it in a comment. Please note that when it says “there are some $a_1,\ldots,a_n$”, there should be “$\in A$” right after. If you have specific doubts, I can try to help. – José Carlos Santos Nov 07 '17 at 22:54
  • The part I don't understand is why there are such neighbourhoods $V_{1},...,V_{n}$ of the neutral element $e$ such that $a_{i}V_{i} \cap B = \emptyset$ and the sets $a_{i}V_{i}^{2}$ cover $A$. I'd really appreciate an explanation for why this is the case, maybe I'm thinking too complicated. – thehardyreader Nov 08 '17 at 08:42
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    @ghhardy For each $a\in A$, since $B$ is closed and $a\notin B$, there is a neighborhood of $A$ which does not intersect $B$. Such a neighborhood is of the form $aV_a$, where $V_a$ is a neighborhood of $e$. Let $O_a$ be an open set such that $e\in O_a$ and that $O_a\subset V_a$. Then ${aO_a,|,a\in A}$ is an open cover of $A$ and, since $A$ is compact, it has a finite subcover ${a_1O_{a_1},\ldots,a_nO_{a_n}}$. Leit $V_i=O_{a_i}$. Then the $V_i$'s are neighborhoods of $e$, ${a_1V_1,a_2V_2,\ldots,a_nV_n}$ is a cover of $A$ and $(\forall i\in{1,2,\ldots,n}):a_iV_v\cap B=\emptyset$. – José Carlos Santos Nov 08 '17 at 09:24
  • Thanks for your reply! I unfortunately misplaced the $.^{2}$ in my comment. Is it obvious that also $a_{i}V_{i}^{2} \cap B = 0$? – thehardyreader Nov 08 '17 at 10:15
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    @ghhardy Start by choosing neighborhoods $W_a$ of $e$ such that $aW_a\cap B=\emptyset$. Then let $V_a$ be a neighborhood of $e$ such that ${V_a}^2\subset W_a$. Then $a{V_a}^2\cap B=\emptyset$. – José Carlos Santos Nov 08 '17 at 10:23