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Consider the action of $\mathbb{R}$ on $\mathbb{R^2}$ given by $t \cdot (x,y) = (x+t, y)$.

I am asked to find closed subsets $A \subseteq \mathbb{R}$ and $B \subseteq\mathbb{R^2}$ such that $A\cdot B$ is not closed.

there is a similar result that stats that

For any topological group $G$ acting on a space $X$, if $A \subseteq G$ is compact and $B \subseteq X$ is closed, then $A \cdot B$ is closed.

Therefore, for my example I am starting from a closed but not compact subset $A$ in $\mathbb{R}$; I have tried with unbounded intervals $[a,\infty)$ or discrete subsets like $\mathbb{N}$ or $\mathbb{Z}$. However; I can't find a suitable $B$ to get what I want.


marked as duplicate by Moishe Kohan, Community Nov 20 '17 at 21:37

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ This question was asked so many times at MSE... $\endgroup$ – Moishe Kohan Nov 20 '17 at 21:36

What if we take $B=\{(1-n,0)\:n\in\Bbb N\}$, and $A=\{n-\frac1{n+1}:n\in\Bbb N\}$? (I'm assuming $0\not\in\Bbb N$.)

Then $A\cdot B$ should include the points $\frac12,\frac23,\frac34\ldots$, but not $1$. Does that work?


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