I'm trying to understand what the closure of $AB$ looks likes...

$AB = \{ab: a \in A, b\in B\}$

So I know the closure of $AB = AB \cup (AB)' = \{ab: a \in A, b\in B\}\cup\{ab: a \in A', b\in B'\} $.

But is this equal to $\{ab: a \in A\cup A', b\in B\cup B'\}$? If yes, is this my properties of sets or just because of the closure?

  • $\begingroup$ $A$ and $B$ are meant to be subsets of … ? $\endgroup$ – Robert Israel Feb 12 at 18:07
  • $\begingroup$ sorry yes, $A$ and $B$ are subsets of a topological group $G$ $\endgroup$ – Sasha Feb 12 at 18:08

Consider the topological group $G = (0,\infty)$ under multiplication (with the usual topology), with $A$ the positive integers and $B = A^{-1}$ the reciprocals of the positive integers. Then $AB$ is the positive rationals, and its closure is $G$. On the other hand, $A$ and $B$ are both closed. So in this case $\overline{AB} \ne \overline{A}\; \overline{B}$.


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