2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $A$ 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
0
$\begingroup$

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 $\overline{AB} \ne \overline{A}\; \overline{B}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.