# Closure of $AB$

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?

• $A$ and $B$ are meant to be subsets of … ? – Robert Israel Feb 12 at 18:07
• sorry yes, $A$ and $B$ are subsets of a topological group $G$ – 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 $$\overline{AB} \ne \overline{A}\; \overline{B}$$.