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> [!proposition] Proposition. ([[openness dominates in internal subspace products in topological groups]])
> Let $G$ be a [[topological group]]. Then for any subsets $A,B \subset G$, if $A$ is open then so are $AB$ and $BA$.
> [!proof]- Proof. ([[openness dominates in internal subspace products in topological groups]])
> First, recall:
>
![[open sets translate to open sets in topological groups#^945a65]]
![[open sets translate to open sets in topological groups#^0e970d]]
>
Then, recognize $AB = \{ ab : a \in A, b \in B \} =\bigcup_{a \in A}^{}aB$
and $BA = \{ ba: b \in B, a \in A \} = \bigcup_{b \in B}^{}bA$
to be open in $G$ as unions of (open, per the above) translates of open sets in $G$.
^a7a2f4
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#### References
> [!backlink]
> ```dataview
> TABLE rows.file.link as "Further Reading"
> FROM [[]]
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> ```
> [!frontlink]
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> ```