If $A$ is an open set and if $B$ is a closed set, then $A + B$ is an open
set.
We know $(A+B)^o \subset A+B$. We need to prove $A+B \subset (A+B)^o$. We need to prove $x$ is an interior point of $A+B$. Let $x\in A+B$, $x=a+b, a\in A $ ,$b\in B$. $ a$ is the interior point of $A$. $\implies$ $\exists$ open set containing $a$ in $A$. $b\in B \implies$ Every neighbourhood of $b$ have a non-empty intersection with $B$.
Using this fact, how to construct a neighbourhood of $x$ which sits in $A+B$? Please help me. Is there any quick alternative way to check the openness?
Best Answer
In topological groups setting? Then for any open set $A$ and any set $B$, $A+B=\displaystyle\bigcup_{x\in B}(x+A)$. That $x+A$ is open by the homeomorphism $\cdot\rightarrow x+\cdot$
So $A+B$ is open.