Given two open sets, show their sum is open

Question

Show that if either $A$ and $B$ is open, then $A + B$ is open.

Attempt: Suppose $A$ is open, then for every $a \in A, $ there exists $\epsilon > 0$ such $B_{\epsilon}(a) \subset A$. Similarly suppose $B$ is open ,then for every $b \in B, $ there exists $\epsilon > 0$ such $B_{\epsilon}(b) \subset B$.

So $A + B$ is open since for every $a + b \in A + B$ there is $\epsilon > 0$ such that $B_{\epsilon}(a+ b) \subset A + B$.

Can someone please verify this is correct? Any better approach or feedback would really help.

Answer 1

You're on the right track, but you could be slightly more clear. We only need to assume one of them is open, so suppose it is $A$. Given $a+b\in A+B$, we have $a\in A$, so there is some $\epsilon>0$ such that $B_\epsilon(a)\subset A$ by openness. But then the translated set $B_\epsilon(a)+b\subset A+b\subset A+B$ is an open ball containing $a+b$. Done.

Answer 2

Somewhat more simply, $A + B$ is a union of translates of an open set, thus a union of open sets, and therefore is open.

Answer 3

You have to show that $B_{\epsilon}(a+b)\subset A+B$. To see this, suppose firstly that $a=0\in A$, $B_c(0)\subset A, B_d(0)\subset B, B_d(b)\subset B_c(0)+B_d(b)$ so $A+B$ contains an open neighborhood of $0+b=b$. So $t_{-a}(A)+B$ contains an open neighborhood of $b$. This implies that $A+B$ contains an open neighborhood of $a+b$ since you have $f:A+B\rightarrow t_{-a}(A)+B$ defined by $f(c,d)=(c-a,d)$ which is invertible.