If $A, B \subset \mathbb{R}^{n}$ such that $A$ is open, is the set $A+B := \{ a+b \in \mathbb{R}^{n} \mid a \in A, b \in B \}$ still open? I strongly suspect that $A+B$ should still be open, since $A+B$ is merely a translation of $A$. But how to give a proof or a counterexample to it?
 A: You're correct; if $A$ is open, then $A+B$ is open. To see this, write:
$$A+B = A + \bigcup_{b \in B} b = \bigcup_{b \in B}(A+b)$$
So $A+B$ a union of open sets of the form $A+b$. I leave it as an exercise to show that the translation of an open set is itself open. Hint: Fix any $b \in \mathbb{R}^n$. Then the function $a \in \mathbb{R}^n \mapsto a+b \in \mathbb{R}^n$ is a homeomorphism.
A: An alternative to the existing answer is to go with more $\mathbb R^n$-centered definitions of open sets.
Take any element $x\in A+B$. Then you know there exists such $a\in A, b\in B$ that $x=a+b$. Now, since $A$ is open, there exists an $\epsilon>0$ such that the ball, centered at $a$ and with a radius of $\epsilon$, i.e. $B(a,\epsilon)$ is a subset of $A$. It is now easy to see that $B(a+b, \epsilon)$ is a subset of $A+B$:

Take any element $y\in B(a+b, \epsilon)$. Then $y=(y-b) + b$, and the distance from $a$ to $y-b$ is equal to the distance from $a+b$ to $(y-b)+b=y$, therefore smaller than $\epsilon$. This means that $y-b\in B(a,\epsilon)\subset A$, so $y-b\in A$, meaning that $y\in A+B$.

meaning that $x$ is an internal point of $A+B$, so $A+B$ has only internal points and is therefore open.
