if $A, B$ are open in $\mathbb R$ then so is $A+B.$

Question

I am trying to find out a counterexample to the problem:

if $A, B$ are open in $\mathbb R$ then so is $A+B.$

But I could not find any such counterexample. Please help me.

Answer 1

Note that

$$ A + B = \bigcup_{a \in A} (a + B). $$

Also note that $a + B$ is open if $B$ is. Why does this help you?

Answer 2

No wonder you could not:

If $x$ is in $A+B$ then there exists $a$ in $A$ and $b$ in $B$ such that $x=a+b$. Since $A$ and $B$ are open (in a metric space), there exists $r\gt0$ such that $B(a,r)\subseteq A$ and $B(b,r)\subseteq B$. Can you prove that $B(x,s)\subseteq A+B$ for some suitable $s\gt0$?

Hence $A+B$ is open.