Sum of two open sets is open Let $A$,$B$ be two open sets in $\mathbb{R}$. Show that $A+B = \{a+b \text{ | } a \in A, b \in B\}$ is an open set.

Let $a\in A$ and $b\in B$. Since $A$,$B$ are open sets, it follows that $a$,$b$ are interior points, therefore $\exists \epsilon_a, \epsilon_b > 0 $ such that $a \in (a - \epsilon_a, a + \epsilon_a) = I_a \subset A$ and $b \in (b - \epsilon_b, b + \epsilon_b) = I_b \subset B$.
Now let's define two closed intervals: 
$$I_{a/2} = [a-\epsilon_a/2, a+\epsilon_a/2] \subset I_a$$
and
$$I_{b/2} = [b-\epsilon_b/2, b+\epsilon_b/2] \subset I_b$$
Now let's define the set: $I = \{x+y \text{ | } x \in I_{a/2}, y \in I_{b/2}\} = [a+b-\frac{\epsilon_a + \epsilon_b}{2} , a+b+\frac{\epsilon_a + \epsilon_b}{2}]$.
Now let $\epsilon = \frac{\epsilon_a + \epsilon_b}{2}$. It follows that $((a+b)-\epsilon , (a+b)+\epsilon) \subset I \subset A+B$. Therefore $A+B$ is an open set.

Can someone please check my work? I've seen a more direct proof using the fact that $A+B = \bigcup_{a \in A}\{a\} + B$. I don't want to prove it that way because it wasn't the way I initially thought about the problem.
Thank you!
 A: Your proof looks fine. As with every proof-verification question, I try to come up with some comments, the pettiness of which is proportional to the correctness of the proof.


*

*The notation $I_{a/2}$ and $I_{b/2}$ could be improved. There's nothing formally wrong with naming them as such, but given the definition of $I_a$ and $I_b$, I would have expected $I_{a/2}$ to be a symmetric interval around $a/2$, not around $a$ with half the width. Maybe call them $J_a$ and $J_b$ respectively?

*I don't see why you need to define such intervals anyway. If you sum $I_a + I_b$, you'll get the open interval $(a + b - (\varepsilon_a + \varepsilon_b), a + b + (\varepsilon_a + \varepsilon_b))$, which is contained in $A + B$.

*I noticed that you brushed over the summing of the two intervals $I_{a/2}$ and $I_{b/2}$. This is quite a mechanical calculation, but it behooves you to know, in principle, how to fill in this detail if ever you need to. In particular, you should be able to do a full set equality proof, by showing each set is a subset of the other.

*I noticed point 3 in light of point 2. I think the main reason that someone would change to closed intervals instead of an open intervals is because they feel they more easily grasp the summing of two closed, bounded intervals, due to the definite maximum and minimum. However, if you write out the full proof that the sets are equal, it becomes quickly evident that closed and open intervals are equally easy to sum. So, if someone changed to closed intervals and brushed over the proof of the sum, then I would tend to assume they haven't written out the proof of the sum of the two intervals, and are simply making assumptions! This may not cost you marks, but it would make me, as a marker, feel slightly less generous in giving out marks for the question (if it were a multi-part question with a loose marking scheme).
Apologies for the pettiness. The proof is good.
