1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Are we working in $\Bbb R$? $\endgroup$ – Robert Shore May 22 at 2:26
  • $\begingroup$ Are you working inside $\Bbb R $? What is $+$ here? $\endgroup$ – Thomas Shelby May 22 at 2:27
  • $\begingroup$ Sorry, I'll edit. Yes, we're in $\mathbb{R}$ $\endgroup$ – Bruno Reis May 22 at 2:30
  • $\begingroup$ Could you adjust your proof to work also when only $B$ is open? $\endgroup$ – Hagen von Eitzen May 22 at 2:51
  • $\begingroup$ @HagenvonEitzen Thank you for commenting. Idk, haven't thought about it since the hypothesis is that $A$ and $B$ are open. For you, what I've done is correct? Thanks $\endgroup$ – Bruno Reis May 22 at 2:55
2
$\begingroup$

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.

  1. 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?

  2. 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 $A + B$.

  3. 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.

  4. 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.

$\endgroup$
  • $\begingroup$ First of all, thanks a lot for your comments on my proof. (1) Totally agree with you. (2) Totally agree with you. (3) What do you mean by "brushed over" ? (4) Can you explain "brushed over the proof of the sum"? Anyway, I still agree with what you said. Thanks a lot for taking some time to help me here! I highly appreciate that :) $\endgroup$ – Bruno Reis May 22 at 4:56
  • $\begingroup$ I mean, you could prove the set equality$$I_{a/2} + I_{b/2} = [a + b - \varepsilon, a + b + \varepsilon]$$using a subset argument. Start by assuming $x \in I_{a/2} + I_{b/2}$, and see if you can conclude that $x$ belongs to the right hand side. Then, assume $x$ lies in the right hand side, and try to find $p \in I_{a/2}$ and $q \in I_{b/2}$ such that $x = p + q \in I_{a/2} + I_{b/2}$. You just stated the sum, rather than proving it. This potentially could be a source of marks lost in an assignment (depending on the rubric used). But, it can be ok to skip mechanical steps like this too. $\endgroup$ – Theo Bendit May 22 at 5:00
  • $\begingroup$ Ohhhh got what you mean by "brushed". When I formally did it, I prove it, but for sake of this forum post, I "brushed" it to make it more direct! In a test or assignment, I would do it with a little more rigor! Thanks anyway for the hints to improve it mate $\endgroup$ – Bruno Reis May 22 at 5:18
  • $\begingroup$ @BrunoReis Great! Then ignore points 3, 4. $\endgroup$ – Theo Bendit May 22 at 5:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.