37
$\begingroup$

Is there a counterexample for the claim in the question subject, that a sum of two closed sets in $\mathbb R$ is closed? If not, how can we prove it?

(By sum of sets $X+Y$ I mean the set of all sums $x+y$ where $x$ is in $X$ and $y$ is in $Y$)

Thanks!

$\endgroup$
  • $\begingroup$ Does this count as a duplicate of this? $\endgroup$ – leo Mar 25 '12 at 6:37
  • $\begingroup$ It's awfully similar, and it's probably quite straightforward to show that it's an equivalent question. However, I feel that the wording is different enough for it not to count as an exact duplicate. $\endgroup$ – user22805 Mar 25 '12 at 6:44
  • 4
    $\begingroup$ As David shows, the answer is no. However, the sum of a closed set and a compact set is closed. $\endgroup$ – Robert Israel Apr 15 '12 at 8:05
74
$\begingroup$

Consider the sets $A=\{ n\mid n=1,2,\ldots\}$ and $B=\{- n+{1\over n}\mid n= 2,3,\ldots\}$. Note that $0$ is not in the sum, but $1\over n$ is for each $n\ge2$.

$\endgroup$
  • 2
    $\begingroup$ Clever, thanks. So $0$ is a limit point that's not in $A+B$, right? $\endgroup$ – ro44 Mar 25 '12 at 3:26
  • 1
    $\begingroup$ @ro44 You're welcome. Yes to your question. $\endgroup$ – David Mitra Mar 25 '12 at 3:32
  • 1
    $\begingroup$ What if one of $A$ or $B$ is compact , will their sum be closed or compact? $\endgroup$ – Bhauryal Jan 23 '14 at 5:00
  • 2
    $\begingroup$ @NeerajBhauryal It will be closed but not necessarily compact. $\endgroup$ – David Mitra Jan 23 '14 at 9:02
  • 6
    $\begingroup$ Perhaps it's too trivial to mention, but if both are compact, then of course their sum will be compact (the continuous image of a compact set is compact). $\endgroup$ – Marcel Besixdouze Jul 26 '14 at 6:49
37
$\begingroup$

consider $\mathbb Z$ and $\sqrt 2 \mathbb Z$ both are closed but the sum is not...:) moreover it is dense on $\mathbb R$

$\endgroup$
1
$\begingroup$

The sum $E +F$ may fail to be closed even if $E$ and $F$ are closed. For instance, set $E = \{(x, y) \in \mathbb R^2 : y > 1/x\text{ and }x > 0\}$ and $F = \{(x, y) \in\mathbb R^2 : y > -1/x\text{ and }x < 0\}$

Then $E$ and $F$ are closed, but $$E + F = \{(x, y) \in \mathbb R^2 : y > 0\}$$ is not closed.

$\endgroup$
  • $\begingroup$ hassan: I've tried to edit your question (add latex formatting for better readability). Please, check whether I did not changed the meaning unintentionally. You can find more about writing math on this site e.g here and here. $\endgroup$ – Martin Sleziak Apr 15 '12 at 8:34
  • 4
    $\begingroup$ No, $E$ and $F$ are not closed; you want $y \ge 1/x$ etc. And the original question was about $\mathbb R$, not ${\mathbb R}^2$. $\endgroup$ – Robert Israel Apr 16 '12 at 7:31
1
$\begingroup$

Take $A=\{(a,0):a\in\mathbb{R}$ and $B=\{(b,\frac{1}{b}):b\in \mathbb{R}-\{0\}\}$. Then both $A,B\subset \mathbb{R}^2$ are closed. But $A+B=\{(a+b,\frac{1}{b}):a\in \mathbb{R},b\in \mathbb{R}-\{0\}\}.$The sequence $\{(0,\frac{1}{n})\}=\{(n-n,\frac{1}{n})\}\subset A+B$ but the limit $(0,0)$ which is not in the sum.

$\endgroup$

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.