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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!

Alex Becker
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ro44
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  • Does this count as a duplicate of this? – leo Mar 25 '12 at 06:37
  • 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. –  Mar 25 '12 at 06:44
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    As David shows, the answer is no. However, the sum of a closed set and a compact set is closed. – Robert Israel Apr 15 '12 at 08:05

5 Answers5

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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 $n\ge2$.

David Mitra
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    Clever, thanks. So $0$ is a limit point that's not in $A+B$, right? – ro44 Mar 25 '12 at 03:26
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    @ro44 You're welcome. Yes to your question. – David Mitra Mar 25 '12 at 03:32
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    What if one of $A$ or $B$ is compact , will their sum be closed or compact? – Mathronaut Jan 23 '14 at 05:00
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    @NeerajBhauryal It will be closed but not necessarily compact. – David Mitra Jan 23 '14 at 09:02
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    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). – Marcel Besixdouze Jul 26 '14 at 06:49
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    What happens if $A$ and $B$ are closed subsets of $[0, \infty)$? Are there such $A$ and $B$ in this case such that $A+B$ is not closed? – Prince Kumar Nov 27 '16 at 08:37
  • @PrinceKumar I don't think so. If $(a_n+b_n)$ converges to $c$, then by comparison, $(a_{n }) $ convergies to some $a\in A$ and $(b_{n }) $ converges to some $b\in B$. But then $c=a+b\in A+B$. – David Mitra Nov 27 '16 at 09:00
  • @DavidMitra I don't understand what you used to deduce the convergence of $(a_n)$ and $(b_n)$ from the convergence of $(a_n+b_n)$. The only way this case is different from the one is question is that we can't choose $(a_n)$ or $(b_n)$ to be negative. Will that be sufficient to deduce the convergence? – Prince Kumar Nov 27 '16 at 09:34
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    @PrinceKumar Oh, sorry, what I have above is off... But, you can choose a "common" subsequece of $(a_n)$ and $(b_n)$, both of which converge (both are bounded and non-negative). – David Mitra Nov 27 '16 at 09:48
  • @DavidMitra They need not have any common subsequence. I will try to find the proof. All I am getting right now is an intuition. :) – Prince Kumar Nov 27 '16 at 10:04
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    @PrinceKumar Yes they do. Take a convergent subsequence $(a_{n_k})k$ of $(a_n)$. Then take a convergent subsequence of $(b{n_k})_k$. – David Mitra Nov 27 '16 at 10:08
  • @DavidMitra Oh, you meant common indices. I thought you are talking about common terms (values) which will not be true for two arbitrary sequences. It will work. Thanks a lot! – Prince Kumar Nov 27 '16 at 10:15
  • @PrinceKumar Yes, sorry for being unclear... – David Mitra Nov 27 '16 at 10:18
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    How to show B is closed in your example? – Myshkin Sep 03 '17 at 12:09
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    @miosaki Notice that $1/n - n$ is decreasing, so if you take the complement of B you get the union of $(0,\infty), (1/2-2,0), (1/3-3, 1/2-2), \ldots$ and each of them are open, so their union is open – Arpit Saxena Feb 04 '20 at 08:18
  • Is it possible to unfold what went through your mind before this counterexample? What are the thought processes? What factored during the construction? A valid counterexample is surely useful but learning from a greater mind matters more to me. – zony_miu May 24 '21 at 01:33
  • The set B can also reach the point 1. – Mariana Jun 14 '21 at 14:14
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consider $\mathbb Z$ and $\sqrt 2 \mathbb Z$ both are closed but the sum is not...:) moreover it is dense on $\mathbb R$

Ross Millikan
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sam
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It's worth mentioning that :

if one is closed + bounded, another one is closed,then the addition is closed

Since closedness can be charaterized by sequence in $\Bbb{R}^n$,if $(x_n) \in A+B$ we need to show limit of the convergence sequence still lies in it.assume $A$ is compact $B $ is closed.

Since $x_n= a_n +b_n \to x$,compactness implies sequential compactness hence $a_{n_k} \to a\in A$ for some subsequence. now $x_{n_k} \to x$ which means subsequence $b_{n_k}\to x-a$ converge,since $B$ is closed,$x-a \in B$ ,hence $x = a+b \in A+B$,which means the sum is closed.

yi li
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    is it also possible to prove that C = {a*b | $a \in A$ and $b \in B$}is closed if A and B are closed by using this method? – luki luk Aug 26 '22 at 17:34
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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.

RAM_3R
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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 $F = \{(x, y) \in\mathbb R^2 : y > -1/x\text{ and }x &lt 0\}$

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

Martin Sleziak
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hassan
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  • 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. – Martin Sleziak Apr 15 '12 at 08:34
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    No, $E$ and $F$ are not closed; you want $y \ge 1/x$ etc. And the original question was about $\mathbb R$, not ${\mathbb R}^2$. – Robert Israel Apr 16 '12 at 07:31