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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!

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  • $\begingroup$ Does this count as a duplicate of this? $\endgroup$
    – leo
    Mar 25, 2012 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, 2012 at 6:44
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    $\begingroup$ As David shows, the answer is no. However, the sum of a closed set and a compact set is closed. $\endgroup$ Apr 15, 2012 at 8:05

5 Answers 5

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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 $n\ge2$.

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    $\begingroup$ Clever, thanks. So $0$ is a limit point that's not in $A+B$, right? $\endgroup$
    – ro44
    Mar 25, 2012 at 3:26
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    $\begingroup$ @ro44 You're welcome. Yes to your question. $\endgroup$ Mar 25, 2012 at 3:32
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    $\begingroup$ What if one of $A$ or $B$ is compact , will their sum be closed or compact? $\endgroup$
    – Mathronaut
    Jan 23, 2014 at 5:00
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    $\begingroup$ @NeerajBhauryal It will be closed but not necessarily compact. $\endgroup$ Jan 23, 2014 at 9:02
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    $\begingroup$ 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). $\endgroup$ Jul 26, 2014 at 6:49
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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 $\mathbb R$

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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.

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    $\begingroup$ 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? $\endgroup$
    – luki luk
    Aug 26, 2022 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.

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    $\begingroup$ the question was about $\mathbb R, $ not $\mathbb R^2$ $\endgroup$ Nov 25, 2019 at 5:31
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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 $F = \{(x, y) \in\mathbb R^2 : y > -1/x\text{ and }x < 0\}$

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

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  • $\begingroup$ 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. $\endgroup$ Apr 15, 2012 at 8:34
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    $\begingroup$ No, $E$ and $F$ are not closed; you want $y \ge 1/x$ etc. And the original question was about $\mathbb R$, not ${\mathbb R}^2$. $\endgroup$ Apr 16, 2012 at 7:31

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