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Let $A$ and $B$ be closed subsets of $[0, \infty)$ then is $A+B$ closed?, I know that the answer is negative when we have $\mathbb{R}$ instead of $[0, \infty)$, but every counterexample in that case exploits the fact that one of A or B can contain negative values.

Note: Using answers given below this can be easily generalised,

If $A$ and $B$ are closed subsets of $\mathbb{R}$ which are bounded below (or above) then $A+B$ is closed.

user276011
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2 Answers2

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Suppose we have $a_n+b_n\to x$ with $a_n\in A$, $b_n\in B$. Then all $a_n$ are in the compact interval $[0,x]$, hence there is a convergent subsequence $a_{n_k}\to a\in A$. Then $b_{n_k}\to x-a\in B$ and so $x=a+b\in A+B$.

Hagen von Eitzen
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    Is there a typo. here, or am I missing something? Say $A=B=[0,2]$ with $a_n=1+\frac1n$ and $ b_n=0+\frac1n$ and $a_n+b_n\to x=1$. That does not mean all $a_n$ are in the (compact) interval $[0,x]=[0,1]$!? – J. W. Tanner Nov 24 '19 at 21:00
  • @J.W.Tanner . You are right. It is not true that all $a_n$ must belong to $[0,x]$.... But $(a_n)_n$ must be a bounded sequence so it has a convergent sub-sequence. – DanielWainfleet Nov 25 '19 at 05:23
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    Thank you very much @DanielWainfleet; that makes sense – J. W. Tanner Nov 25 '19 at 05:27
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Yes, it is closed. Suppose otherwise. That is, suppose that there are sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ of $A$ and $B$ respectively such that the sequence $(a_n+b_n)_{n\in\mathbb N}$ converges to some $x\in\mathbb R\setminus(A+B)$. Then the sequence $(a_n+b_n)_{n\in\mathbb N}$ is bounded and therefore both sequences $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are bounded too. Take $K>0$ such that$$(\forall n\in\mathbb N):a_n,b_n\in[0,K].$$Then $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ are sequences of elements of $A\cap[0,K]$ and of $B\cap[0,K]$ respectively, which are compact sets. But the sum of two compact subsets of $\mathbb R$ is compact and therefore closed.

José Carlos Santos
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