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Defining for $A,B \subseteq \Bbb R$, $A+B = \{ a+b : a \in A,b \in B \}$. If A and B are two bounded closed subsets, must A+B be closed?

The question arose from the previous case where A and B were not given to be bounded and the answer came false with the example of $A = \{ 1,2,...\}\ and \ B = \{ 1-1,1-(1/2),1 - (1/3),... \}$

However, how do we prove/counter in this case?

EDIT : - I had written 'bounded' instead of 'closed'

EDIT 2 : - I saw that there are proofs available using properties of arbitrary metric spaces but since I haven't studied them, they are beyond me. I would like to know if a proof is possible using the properties of $\Bbb R$

Lelouch
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  • @Adam Hughes - Yes that does answer my question but that proof is beyond me since I haven't studied Metric Spaces yet. I have added the 'Alternate proof' tag, is it still a duplicate? – Lelouch Dec 10 '16 at 18:24
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    @Lelouch The statement strongly depends on compactness. – egreg Dec 10 '16 at 18:34
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    Yes, that's the way. The answer http://math.stackexchange.com/a/1093639/62967 to the linked question tells you how. – egreg Dec 10 '16 at 18:42
  • @egreg - Thanks, I was just trying to write something similar after your comment. The first paragraph is all that's needed for $\Bbb R$, right? – Lelouch Dec 10 '16 at 18:44

1 Answers1

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HINT: Suppose that $x$ is a limit point of $A+B$. Then there is a sequence $\sigma=\langle a_n+b_n:n\in\Bbb N\rangle$ such that $a_n\in A$ and $b_n\in B$ for each $n\in\Bbb N$, and $\sigma$ converges to $x$. The set $A$ is bounded, so the sequence $\langle a_n:n\in\Bbb N\rangle$ has a convergent subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$. Let $a$ be the limit of this subsequence; $A$ is also closed, so $a\in A$.

The subsequence $\sigma'=\langle a_{n_k}+b_{n_k}:k\in\Bbb N\rangle$ of $\sigma$ converges to $x$. (Why?) Now you can proceed in either of two ways.

  • You can use the fact that $\langle a_{n_k}:k\in\Bbb N\rangle$ converges to $a$ to show that $\langle b_{n_k}:k\in\Bbb N\rangle$ converges to $x-a$, and then use the fact that $B$ is closed to show that $x-a\in B$. Then $$x=a+(x-a)\in A+B\;,$$ as desired.

  • Or you can apply the original idea to the sequence $\langle b_{n_k}:k\in\Bbb N\rangle$ and argue that this sequence in the bounded set $B$ must have a convergent subsequence $\langle b_{n_{k_i}}:i\in\Bbb N\rangle$, and that since $B$ is closed, the limit $b$ of this subsequence must be in $B$. Then you can show that $$\left\langle a_{n_{k_i}}+b_{n_{k_i}}:i\in\Bbb N\right\rangle$$ is a subsequence of $\sigma$ converging to $a+b$ and hence that $x=a+b\in A+B$.

Brian M. Scott
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