If C and D are closed and C is bounded, then C+D is closed.

Question

Let $C,D\subset R^n$ are both closed. And C is bounded. Prove $C+D $ is closed.

I am having a hard time understanding why C being bounded matters. Also, I have no idea how to start.

I would really appreciate your help.

You may try using the sequencial criterion to prove that $C+D$ is closed : take a converging sequence in $C+D$ and prove that the limit stays in $C+D$. You will need to use boundedness of $C$ somewhere. — Suzet, Feb 18 '20 at 05:32
First consider examples where $C+D$ is not closed for $C,D$ closed. There are plenty on this site alone. — Henno Brandsma, Feb 18 '20 at 07:12
One amusing example with $n=1$ where $C,D$ are both closed and unbounded: Let $\Bbb Q\cap [0,1)={q_n: n\in \Bbb N}$ and let $C={n+q_n: n\in \Bbb N}.$ Let $D=\Bbb Z.$ Then $C+D=\Bbb Q.$ — DanielWainfleet, Feb 18 '20 at 14:54

score 1 · Answer 1 · answered Feb 18 '20 at 05:38

Consider a Cauchy sequence $ \{ x_n \} $ in $ C + D $. Our goal is to show that its limit is in $ C + D $. By definition, $ x_n = c_n + d_n $. Now by the boundedness of $ C $... do you see what you can accomplish? If you still need help, I'll finish this in the morning.

Kavi Rama Murthy · Answer 2 · 2020-02-18T06:13:43.450

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In a closed and bounded set any sequence has a convergent subsequence. Suppose $x_n=c_n+d_n \to z$ with $c_n \in C$ and $d_n \in D$ for all $n$. There exists $(n_k)$ such that $c_{n_k} \to$ some point $x \in C$. Since $c_{n_k}+d_{n_k} \to z$ we see that $d_{n_k} \to z-x$. Since $D$ is closed, $z-x \in D$. Hence $z =x+(z-x) \in C+D$.

I have used Bolzano -Weierstrass Theorem in the proof. See https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem

edited Feb 18 '20 at 06:13

answered Feb 18 '20 at 05:35

Kavi Rama Murthy

311,013

Thank you. However, where do we need Boundedness? – measurehell Feb 18 '20 at 06:04
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If $C$ is not bounded the the subsequence $(c_{n_k})$ may not exist. [ For example $C$ could be ${1,2,3,...}$]. @measurehell – Kavi Rama Murthy Feb 18 '20 at 06:07
Also, if $c_{2n}=1=d_{2k+1}$ and $c_{2n+1}=-1=d_{2n}$, we get $x_n=0$. However, $c_n $ and $d_n$ do not converge. How do we cover this kind of special case? – measurehell Feb 18 '20 at 06:09
Thank you for your comment. Would you please explain why boundedness is related to the subsequence? or would you please let me know some internet link that will help me understand it? – measurehell Feb 18 '20 at 06:11
You do to the subsequences $(c_{2n})$ and $(d_{2n})$ in this case, To prove that $C+D$ is closed you need not use the entire sequence $(c_n)$ and $(d_2)$. – Kavi Rama Murthy Feb 18 '20 at 06:11
@measurehell See my revised answer . – Kavi Rama Murthy Feb 18 '20 at 06:14

If C and D are closed and C is bounded, then C+D is closed.

2 Answers2