Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let A and B be subsets of R. Define C = {a + b : a belongs to A, b belongs to B}. Pick out the true statements. (a) C is closed if A and B are closed. (b) C is closed if A is closed and B is compact. (c) C is compact if A is closed and B is compact.

i think a is not true but cant find any counter example. no idea for b and c

share|improve this question
You might want to see en.wikipedia.org/wiki/Interval_arithmetic – Jayesh Badwaik Sep 1 '12 at 3:55
False, True, False. – Kevin Sep 1 '12 at 4:00
@Kevin: answering in a comment doesn't really help. – Ben Millwood Sep 1 '12 at 9:43

2 Answers

For the first problem, let $A$ be the set of negative integers. Let $B$ be the set of all numbers of the form $n+\frac{1}{2^n}$, where $n$ ranges over the positive integers. Then $A$ and $B$ are closed. The set $A+B$ contains numbers arbitrarily close to $0$, but does not contain $0$. So $A+B$ is not closed.

There is a very easy counterexample for the third assertion. Let $A$ be the integers, and $B$ almost anything.

share|improve this answer

HINTS:

(a) What if $A=\left\{n+\frac1{n+1}:n\in\Bbb Z^+\right\}$ and $B=\Bbb Z$? Is $0\in A+B$? Is $0$ a limit point of $A+B$?

(b) Let $\langle a_n+b_n:n\in\Bbb N\rangle$ be a sequence in $A+B$ converging to some real number $x$, where each $a_n\in A$ and $b_n\in B$. $B$ is compact, so $\langle b_n:n\in\Bbb N\rangle$ has a convergent subsequence $\langle b_{n_k}:k\in\Bbb N\rangle$. Let $b=\lim_{k\in\Bbb N}b_{n_k}$; then $b\in B$, and $\langle a_{n_k}+b_{n_k}:k\in\Bbb N\rangle$ converges to $x$. Then $\lim_{k\in\Bbb N}a_{n_k}=x-b$. (Why?) Is this limit in $A$?

(c) What if $A=\Bbb Z$ and $C=\{0\}$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.