Explore BrainMass

Convergent Sequences and Subsequences, Compact Set and Accumulation Points

This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!

Prove that a set A, a subset of the real numbers, is compact if and only if every sequence {an} where an is in A for all n, has a convergent subsequence converging to a point in A.

For the forward direction, I know that a compact set is closed and bounded, thus every sequence in A is bounded, and so has a convergent subsequence. Also, since A is closed, it contains all of it's accumulation points, and a point s is an accumulation point iff there exists some sequence {bn} in A a such that bn is not = s and {bn} --> s. Is this enough for this proof?

I also need help proving the other direction. Any help will be greatly appreciated. Thanks.

© BrainMass Inc. brainmass.com October 16, 2018, 5:26 pm ad1c9bdddf

Solution Preview

"=>" If A is compact, then A is closed and bounded. So every sequence {a_n} in A is bounded and thus it has a convergent subsequence. Since A is closed, then every accumuation point of A is in A. So the limit of the convergent subsequence is in A. We are done.
"<=" If every sequence {a_n} in A ...

Solution Summary

Convergent Sequences and Subsequences, Compact Set and Accumulation Points are investigated.