- Calculus and Analysis
Proving for Compactness and Convergence of Sequences
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
Prove that [0,1]^n is compact for any number (n e N) by using theorem 2. (see attached file)
Theorem 2: A subset S of a metric space X is compact if, and only if, every sequence is S has a subsequence that converges to a point in S.
© BrainMass Inc. brainmass.com October 10, 2019, 7:04 am ad1c9bdddf
The solution proves [0,1]^n is compact through a given theorem.