Let (an) be a real sequence in the standard metric. Prove that (an) is bounded if...

Let (an) be a real sequence in the standard metric. Prove that (an) is bounded if and only if every subsequence of (an) has a convergent subsequence.

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Colby Messinger answered 2 years ago

Let {an} be a bounded sequence. In this question, you will prove that there exists a...

Let {an} be a bounded sequence. In this question, you will prove that there exists a convergent subsequence. Define a crest of the sequence to be a term am that is greater than all subsequent terms. That is, am > an for all n > m (a) Suppose {an} has infinitely many crests. Prove that the crests form a convergent subsequence. (b) Suppose {an} has only finitely many crests. Let an1 be a term with no subsequent crests. Construct a...

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Let A be a subset of all Real Numbers. Prove that A is closed and bounded...

Let A be a subset of all Real Numbers. Prove that A is closed and bounded (I.e. compact) if and only if every sequence of numbers from A has a subsequence that converges to a point in A. Given it is an if and only if I know we need to do a forward and backwards proof. For the backwards proof I was thinking of approaching it via contrapositive, but I am having a hard time writing the proof in...

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Let (an) be a real sequence in the standard metric. Prove that (an) is bounded if...

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