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 a way that is understandable to the readers.

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Colby Messinger answered 1 week ago

If a set K that is a subset of the real numbers is closed and bounded,...

If a set K that is a subset of the real numbers is closed and bounded, then it is compact.

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

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