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If it is, how would one go about proving it...?

Daniel.

- Thread starter dextercioby
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If it is, how would one go about proving it...?

Daniel.

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George Jones

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I found a proof in Kreyszig.dextercioby said:

If it is, how would one go about proving it...?

Daniel.

Every Banach space is a metric space, and in a compact subset of a metric space, every sequence has a convergent subsequence.

Kreyszig assumes that the closed unit ball of an infinite-dimensional Banach space is compact. He then uses Riesz's Lemma (which isn't the Riesz Representation Theorem) to construct a sequence that doesn't have convergent subsequence.

Conclusion: the closed unit ball of a Banach space is compact if and only if the Banach space is finite-dimensional.

Regards,

George

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Thank you, George.

Daniel.

Daniel.

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