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Is this true?

Closed and bounded ⇒ compact

No — it is false.

The counterexample

Closed unit ball in $ℓ^{2} ($ or in C[0,1] with sup norm)

$e_{1}, e_{2}, \dots$ has no convergent subsequence since ‖ $e_{n} - e_{m}$ ‖ $= 2$ . Heine–Borel is $R^{n} -$ only.

The kind of mistake this is

Hypothesis dropped

A theorem applied without checking one of its preconditions.

Drill statements like this

compactness metric spaces

Others that fail the same way

From Metric Spaces › Compactness: open covers, sequential, Heine–Borel

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