Proving [0,1] Compact

Real Analysis · Axiom Academy

EXAMPLE Proving [0,1] is Compact Using sequential compactness and the Bolzano–Weierstrass theorem to prove the closed unit interval is compact. Prove that the closed unit interval [0,1] is compact . We work in with the usual metric, where a set is compact exactly when it is sequentially compact — so it suffices to show every sequence in [0,1] has a subsequence converging to a point of [0,1] . Because [0,1] is closed , the limit L of a convergent subsequence is forced to lie inside [0,1] . On the open interval (0,1) the sequence has its only limit at 0 , which is excluded — so (0,1) is not compact. Closedness is exactly what makes the difference. Nice work — you proved that [0,1] is compact by showing it is sequentially compact. The essential points: Sequential compactness: a set is sequentially compact when every sequence in it has a convergent subsequence whose limit is also in the set. Bolzano–Weierstrass is the engine: it turns boundedness into a convergent subsequence — every sequence in [0,1] is bounded, so a limit always exists. Closedness traps the limit: because and limits preserve , the limit satisfies , so . On the open interval (0,1) the limit can escape (e.g. ), which is why (0,1) is not compact. In a metric space (like ), sequential compactness is equivalent to compactness.

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