Closed Intervals are Compact

Topology · Axiom Academy

EXAMPLE Closed Intervals are Compact The classic supremum proof that [a,b] satisfies the Heine-Borel property Let be a closed bounded interval. If is any open cover of , then there exists a finite subcover. The strategy: We'll use proof by contradiction. Assume no finite subcover exists, then construct a set of points that can be covered by finitely many sets, and examine its supremum. What happens at the supremum ? We'll show that assuming leads to contradiction. Let . Since is in , it must be in some open set from our cover. This open set contains an entire interval around . We must show that our assumption (no finite subcover exists) leads to contradiction in all cases. We've shown that if , we can extend the finite cover beyond . The only remaining case is . This completes the proof. Completeness is Essential: We needed sup(A) to exist, which relies on the completeness of the real numbers. Open Sets Give Breathing Room: Because U is open, we get an interval around c, allowing us to extend the cover. Supremum Forces a Decision: Either c < b (and we extend, contradiction) or c = b (and we're done). The Method Generalizes: This supremum approach works for proving many other properties of closed bounded sets. Heine-Borel Theorem: This result is one direction of the famous theorem: in , compact if and only if closed and bounded.

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