Proving Completeness

Topology · Axiom Academy

Unit 7 - Metric Spaces: Three Essential Completeness Proofs 1 The Real Numbers ℝ are Complete Let be a Cauchy sequence in ℝ. By definition, for , there exists such that: In particular, for , all terms beyond some lie within distance 1 of . Therefore: So the sequence is bounded by . Since is bounded, by the Bolzano-Weierstrass Theorem , it has a convergent subsequence. Let be a subsequence that converges to some . Let . We need to show there exists such that for all . Since is Cauchy, choose so that: Since converges to , choose large enough so that and: For any , by the triangle inequality: Therefore, every Cauchy sequence in ℝ converges, so ℝ is complete. 2 The Closed Interval [0,1] is Complete Let be a Cauchy sequence in [0,1]. This means: For all , there exists such that for all Since is also a Cauchy sequence in ℝ, and we know ℝ is complete (Example 1), the sequence must converge to some limit : We must show that . Since for all , and limits preserve inequalities: Therefore, , and the sequence converges to a point in [0,1]. Every Cauchy sequence in [0,1] converges to a limit in [0,1], so [0,1] is complete. 3 The Open Interval (0,1) is NOT Complete Consider the sequence defined by: Clearly, for all , so this sequence lives entirely within (0,1). Given , choose . Then for all : Therefore, is a Cauchy sequence in (0,1). However, in ℝ, this sequence converges to: But ! The limit exists in ℝ but lies outside the open interval (0,1).

This is the written version of the interactive lesson above. See the full Topology course.