Bolzano-Weierstrass Theorem

Real Analysis · Axiom Academy

LESSON The Bolzano-Weierstrass Theorem Every bounded sequence has a convergent subsequence — and a repeated bisection traps the limit point that proves it. Call a sequence (a_n) bounded when every term lives inside a single closed interval: there is some [a,b] with for all n . No term can escape — and yet there are infinitely many of them crowded into that finite stretch. Start with I_0=[a,b] , which holds every term. Cut it in half. At least one half still contains infinitely many terms (two halves can't split an infinite set into two finite ones) — keep that half and call it I_1 . Halve again, keep an infinite half I_2 , and repeat forever. nested intervals pin down one point L 3. Building the Convergent Subsequence The trap pins down L , but a subsequence is a choice of terms . Walk down the nested intervals and pick one term out of each, always choosing a later index than before — this is possible precisely because every I_k holds infinitely many terms, so an unused index is always waiting. Since a_ n_k and L both sit inside I_k , their gap is at most the interval's width: . So the chosen terms march in on L , and . The theorem is proved. Take a_n=(-1)^n : bounded in [-1,1] but divergent. Bolzano-Weierstrass only promises a convergent subsequence, and here there are two — the even-index terms and the odd-index terms . A limit point need not be the limit of the whole sequence.

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