(Cauchy’s Criterion)

A sequence of real numbers is convergent if, and only if, it is a Cauchy sequence.

Question

(Cauchy’s Criterion)

A sequence of real numbers is convergent if, and only if, it is a Cauchy sequence.

Accepted Answer

Let [latex](x_{n})[/latex] be a Cauchy sequence and [latex]A = \left\{x_{n} : n ∈ \mathbb{N}\right\}.[/latex] We have two cases:

(i) The set A is finite. In this case one of the terms of the sequence, say x, is repeated infinitely often. We shall prove that [latex]x_{n} → x.[/latex] Given ε > 0, there is an integer N such that

[latex]m, n ≥ N ⇒ |x_{n} − x_{m}| < ε.[/latex]

Since the term x is repeated infinitely often in the sequence, there is an m > N such that [latex]x_{m} = x.[/latex] Hence

[latex]|x_{n} − x| = |x_{n} − x_{m}| < ε[/latex] for all n ≥ N,

which just means [latex]x_{n} → x.[/latex]

(ii) A is infinite. In view of Lemma 3.1, the set A is bounded and therefore, by the Bolzano-Weierstrass Theorem, it has a cluster point x. We shall prove that [latex]x_{n} → x.[/latex] Given ε > 0, there is an integer N such that

[latex]|x_{n} − x_{m}| < ε[/latex] for all n, m ≥ N.

Since [latex]x ∈ \hat{A},[/latex] the interval (x − ε, x + ε) contains an infinite number of terms of the sequence [latex](x_{n}).[/latex] Hence there is an m ≥ N such that [latex]x_{m} ∈ (x − ε, x + ε),[/latex] i.e., such that [latex]|x_{m} − x| < ε.[/latex] Now, if n ≥ N, then

[latex]|x_{n} − x| ≤ |x_{n} − x_{m}| + |x_{m} − x| < ε + ε = 2ε,[/latex]

which proves [latex]x_{n} → x.[/latex]

Question 3.T.9: (Cauchy’s Criterion) A sequence of real numbers is convergen...

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