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Let {$x_n$} be a Cauchy sequence of real numbers. The proof I am reading uses the completeness axiom to prove the hypothesis. The proof starts with a lemma stating that this sequence must be bounded, whose proof I understand. The next part of the proof is written in the textbook exactly as I write below:

Since {$x_n$} is bounded by the lemma, there is a greatest lower bound $$b_n=g.l.b.\{x_n,x_{n+1},x_{n+2},...\}.$$ Then {$b_n$} is an increasing sequence, bounded.

Then the proof goes on to show that the least upper bound of sequence {$b_n$} is the limit of the Cauchy sequence. I don't understand how {$b_n$} is increasing and bounded.

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2 Answers 2

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Any lower bound for $\{x_n,x_{n+1},...\}$ is a lower bound for $\{x_{n+1},x_{n+2},...\}$. In particular g.l.b.$\{x_n,x_{n+1},...\}$ is a lower bound for $\{x_{n+1},x_{n+2},...\}$. Hence $b_n \leq b_{n+1}$. If $c\leq x_n \leq d$ for all $n$ then $c$ is a lower bound for $\{x_n,x_{n+1},...\}$ so $c \leq b_n $ for all $n$. On the other hand $b_n \leq x_n \leq d$, so $(b_n)$ is bounded.

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$(b_{n})$ is increasing sequence as: Let $n\in\mathbb{N}$ then $$g.l.b.(x_{n},x_{n+1},x_{n+2},\dots)\leq g.l.b.(x_{n+1},x_{n+2},\dots)$$ $$b_{n}\leq b_{n+1},\forall \ n\in\mathbb{N}$$

$(b_{n})$ is bounded sequence as: Since $(x_{n})$ is bounded, therefore $\exists\ M>0$ such that $$|x_{n}|\leq M,\ \forall \ n\in \mathbb{N}$$ $\because M$ act as an upper bound for every set of the form $(x_{n},x_{n+1},x_{n+2},\dots)$ $$\therefore b_{n}\leq M,\forall \ n\in\mathbb{N}$$

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