# Problem understanding a proof for the existence of limit of a real Cauchy sequence.

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.

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.
$$(b_{n})$$ 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 $$(x_{n},x_{n+1},x_{n+2},\dots)$$ $$\therefore b_{n}\leq M,\forall \ n\in\mathbb{N}$$