I'm having trouble understanding the proof of the completeness of $\mathbb{R}$ with respect to Cauchy sequences.
5 Theorem $\mathbb{R}$ is complete with respect to Cauchy sequences in the sense that if $(a_n)$ is a sequence of real numbers which obeys a Cauchy condition then it converges to a limit in $\mathbb{R}$.
Proof. First we show that $(a_n)$ is bounded. Taking $\varepsilon$ = 1 in the Cauchy condition implies there is an N such that for all n,k ≥ N we have |$a_n$−$a_k$| < 1. Take K large enough that −K ≤ $a_1$,...,$a_N$ ≤ K. Set M=K +1. Then for all n we have −M<$a_n$<M, which shows that the sequence is bounded. Define a set X as X ={x∈R:∃ infinitely many n such that $a_n$ ≥ x}. −M ∈X since for all n we have $a_n$ > −M, while M/ ∈ X since no $x_n$ is ≥ M.Thus X is a nonempty subset of R which is bounded above by M. The least upper bound property applies to X and we have b =l.u.b.X with −M ≤ b ≤ M.
We claim that an converges to b as n →∞.Given >0 we must show there is an N such that for all n ≥ N we have |$a_n$ −b|<$\varepsilon$ Since $(a_n)$ is Cauchy and $\varepsilon$/2 is positive there does exist an N such that if n,k ≥ N then |$a_n$ −$a_k$| < $\varepsilon$/2 Since b −$\varepsilon$/2 is less than b it is not an upper bound for X,so there is x ∈ X with b − $\varepsilon$/2 ≤ x. For infinitely many n we have $a_n$ ≥ x.Sinceb b+$\varepsilon$/2 >b it does not belong to X, and therefore for only finitely many n do we have an $a_n$>b+ $\varepsilon$/2. Thus, for infinitely many n we have b − $\varepsilon$/2 ≤ x ≤ $a_n$ ≤ b+$\varepsilon$/2 Since there are infinitely many of these n there are infinitely many that are ≥ N. Pick one, say $a_n0$ with $n_0$ ≥ N and b− $\varepsilon$/2 ≤ $a_n0$ ≤ b+ $\varepsilon$/2. Then for all n ≥ N we have |$a_n$ −b|≤|$a_n$−$a_n0$ |+|$a_n0$ −b| < 2 = which completes the verification that $(a_n)$ converges.
I understand (I think) everything in this proof except for one thing
So my question is: There are three parts to this proof (1) proving that $a_n$ is bounded, (2) proving that it has a least upper bound and (3) showing that the sequence converges to that least upper bound. In the final part,
We claim that an converges to b as n →∞.Given >0 we must show there is an N such that for all n ≥ N we have |$a_n$ −b| <$\varepsilon$ Since $(a_n)$ is Cauchy and $\varepsilon$/2 is positive there does exist an N such that if n,k ≥ N then |$a_n$ −$a_k$| < $\varepsilon$/2
Why did the author simply not set |$a_n$−b|<2|$a_n$−$a_k$|<$\varepsilon$/2 + $\varepsilon$/2=$\varepsilon$?