Trouble with the proof of Cauchy convergence criterion While reading the Analysis 1 textbook by Vladimir A. Zorich I encountered a proof which has this one conclusion I fail to understand.
The theorem and proof :
(Cauchy’s convergence criterion) A numerical sequence converges if and only if it is a Cauchy sequence.
Proof.
$\implies$:(I skipped this part of the proof since I have no issues with it.)
$\impliedby$:
Let ${x_k}$ be a fundamental sequence. Given $\epsilon > 0$, we find an index $N$ such that $|x_m − x_k| < \frac{\epsilon}{3}$ when $m ≥ N$ and $k ≥ N$. Fixing $m = N$, we find that for any $k >N$
$$x_N − \frac{\epsilon}{3}<x_k <x_N + \frac{\epsilon}{3}\ \text{,} \ \ \ \ \ \text{(3.1)}$$
but since only a finite number of terms of the sequence have indices not larger than $N$, we have shown that a fundamental sequence is bounded.
$$\text{For}\ n \in \mathbb{N}\ \text{we now set } a_n := \inf_{k≥n} x_k ,\ \text{and }\ b_n := \sup_{k≥n} x_k \ \text{.}$$
It is clear from these definitions that $a_n ≤ a_{n+1} ≤ b_{n+1} ≤ b_n$ (since the greatest lower bound does not decrease and the least upper bound does not increase when we pass to a smaller set). By the nested interval principle, there is a point A common to all of the closed intervals $[a_n, b_n]$. Since
$$a_n ≤ A ≤ b_n$$
for any $n \in \mathbb{N}$ and
$$a_n = \inf_{k≥n} x_k ≤ x_k ≤ \sup_{k≥n} x_k = b_n$$
for $k ≥ n$, it follows that
$$|A − x_k| ≤ b_n − a_n\ \text{.}\ \ \ \ \ \text{(3.2)}$$
But it follows from Eq. $\text{(3.1)}$ that
$$\underbrace{x_N − \frac{\epsilon}{3}≤ \inf_{k≥n} x_k = a_n ≤ b_n = \sup_{k≥n} x_k ≤ x_N + \frac{\epsilon}{3}}_{\text{The problematic part}}$$
for $n>N$, and therefore
$$b_n − a_n ≤ \frac{2\epsilon}{3} < \epsilon \ \ \ \ \ \text{(3.3)}$$
for $n>m$. Comparing Eqs. $\text{(3.2)}$ and $\text{(3.3)}$, we find that $|A −x_k| < \epsilon$, for any $k > N$, and we have proved that $\lim_{k \to \infty}x_k = A$.
End of proof.

The underbraced part doesn't make sense to me, because from the stated it could happen that $$x_N − \frac{\epsilon}{3} = a_n $$ and since $a_n≤x_k$ it is possible that $a_n=x_k$ and if those equalities hold, then $x_N − \frac{\epsilon}{3} = a_n=x_k$, but this contradicts what was stated before in $\text{(3.1)}$, $x_N − \frac{\epsilon}{3}<x_k $.
Why does the problematic part hold despite this ?
Thanks
 A: If $x_N − \frac{\epsilon}{3}<x_k <x_N + \frac{\epsilon}{3}$ holds for all $k>N$, then $ x_N-\frac{\epsilon}{3}≤ \inf_{k≥N} x_k$ and $ \sup_{k≥N} x_k ≤ x_N + \frac{\epsilon}{3}.$ Combine these two, to get the underbraced inequality and then the rest of the proof go through. Note, you hve used the generic $n$, which is not quite right. The inequalities follow one after the other, for $k>N$, which you fixed previously.
Maybe a more detailed proof will help with the ideas:
$(1).$ Since $(x_n)$ is Cauchy, it is bounded. And it's much easier to see this simply by taking $\epsilon=1$ and finding an integer $N$ such that $n,m>N\Rightarrow |x_n-x_m|<1$. Then, $|x_n|<\max\{|x_1|,\cdots |x_N|,|x_{N+1}|+1\}.$
$(2).$ Since $(x_n)$ is bounded, it has a convergent subsequence (I will give a proof of this at the end). So,
$(3).$ Let $x_{n_k}\to x$. Then, there is an integer $K$ such that if $k>K,\ |x-x_{n_k}|<\epsilon/2$. And there is an integer $N$ such that if  $n>N, |x_m-x_n|<\epsilon/2.$ Now, choose an integer $l>K$ such that $n_l>N$. Then, if $n>N, |x_n-x|\le |x-x_{n_l}|+|x_n-x_{n_l}|<\epsilon.$
Proof of $(2):$ Let $(x_n)$ be a bounded sequence. Then there exists some $M > 0$ such that $|x_n| \le M$ for all integers $n$. Bisect the interval $[−M, M]$ into two closed intervals of equal length. One of these intervals must contain infinitely many $x_n.$ Let $I_1$ be that interval, and choose $x_{n_1}\in I_1$. Now, bisect $I_1$ into two closed intervals. Let $I_2$ be the subinterval of $I_1$ that contans infinitely many $x_n$, and choose one, $x_{n_2}\neq x_{n_1}$ such that $n_2>n_1$ (why is this possible?). In general, having constructed $\{I_j:I_j\subset I_{j-1},\ 1\le j\le k-1\}$, bisect $I_{k−1}$ into two closed intervals, one of which must contain infinitely many terms of $(x_n).$ Let $I_k$ be this closed interval, and choose $x_{n_k}\in I_k$ such that $n_k > n_{k−1}.$ Therefore, the induction proceeds and we obtain a subsequence $(x_{n_k})$ and  a sequence of nested intervals $\{I_k\}_k$ whose diameters $|I_k|=M\cdot \left(\frac{1}{2}\right)^{k-1}$ tend to $0$ as $k\to \infty.$ This fact, and the Nested Interval Property, imply that the intersection $\bigcap I_k$ contains exactly one point $x$. Now let $\epsilon>0$ and choose $K$ such that $M\cdot \left(\frac{1}{2}\right)^{K-1}<\epsilon.$  Then, if $k>K,\ x_{n_k}\in I_k$ and $|I_k|<\epsilon.$ But, $x$ is contained in $\textit{every}\ I_k$ so $|x_{n_k}-x|<\epsilon.$ It follows that $x_{n_k}\to x.$
A: I think it is better to extract out the real issue (which is bothering you) from all the other irrelevant details given in the proof.
Let $M=x_N-(\epsilon /3)$ and then we are given that $$x_k>M\tag{1}$$ for all $k\geq N$. This clearly implies that $$a_N=\inf _{k\geq N}x_n\geq M\tag{2}$$ and since $a_n$ is increasing (another equivalent term is non-decreasing) we have $$a_n\geq a_N\geq M\tag{3}$$ whenever $n\geq N$.
It is quite possible that for some value $n_0\geq N$ we have $a_{n_0}=M$ in which case from $(3)$ we have $$M=a_N=a_{N+1}=a_{N+2}=\dots=a_{n_0}$$ Now your argument is that $$a_{n_0}=\inf_{k\geq n_0}x_k$$ and hence there could be some value $x_k=a_{n_0}=M$ and this would contradict $x_k>M$.
Well this can't be the case because the values of $x_k$ are given beforehand to satisfy $x_k>M$. Their infimum $a_n$ may equal $M$ but when this happens the infimum will not be equal to any $x_k$ it will rather be less than all $x_k$.
A typical example is $x_k=1/k$ and $\inf x_k=0$ but you clearly see that the infimum is not equal to any value of sequence.

To sum up, the bounds of the sequence are given apriori and they can't be invalidated at a later stage. If the sequence values are always greater than some lower bound and infimum equals that lower bound it necessarily and trivially means that the sequence values are greater than the infimum.
A: Consider the two possibilities
i) $x_N-\epsilon/3 =a_n$
ii) $\,a_n = x_{k_0}$ for some $k_0\ge n.$
Each of these cases can arise, but not simultaneously. For if i) holds, then $a_n$ can't equal $x_{k_0},$ simply because $x_{k_0} >x_N-\epsilon/3.$ Thus if i) holds, ii) cannot hold. And if ii) holds, then $a_n=x_{k_0}> x_N-\epsilon/3,$ which implies i) can't hold
So the "problematic part" is not really problematic.
A: You just reversed cause and result: as we have for $x_N − \frac{\epsilon}{3}<x_k <x_N + \frac{\epsilon}{3}$ for $k>N$ (3.1), then from $x_N − \frac{\epsilon}{3}≤ \inf_{k≥n} x_k = a_n$ it is impossible to have $x_N − \frac{\epsilon}{3} = a_n=x_k$.
