Uniqueness of a Point in Cauchy- Cantor Lemma Zorich

Question

This a picture from Zorich's Mathematical Analysis I

I am having trouble understanding the uniqueness part of this theorem(last paragraph), in particular the last sentence.

First,

why should $c$ be unique when by the completeness axiom a second time between $a_n \le c$ we get $$a_n \le c \le b_n \implies a_n \le c' \le c \ \le b_n $$ Also I do not see uniqueness mentioned on the Wikipedia page: https://en.wikipedia.org/wiki/Cantor%27s_intersection_theorem

Second,

I want to make sure I understand what the author is implying because it's vague to me and I also don't feel as if the last line makes a rigorous argument because it's not defined what it means to be arbitrarily small. Shouldn't a variable be 'arbitrarily small' with respect to another variable?? But going back to the argument at an earlier point, I see we can have an 'arbitrarily small' $\varepsilon >0$ such that $$a_n \le c_1 \lt c_2 \le b_n \lt \varepsilon \implies 0 \lt c_2-c_1 \lt b_n-a_n < \varepsilon$$

I can't get a contradiction even if I assume the there is an interval inside and smaller than $[a_n,b_n]$, for suppose that there is an interval $[a_k, b_k ]$ that has a length less that of $[a_n,b_n]$ then we write $$a_n< a_k \le c_1 \lt c_2 \le b_k <b_n $$ $$\implies 0 \lt a_n -a_k \le c_1- a_k< c_2 -a_k \le b_k -a_k< b_n-a_k $$

$$\implies 0 \lt 0 \le c_1- a_k -(a_n -a_k)< c_2 -a_k -(a_n -a_k)\le b_k -a_k -(a_n -a_k)< b_n-a_k -(a_n -a_k)$$

$$\implies 0<c_1-a_n < c_2- a_n \le b_k - a_n < b_n - a_n < \varepsilon$$

but like I said this still does not lead to a contradiction in order to show that $c_1=c_2.$ Zorich has not introduced the idea of a limit point yet,if that is relevant.

Thanks in advance and if you wish to down-vote do let me know why.

Answer 1

For $I_k = [a_k, b_k]$, let $|I_k| = |b_k - a_k|$

What is meant by intervals of arbitrary small lenght is the part of the hypothesis that states that for any $\varepsilon > 0$ there exists an interval $I_k = [a_k, b_k]$ whose lenght is smaller than $\varepsilon$. This is, $\forall \varepsilon > 0 \, \, \exists k \in \mathbb{N}: |b_k - a_k| < \varepsilon$.

Regarding the uniqueness of $c$, assume there are two distinct points $p, q \in \bigcap_{n \in \mathbb{N}} I_n$. Without lose of generality, assume $p < q$. This means that $p, q \in I_n$ for all $n$. So $\forall n: a_n \leq p < q \leq b_n$, which implies $|p - q| \leq |b_n - a_n| = |I_n|$.

But by our hypothesis, for $r = |p-q|/2$ there exists a $k$ such that $|I_k| < r$, i.e., $|I_k| < |p-q|$. A contradiction to what we found in the previous paragraph.

Answer 2

If the second condition holds then there cannot exist $c_1$ and $c_2$ in $\cap_{n=1}^{\infty} I_n$ with $c_1\ne c_2 .$ Otherwise there exists $k$ such that $|I_k|<|c_1-c_2|$ and $\{c_1,c_2\}\subset I_k,$ which is impossible.