My attempt:
I encountered this problem while reading Understanding Analysis by Stephen Abbott.
My Attempt: We want to show that $[a_n, b_n]$ contains one element $x$, for all $n \in \mathbb N$.
Consider the sequence of all the left end points of the nested intervals. Clearly, $(a_n)$ form an increasing sequence which is bounded above by every $b_n$.
So, by Monotone convergence theorem, $(a_n)$ converges. Suppose $(a_n)$ converges to $x$. We now want $a_n \leq x \leq b_n$, for all $n$.
We have $a_n \leq x$, for all $n$ as $\lim_{n\to \infty} a_n = x$ (I am not sure about this step)
If $x \leq b_n$ for all $n$ is not true, then $x > b_n$ for some $n$. Let $x - b_n = \epsilon$. As $x$ is the limit of $(a_n)$, we will have some $a_n$ in the $\epsilon/2$ - nbd of $x$ which will imply that $a_n > b_n$ for some $n$, which is contradictory. So we have $x \leq b_n$ for all $n$.
Finally, $a_n \leq x \leq b_n$ implies the Nested Interval Property.
I am confused about the step where I wrote $a_n \leq x$ for all $n$. The confusion arises because I cannot use Axiom Of Completeness. Without AoC, I cannot claim the existence of supremum. So, even though $(a_n)$ converges, I cannot claim that it converges to its supremum. Had that been the case, $a_n \leq x$ would follow easily as $x = \lim_{n\to\infty} a_n$. Since the limit of $(a_n)$ is not its supremum, can it not be the case that $a_n > x$ for some $n$? That does not violate $x$ being the limit anyway. Or can I say that $a_n \leq x$ for all $n$ follows from the fact that $(a_n)$ converges to some upper bound?
Thanks in advance :)
You can prove that $a_n \leq x$ by contradiction, using the fact that $a_n$ is monotone non-decreasing ($a_n \leq a_{n+1}$). So, assume that $\exists n_o:x>a_{n_o}$, hence, $a_{n_o} - x = k>0$. But for $$k>\epsilon >0 \exists N: n>N \implies |a_n - x|<\epsilon$$ Since the sequence is monotone non-decreasing, $\forall n>n_o$ and $n>N$ $$\epsilon >a_n - x \geq a_{n_o} -x>\epsilon$$ Which is a contradiction, so indeed, $a_n \leq x$.