Monotone convergence theorem implies Nested Interval property

Question

Assuming MCT holds we need to prove the NIP

Monotone Convergence theorem: A monotone and bounded sequence is convergent.

Nested Interval Property: If for all $n\in \mathbb{N}, I_n=[a_n, b_n]$ and $I_{n+1} \subset I_n$ then $\bigcap_{n=1}^\infty I_n\neq \emptyset $

Proof: $(a_n)$ is a monotone a bounded sequence and increasing then it converges to some $a$. We say that $I_n=[a_n, a]$, then $I_{n+1}\subset I_n$, as $a_n<a_{n+1}$. Now the intersection $\bigcap_{n=1}^k I_n=[a_k,a]$, now we can take the limit $$\lim_{k\rightarrow \infty}\bigcap_{n=1}^k I_n=a\neq\emptyset$$This proves the NIP.

Does this proof look ok?

Answer 1

Your proof is not O.K. You wrote $I_n=[a_n, a]$. But we have $I_n=[a_n, b_n]$ !

We have:

$a_1 \le .... \le a_n \le a_{n+1} \le b_{n+1} \le b_b \le .... \le b_1$. Hence the sequences $(a_n)$ and $(b_n)$ are bounded and monotonic, hence convergent.

Let $a= \lim a_n$ and $b= \lim b_n$. Then we have $a \le b$. Now show that $[a,b ] \subseteq \bigcap_{n=1}^\infty I_n.$

Answer 2

Correct me if wrong.

A missing detail:

The OP wants to prove MCT $\rightarrow$ NIP.

As User Fred points out for all $I_n =[a_n,b_n]$ with $I_{n+1} \subset I_n$, $n\in \mathbb{N}$:

$\lim_{n \rightarrow \infty} \displaystyle \bigcap_{n=1}^{\infty} I_n \not = \emptyset.$

Recall:

$a_1\le a_2\le...\le a_n \le a \le $

$b \le b_n\le...\le b_2\le b_1$,

where $a=\lim a_n,$ and $b =\lim b_n.$

For $x \in [a,b]$ , i.e. $a \le x\le b$ we have:

$a_n \le a \le x \le b \le b_n$ for $n \in \mathbb{N}$, i.e.

$x \in I_n$ for all $n$, hence

$[a,b] \subset \displaystyle \bigcap_{n=1}^{\infty}I_n $.

Note: $[a,b] \not = \emptyset$

(Has at least one element).