# Proof of the Monotone Convergence Theorem using Nested intervals Theorem?

Nested Intervals Theorem: If $I_{n}=\left [ a_{n},b_{n} \right ]$ and $I_{1}\supseteq I_{2}\supseteq I_{3}\supseteq ...$ then $\bigcap_{n=1}^{\infty}I_{n}\neq \varnothing$ In addition if $b_{n}-a_{n}\rightarrow 0$ as $n \to \infty$ then $\bigcap_{n=1}^{\infty}I_{n}$ consists of a single point

Monotone Convergence Theorem: If $a_{n}$ is a monotone and bounded sequence of real numbers then $a_{n}$ converges.

How can you prove the second theorem using only the first(WITHOUT using the least upper bound property, the Bolzano Weierstrass etc.)?

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@nick: given a monotone and bounded sequence, can you construct a nested sequence of intervals, so that you could apply the first Theorem? – student Feb 15 '12 at 15:12
Let us assume that $a_n$ is increasing (if not, replace $a_n$ with $-a_n$). Intuitively, $a_n$ should accumulate at the least upper bound of $\{a_n\}_{n\in\mathbb{N}}$. So set $b = \sup_{n\in\mathbb{N}}\{a_n\}$. Now consider $I_n = [a_n, b]$. What can you conclude about the sequence $\{I_n\}_{n\in\mathbb{N}}$ in the context of the Nested Intervals Theorem? – William Feb 15 '12 at 15:13
@WNY How do you know that b exists? – nick Feb 15 '12 at 15:13
Because the hypothesis says "if $a_n$ is monotone and bounded" (so $\{a_n\}$ is bounded). – William Feb 15 '12 at 15:15
@nick: If this isn't direct, I am not sure how more direct one can make it. When you find an acceptable answer, please let us know - I'm curious now. I don't want to go into an extended discussion, but LUB is in a sense equivalent to the theorem you're trying to prove :) – William Feb 15 '12 at 15:24

The examples I can think of involve non-standard models of analysis. For example, let $\mathbb{N}$ be the set of natural numbers, and let $D$ be a non-principal ultrafilter on $I$. Then the ultrapower $\mathbb{R}^{\mathbb{N}}/D$ has the nested interval property but does not satisfy Monotone Convergence.
Briefly, one constructs the ultrapower by first considering the product $\mathbb{R}^{\mathbb{N}}$, that is, the set of all sequences of reals. Two such sequences $(x_n)$ and $(y_n)$ are equivalent modulo $D$ is the set of $i$ such that $u_i=v_i$ is an element of the ultrafilter $D$. On the ultrapower, one puts a ring structure by defining addition and multiplication coordinatewise modulo $D$. And if $(u_n)/D$ and $(v_n)/D$ are elements of $\mathbb{R}^{\mathbb{N}}/D$, we say that $(u_n)/D < (v_n)/D$ if the set of $n$ such that $u_n<v_n$ is an element of $D$. It turns out that the ultrapower just defined is a real-closed ordered field. The reals can be embedded in the ultrapower via equivalence classes of constant sequences. The ordering is non-Archimedean, since if $v_n=n$, then $(v_n)/D$ is larger than any $(u_{k,n})/D$, where $u_{k,n}$ is the integer $k$ for all $n$.
If in addition to Nested Intervals, we ask that our field have the Archimedean Property, then Nested Intervals does imply Monotone Convergence. Let the sequence $(c_i)$ be say non-decreasing. For any $n$, let $a_n=c_n$. Let $b_1$ be an upper bound for the sequence $(c_i)$, and define $b_n$ as follows. If nothing below $b_{n-1}$ is an upper bound for $(c_i)$, let $b_n=b_{n-1}$. Otherwise, let $b_n=b_{n-1}-2^{k}$, where $k$ is the largest integer (possibly negative) such that $b_{n-1}-2^k$ is an upper bound for $(c_i)$. From the Nested Interval Property for the sequence of intervals $(a_n,b_n)$, we can deduce the convergence of the sequence $(c_i)$.