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.)?
Best Answer
It is not possible to prove that the Nested Interval Property implies the Monotone Convergence Theorem. By that I mean that there are ordered fields with the Nested Interval Property that do not satisfy Monotone Convergence.
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 $(c_i)$.
Remark: There is a fair literature on the subject. A good survey, at least from the non-standard analysis side, can be found here.