Is Nested Interval Property $\iff$ Monotone Convergence Property in an ordered field?

Question

Given any ordered field $R$
Can we conclude Monotone convergence Property $\iff$ Nested Interval Property?

Monotone Convergence Property: Any monotone increasing sequence bounded above in $R$ is convergent in $R.$

Nested Interval Property: If $I_1,I_2,...,I_n,...$ be a collection of nested closed intervals, i.e $I_1 \supseteq I_2 \supseteq ... \supseteq I_n \supseteq ...$ then $\bigcap_{1}^{\infty} I_i \neq \phi$.

Monotone convergence Property $\implies$ Nested Interval Property.

Let $\{I_n:=[a_n,b_n]:n \in \mathbb{N}\}$ be a nested sequence of closed intervals in $R.$ Then the sequence $\{a_n\}$ is bounded above by $b_1$. Using MCP one concludes $\{a_n\} \to a \in R$.

Claim: $a \in \cap_1^\infty I_i$

I must show that $a_n\leq a$ for all $n \in \mathbb{N}$. If not then there exists $p \in \mathbb{N}$ such that $a<a_p \implies$ $a<a_n$ for all $n\geq p.$ Choose $\epsilon=(a_p-a)/2_{N_{R}}<(a_p-a)$. For this $\epsilon$ there is no $k \in \mathbb{N}$ such that $|a_n-a|_R < \epsilon$ for all $n \geq k$ . Contradiction. Hence we can conclude $a_n\leq a$ for all $n \in \mathbb{N}$. i.e $a$ is an upperbound of the sequence $\{a_n\}$.

Want to show: $a\leq b_n$ for all $n \in \mathbb{N}$. If not then there exists $q \in \mathbb{N}$ such that $b_q<a \implies a_n<b_n\leq b_q<a$ for all $n\geq q.$ i.e There are no elements of $\{a_n\}$ but finitely many in between $b_q$ and $a$, contradicting $(a_n) \to a$. Hence $a\leq b_n$ for all $n \in \mathbb{N}$.

So we conclude that $a \in \cap_1^\infty I_i$ $\implies$ $\cap_1^\infty I_i \neq \phi.$ i.e Nested Interval Property.

My questions are the following:

1. Please tell me if the above proof is correct. i.e Is the above proof valid in an arbitrary ordered field with MCP?

2. I am not sure about the converse. If the converse is also true, how to prove it. If not, counterexample? Is the converse true in an archimedean ordered field?

Answer 1

The proof is fine. You can make the proof that $a$ is an upper bound for the sequence of left endpoints very slightly simpler by letting $\epsilon=a_p-a$: then $|a_n-a|_R\ge\epsilon$ for all $n\ge p$, and you get the same contradiction.

The converse is false, but the counterexamples that demonstrate this are not elementary. One is discussed on page $13$ of this PDF, and another is briefly outlined in this answer to an earlier question on the subject.

Answer 2

1. Your proof is fine.

2. If an ordered field satisfies MCP, then it is Dedekind complete (see page 11 here), thus isomorphic to $\mathbb R$ and hence has every nice property the reals enjoy.