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

ordered-fields

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?

Best Answer

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.

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