Nested Interval Property and Ordered Fields I have been given the following as a general version of the Nested Interval Property, which is a predicate whose free-variable F ranges through the universe of all ordered fields :   

Every decreasing sequence of closed and bounded intervals in the ordered field F
  has a nonempty intersection.

This is a generalized version of the Nested Interval Theorem related to the real numbers.
I'm reading in a book that this suggests that the meaning of the Nested Interval Property is that the field has decimal expansions and that each decimal expansion corresponds to an element in the field.    
But i'm confused if this is the best generalization possible, because to me the notions of decreasing sequence, closed and bounded intervals only require the concept of a totally ordered set (we can define closed and bounded interval in any totally ordered set), we don't need to introduce arithmetic in any way (which would require the concept of a field).
For example, the natural total order on Z would satisfy this Nested Interval Property (because it satisfies LUB).
So, is the nested interval property intrinsic to totally ordered sets and then whenever we introduce the arithmetic in such totally ordered sets, the nested interval property gives the extra information that the field has decimal expansions representing uniquely each element in the field ?     
I don't know how the concept of Field must enter into the picture for this property.   
 A: Any totally ordered set $(M,<)$ has a natural topology related to the order and which is a good way to express its second order properties. This is the order topology, of which open intervals form a basis $\mathcal{I}$. Let's assume for simplicity that $M$ has no extrema.
If there is a countable sub-basis $\mathcal{N}$ of $\mathcal{I}$, then the Nested Interval Property implies local compacity for $(M,\mathcal{T}_{order})$.
Indeed, with this condition, any closed set is a countable intersection of closed intervals $]-\infty;a]$, $[b;+\infty[$,  where $\exists c \in E$, $]a;c[$ or $]c;b[$ or $]-\infty;b[$ or $]a;+\infty[$ is in $\mathcal{N}$. Let $\mathcal{F}$ denote this countable set of closed intervals.
If $I$ is a closed interval, and $(F_j)_{j \in J}$ is a family of closed subspaces of $I$ whose intersection is empty, then there is a countable family of closed intervals $(I_n)_{n \in \mathbb{N}}$ such that $I \cap \bigcap \limits_{n \in \mathbb{N}} I_n =\varnothing$ and $\forall n \in \mathbb{N}, \exists j \in J, F_j \subset I_n \cap I$. 
This is because each $F_j$ is an intersection of $I$ and countably many closed intervals in $\mathcal{F}$, so $\{A \in \mathcal{F} \ | \ \exists j \in J(F_j \subset A \cap I)\}$ is a countable set of closed intervals whose intersection is disjoint from $I$.
Then, considering $(J_n)_{n \in \mathbb{N}}$ where $J_0 = I \cap I_0$ and $\forall n \in \mathbb{N}, J_{n+1} = J_n \cap I_{n+1}$, you get a decreasing sequence of closed intervals, of empty intersection.
Therefore, if $(M,<)$ has the NIP, one of those intervals must be empty, and one can find $p \in \mathbb{N}$ such that $\bigcap \limits_{k=0}^p I \cap I_p = \varnothing$. 
Now, each $I_p \cap I$ contains some $F_{j_p}$, so $\bigcap \limits_{k=0}^p F_{j_p} \subset \bigcap \limits_{k=0}^p I \cap I_p = \varnothing$.
This proves that $I$ is compact, and with little work, that $(M,\mathcal{T}_{order})$ is locally compact.

As for decimal expansion in ordered fields, I'm not sure that the NIP implies the archimedean property. Decimal expansion exists only for decimal numbers in non archimedean fields with the order topology. Does the author of the book you're reading prove decimal expansion for ordered fields with NIP?
