Let $\mathfrak{A}$ be a ordered field and $\mathfrak{X}$ be the smallest subfield of $\mathfrak{A}$. Is $X$ dense in $A$?

Question

Let $\mathfrak{A}=\langle A,<,+,\cdot,0',1' \rangle$ be an ordered field and $\mathfrak{X}=\langle X,<,+,\cdot,0',1' \rangle$ be the smallest subfield of $\mathfrak{A}$.

It is well-known that $\mathfrak{X}$ is uniquely isomorphic to $\langle \Bbb Q,<,+,\cdot,0,1 \rangle$ where $\Bbb Q$ is the set of rationals.

I would like to ask if $X$ is dense in $A$.

Thank you for your help!

Answer 1

Well, no. In brief, we can make an ordered field out of rational functions with real coefficients. A rational function is called positive if its value is positive for all sufficiently large positive $x$

I'm looking up some details, give me a minute.

Added: yes, I got it right, Proposition 18.1, pages 158-159 in Hartshorne, Geometry: Euclid and Beyond

Answer 2

No.

1. There are ordered fields of arbitrary high cardinality.

2. Every field has a countable subfield. (Therefore, its smallest subfield is countable.)

3. If a totally ordered set has a countable dense subset, then its cardinality is at most $2^{\aleph_0}$. (Call the ordered set $A$ and the countable dense subset $Q$; then the map $a\mapsto\{q\in Q:q\lt a\}$ is an injection from $A$ to the power set of $Q$.)

Hence any ordered field of cardinality greater than $2^{\aleph_0}$ is a counterexample.