Density of rationals in an ordered field

Question

Given an ordered field $F$, is it equivalent to say that $\mathbb{Q}$ is dense in $F$ with respect to the order topology and to say that $\forall x\in F,\forall y\in F,x<y\Rightarrow\exists q\in \mathbb{Q},x<q<y$ ?

Answer 1

Yes.

In fact, for any linearly ordered set $(X,\leq)$, a subset $Y \subset X$ is called order-dense if for all $x_1 < x_2 \in X$ there is $y \in Y$ with $x_1 < y < x_2$. Suppose $X$ has at least two points and is order-dense in $X$. Then a subset $Y$ of $X$ is order-dense in $X$ if and only if $Y$ is dense in $X$ for the order topology: if $Y$ is order dense in $X$ and $U \subset X$ is nonempty open, then $U$ contains a nonempty interval of the form $(a,b)$ (for $a < b$) $[a,b)$ (for $a$ the least element of $X$) or $(a,b]$ (for $b$ the greatest element of $X$). The hypotheses imply there is $c \in Y$ with $a < c < b$. If $Y$ is dense in the order topology and $a < b \in X$, then $(a,b)$ contains an element $c$ of $Y$.

The hypothesis that $X$ has at least two points and is order-dense in $X$ applies to every ordered field $F$: $0 < 1 \in F$, and for $x < y \in F$, we have $x < \frac{x+y}{2} < y$. (This uses that every ordered field has characteristic $0$, or at least characteristic different from $2$. We could get around this by also using $\frac{2x+y}{3}$.)

Answer 2

Yes for ordered fields. We say that a set $S\subset K$ is order-dense in $K$ if and only if $\forall x,y\in F,x<y \Rightarrow \exists q\in S,x<q<y$. The order topology is the one having as base the sets : $]x,\infty[, ]x,y[, ]\infty,x[$ with $x,y\in F$. To prove that an order-dense set is dense is straightforward. Let us call this set $S$. Given a neighborhood $V$ of $x$, we have a set $B$ of the base with $x\in B\subset V$ and we can see that in the three cases for $B$, there is a $q\in S$ that is in $V$.

The converse requires that the ordered set is a ordered field. First, recall that a ordered field has characteristic $0$. So we have a set $S$ dense in $F$. If $x,y\in F$ such that $x<y$, necessarily, $]x,y[$ is not empty, since $\frac{x+y}{2}$ is in $]x,y[$ and thence $]x,y[$ is a neighborhood of $\frac{x+y}{2}$. So there is $q\in S$ such that $x<q<y$.