*Inspired by [this question](https://mathoverflow.net/questions/344775/free-ordered-field) I was wondering whether there is a natural topology on the set of all orders on a field (that extend a given order on a subfield)?*


For example for the function field $\mathbb{Q}(X)$ there are the following examples of total orders:


1. The convention $X>q$ for all $q\in \mathbb{Q}$;
2. the convention $0<X<1/n$ for all $n\in \mathbb{N}$;
the pullback of
3. the standard-order on the reals along an embedding $\mathbb{Q(X)}\rightarrow \mathbb{R}$ sending $X$ to any transcendental element;
	4. the pullback of the total order on the surreal numbers along an embedding.


I suspect that different embeddings in (4) still can give the same total order, but maybe all orders arise that way.


*Apart from the question how to classify these field orderings as a set, it seems interesting to ask how to topologize them.*



What is the topology on the set of field orders
The topology you are looking for is called the *Harrison topology*. If we denote the set of ordering of a field $F$ with $\mathrm{Sper}\,F$ (more on that in a moment), this is the subspace topology given by the embedding
$$\mathrm{sign}:\mathrm{Sper}\,F\to \prod\_{x\in F^\times}\{+1,-1\}$$
sending an order to the collection of signs of the elements of $F^\times$. This identifies $\mathrm{Sper}\,F$ as a closed subset of that product, and hence as totally disconnected compact Hausdorff space. A basis of clopen sets is given by $U\_x:=\{\le\mid x> 0\}$ for every $x\in F^\times$.


This is a special case of the construction of the *real spectrum* of a ring $A$. That is the collection of the ``partial orders'' of $A$, which are essentially couples $(p,\le)$ where $p\in\mathrm{Spec}\,A$ is a prime ideal and $\le$ is an order on the residue field $k(p)$. We can topologize it as the closed subspace of
$$\mathrm{sign}:\mathrm{Sper}\,A\to \prod\_{x\in A}\{+1,0,-1\}$$
where $\{+1,0,-1\}$ has the coarsest topology such that $\{0\}$ is closed and $\{+1\}$ and $\{-1\}$ are open.


To give a concrete example, the points of $\mathrm{Sper}\,\mathbb{Q}(x)$ are of one of the following forms


* $a^+,a^-$ where $a$ is a real algebraic number. The order is given by $f\ge 0$ iff $f$ is non-negative in $(a,a+\epsilon)$, resp. $(a-\epsilon,a)$ for some $\epsilon>0$
* $r$ where $r$ is a trascendental real number. Here $f\ge0$ iff $f(r)\ge0$
* $+\infty,-\infty$. Here the order is given by $f\ge0$ iff $f$ is non-negative on $(M,\infty$), resp. $(-\infty,-M)$ for some $M>0$. Equivalently, this picks the sign of the leading coefficient (resp. $(-1)^{\deg f}$ times the sign of the leading coefficient).


These are all distinct, as one can easily show.


The topology of $\mathrm{Sper}\,\mathbb{Q}(x)$ has a basis of clopen subsets of the form $(a,b)$ where $a,b$ are either real algebraic numbers or $\pm\infty$. This is the set of orders $\ge$ such that $f\ge0$ if $f$ is non-negative on $(a,b)$. Note that $a^+\in (a,b)$, but $a^-\not\in(a,b)$.


There is a more comprehensive treatment of this, with more examples, in chapter 7 of



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> *Bochnak, Jacek; Coste, Michel; Roy, Marie-Françoise*, Real algebraic geometry. Transl. from the French., Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge. 36. Berlin: Springer. ix, 430 p. (1998). [ZBL0912.14023](https://zbmath.org/?q=an:0912.14023).
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