Saturation of real closed fields In this MO-Post I found the following theorem

A real closed field $(F,+,\cdot,1,0,<)$ is $\kappa$-saturated if and only if the order $(F,<)$ is $\kappa$-saturated, i.e. if the order is $\eta_\alpha$ for $\kappa=\aleph_\alpha$.

together with a proof given by Simpson (here with adjusted notation)

By Tarski's result on quantifier elimination for real closed
  ordered fields, any subset of $F$ which is definable over $F$ allowing
  parameters from $F$ is a finite union of intervals, all of whose
  endpoints are in $F\cup\{\pm\infty\}$.  But then Tychonoff's theorem plus $\kappa$-saturation of $(F,<)$ implies that any family of $<\kappa$ such sets has the finite
  intersection property.  Hence $F$ is $\kappa$-saturated.

But I don't really understand this proof. First of all for me $\kappa$-saturation means that for all $A\subseteq F$ with $|A|<\kappa$ all $1$-types in $S^F_1(A)$ are realized in $F$ and I think this is equivalent to the statement that for any family of definable subsets $(D_\alpha)_{\alpha<\lambda}$ of size $\lambda<\kappa$ with the finite intersection property $\bigcap_{\alpha<\lambda} D_\alpha\neq\emptyset$.


*

*So shouldn't the conclusion in the proof read "any family of $<\kappa$ such sets with the finite intersection property has nonempty intersection"? and

*Could you explain how exactly this follows from Tychonoffs Theorem and $\kappa$-saturation of the order?

 A: First of all, this statement is only true for $\kappa>\aleph_0$. For any real closed field $F$, the reduct $(F,<)$ is a dense linear order without endpoints, and every such order is $\aleph_0$-saturated. But there are certainly real closed fields which are not $\aleph_0$-saturated. For example, $\mathbb{R}$ does not realize the partial type  $\{x>n\in \omega\}$ over the empty set (where each natural number $n$ is $1+1+\dots+1$). So let's assume $\kappa$ is uncountable. 
To your questions:


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*My guess is that Simpson just meant to write "any family of $<\kappa$ such sets with the finite intersection property has nonempty intersection."

*I have no idea how Tychonoff's Theorem is relevant here (is there another statement called Tychonoff's Theorem other than the fact that a product of compact spaces is compact?).
Here's how I would write the argument: Clearly if $(F,+,\cdot,0,1,<)$ is $\kappa$-saturated, then its reduct $(F,<)$ is $\kappa$-saturated. For the converse, let $p$ be a type in one free variable in the language of fields over $|A|$ with $|A|<\kappa$. So $|p|<\kappa$. By Tarski's theorem (i.e. by o-minimality), every formula $\varphi$ in $p$ is equivalent to a formula $\varphi'$ in the language of order, possibly with other parameters. Let $p' = \{\varphi'\mid \varphi\in p\}$, and let $A'$ be the set of all parameters appearing in $p'$. Then $|A'|<\kappa$. $p$ is consistent (has the finite intersection property), and the formulas in $p'$ are equivalent to the formulas in $p$, so $p'$ is consistent too, so $p'$ is realized in $(F,<)$, by $\kappa$-saturation, and hence $p$ is realized in $F$ as well. 
Where did we use the assumption that $\kappa>\aleph_0$? Hint: it was in the computation that $|p|<\kappa$.
Closely connected to this is your comment:

I think [$\kappa$-saturation] is equivalent to the statement that for any family of definable sets $(D_\alpha)_{\alpha<\lambda}$ of size $\lambda<\kappa$ with the finite intersection property, $\bigcap_{\alpha<\lambda}D_\alpha\neq \emptyset$.

The property you described in the quote above is called "$\kappa$-compactness". $\kappa$-compactness is equivalent to $\kappa$-saturation when $\kappa>|L|$. The problem is that when $\kappa\leq |L|$, it's possible to find complete types $p$ over $<\kappa$ parameters with $|p| = |L| \geq \kappa$. So $\kappa$-saturated is strictly stronger than $\kappa$-compact for small $\kappa$. Here $L$ is the size of the language (the cardinality of the set of all formulas). So for real closed fields, $|L| = \aleph_0$.
