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Let's say that an ordered Field is a class (proper or not) which satisfies the axioms of ordered fields. We work in NBG set theory with global choice.

Let's say that an ordered Field is real closed if its polynomial functions satisfy the Intermediate Value Theorem.

Let's say that an ordered Field is complete if it has no proper dense extension.

-Every ordered field $F$ has a real closure, that is a unique real closed (and algebraic) extension $(\mathcal{R}(F),\rho)$ such that whenever $F$ embeds in a real closed extension $(F',\lambda)$, there is an embedding $\varphi: \mathcal{R}(F) \rightarrow F'$ such that $\varphi \circ \rho = \lambda$.

-Every ordered field $F$ has a completion, that is a unique complete (and dense) extension $(\widetilde{F},\mu)$ such that whenever $F$ embeds in a cofinal complete extension $(F',\lambda)$, there is an embedding $\varphi: \widetilde{F} \rightarrow F'$ such that $\varphi \circ \mu = \lambda$.

I don't think this is the case for ordered Fields, but I am unsure whether I have good reasons to think that.


To explain myself, here is one more definiion: Let $A,B$ be subclasses of a real closed Field $F$ such that $A < B$ and such that there is no element of $F$ lying between $A$ and $B$. We say that $(A,B)$ is a cut in $F$.

Let $F(X)$ be the field of fractions in one indeterminate $X$ over $F$.

One can define an ordered field $F(A \ | \ B)$ by setting $0<f(X)$ iff $(\forall a \in A \exists a' \in A (a < a' \wedge 0 < f(a'))) \wedge (\forall b \in B \exists b' \in B (b' < b \wedge 0 < f(b')))$.

This works because $F$ being real closed, $f$ can only make a finite amount of sign shifts. If $B$ is empty, this yields the classical field $F(X)$ where $X > F$, if $A$ is empty, this is the same except $X < F$. If points in $A$ can be as close to points in $B$ as one wants then $(A,B)$ is said to be a good cut, and $F(X)$ is a dense extension of $F$ where $X$ is the only element lying between $A$ and $B$ (seen as parts of $F(X)$). I don't have much to say about other type of cuts, apart from the fact that they occur iff $F$ is non archimedean.

Any simple extension $F(\alpha)$ of a real close Field is isomorphic to $F(A \ | \ B)$ where $A$ and $B$ are respectively the classes of elements of $F$ lower and larger than $\alpha$ in $F(\alpha)$, but this is irrelevant here. What matters is that this construction can be done without semantic problems in NBG: elements in $F[X]$ are just maps $\mathbb{N} \rightarrow F$ with finie support, and fractions of such elements can be defined as pairs $(a,b)$ where $a,b$ have no common factor and $b$ is monic. Operations and order are also definable with parameters in $\{F;<;+;.;A;B\}$ so there doesn't seem to be a problem.

On the contrary, the proofs I know of the existence of the real closure don't work for class-sized Fields. For instance, using transfinite induction to add roots of polynomials one by one can't be done if the original Field is a proper class simply because the concept of a map (let alone class-sized maps) whose values are classes isn't valid. You can't use Cauchy sequences indexed by $Ord$ because they don't fit in a class. You can't take the whole collection of good cuts (plus cuts in between which there already lies a point) for the same reason.

So I wonder if it actually can't be done in general.

Let's consider the Field $No$ of surreal numbers, and its extensions $N_1 := No(Ord \ | \ \varnothing); \ N_2 := No(A \ | \ B)$ where $A,B$ are the class of surreals $a_{\alpha} = \{0; a_{\beta} \ | \ \beta < \alpha \} \ | \ \{1;b_{\beta} \ | \ \beta < \alpha\},b_{\alpha} = \{a_{\alpha}\} \ | \ \{1;b_{\beta} \ | \ \beta < \alpha\}$,. where $\alpha \in Ord$.

-I claim that $N_1$ doesn't have a real closure. Let's assume it does. Then, by theorem 19 in $[1]$, this real closure, and therefore $N_1$ itself, embeds in $No$. Let $\sigma$ be such an embedding. $\sigma(N_1)$ must have an upper bound in $No$ because else $\sigma(X)$ would be a surreal whose sequence of powers is cofinal in $No$, which is impossible for various reasons. So $\sigma(N_1)$ is a subset of the convex class $C$ generated by $No({\kappa}^+)$ for some cardinal $\kappa$ which is a strict upper bound of $\sigma(N_1)$. We can cut $C$ in less than $2^{{\kappa}^+}$ many disjoint intervals of diameter $\frac{1}{\kappa}$ using the classical integer part of $No({\kappa}^+)$ (Conway ingeters): write $C = \bigsqcup \limits_{a \in Oz \cap No({\kappa}^+)} [\frac{1}{\kappa}a;\frac{1}{\kappa}(a+1)[_{No}$. Now, taking a subset of $N_1$ of size $\geq (2^{{\kappa}^+})^+$ and using the pigeonhole principle with $\sigma$, we see that one of those intervals must contain two elements of $\sigma(N_1)$, but then the inverse of their absolute difference isn't in $\sigma(N_1)$ because it is greater than $\kappa$: a constradiction.

-$No$ is not complete since $N_2$ is a proper dense extension. [What follows is pure speculation] I have a hard time figuring out if $No$ can be completed. If it could, it seems to me that the resulting Field would be an interesting one since it would contain some missing points, and it could be equipped with any interesting map that is locally uniformy continuous on $No$. Now it is known that the completion of a real closed field is real closed. This can be proved using the continuity of roots and Cauchy sequences so it should at least still work for fields whose uniform structure can be reduced to the study of Cauchy sequences indexed by ordinals or Ord, including $No$. So $\widetilde{No}$ should be real closed, and as mentionned erlier, should embed in $No$. I am not so sure that it is a contradiction for the completion of an incomplete field to embed in said field. What may contradict this is theorem 5 in $[1]$, which I don't want to explain in detail but says that $No$ has no proper extension which has a simplicity relation like that of $No$. I feel that it would not be that difficult to define such a simplicity relation for elements in $\widetilde{No}$: just take $<_s$ in $No$ and add $x \widetilde{<_s} y$ when $y$ is in $\widetilde{No}$, $x$ is in $No$ and it is simpler than any surreal (in $No$) close enough to $y$. At the same time, I suppose this would also yield a simplicity relation for $N_2$, so I must be wrong somewhere...


My questions are:

**-Do you spot mistakes in what I jut said?

-Does every ordered Field have a real closure / completion?**


$^{[1]}$ Philip Ehrlich, 2001, Number Systems with Simplicity Hierarchies: A Generalization of Conway's Theory of Surreal Numbers

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    $\begingroup$ Have you defined the real closure properly? You say only that $F'$ is an algebraic extension of $F$, and nothing seems to prevent $F=F'$, even when $R(F)$ is larger. $\endgroup$ – Joel David Hamkins Apr 30 '16 at 23:27
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    $\begingroup$ It seems to me that much of what is going on here has little to do with the field structure, as opposed to the order structure. The surreal line as an order is not "complete" in the sense that it admits unfilled proper class cuts. There is no class order that completes it, in the sense that No is dense in it and that order has no unfilled proper class cuts. Indeed, I think there is no proper class dense linear order at all that is complete in that sense. $\endgroup$ – Joel David Hamkins May 1 '16 at 0:13
  • $\begingroup$ You're right, I edited my definitions. Don't you think the same computation (as suggested by Philip Ehrlich) done in Ackermann's set theory of such a linear order's cardinal would prove your point? $\endgroup$ – nombre May 1 '16 at 9:34
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I suspect the following should address the above stated queries.

Let $A$ be an ordered field whose universe is a proper class of NBG (which I take to include Global Choice).

(i) A real-closure of $A$ exists in NBG.

Proof. In virtue of Global Choice, A is the union of a continuous chain $A_{\alpha}$ of ordered fields indexed over all $\alpha < On$ where each $A_{\alpha}$ is a set. One can now construct a continuous chain $B_{\alpha}$ of ordered fields indexed over all $\alpha < On$, where each $B_{\alpha}$ is a real-closure of $A_{\alpha}$. The union of the latter chain is a real-closure of $A$ in NBG.

(ii) The Dedekindean completion of No does not exist in NBG.

Proof. The Dedekindean completion of No arises by supplementing No (whose elements we may assume to be written in normal form) with all entities of the form $${\sum\limits_{\alpha < \, On } {\omega ^{y_\alpha } .r_\alpha }} $$

where $\left( {y_\alpha} \right)_{\alpha < On } $ is a strictly decreasing sequence of surreals that is coinitial with the negative surreals, and $\left( {r_\alpha } \right)_{\alpha < On } $ is a sequence of nonzero real numbers that are neither ultimately positive nor ultimately negative.

While this real-closed ordered field exists in Ackermann’s set theory (as formulated by Reinhardt in Ackermann’s Set Theory Equals ZF 1970, Annals of Math Log]), it does not exist in NBG since it contains $2^{\aleph_{On}}$ many members.

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  • $\begingroup$ Thank you for this, I had not though of the first method. Do you happen to see what's wrong with my proof that $N_1$ lacks a real closure? I actually proved that in NBG with global choice, no class-sized field can embed "boundedly" in $No$. As for Ackermann's set theory, I didn't know about it and it will take some time for me to undestand your computation and the limitation of cardinal for classes in NBG you seem to point. $\endgroup$ – nombre May 1 '16 at 9:38
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    $\begingroup$ While I haven't had the opportunity to analyze your argument, in Theorem 19 of the paper of mine you cited in your question it is shown that: every real-closed ordered field (whose universe is a set or proper class) is isomorphic to an initial subfield of $\bf{No}$, i.e. a subfield of $\bf{No}$ that is an initial subtree $\bf{No}$. This contradicts your contention unless you are appealing to sense of "boundedly" I do not understand. $\endgroup$ – Philip Ehrlich May 1 '16 at 14:23
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    $\begingroup$ By writing that a real closed Field $F$ embeds boundedly in $No$ I meant there was an embedding $\varphi: F \rightarrow No$ such that $\varphi(F)$ was bounded in $No$, so this doesn't directly contradict the theorem but still. $\endgroup$ – nombre May 1 '16 at 16:56

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