I am currently reading up on ordered fields and I think I found out a problem in the class notes. I'm reading a theorem whose corollary proves the existence of the real closure of a (formally) real field, I've got a couple of problems with this. The main problem is with the usage of Zorn's Lemma in the proof of the aforementioned theorem. The other one is with the statement of the theorem.

The Problem

The theorem states:

For a real field $K$ and $F$ a formally positive set of $K$ then $K$ admits an algebraic extension $L$ which is maximal with respect to the property "every order of $K$ which contains $F$ extends to an order of $L$".

My first doubt is: Can the statement of the theorem be changed to

Let $K$ be a real field and $F$ a positive set in $K$. Then for every order $P$ of $K$ which contains $F$, there exists $L$ a maximal field such that $L|K$ is an algebraic extension whose order extends $P$.

As a quick definition, a field is said to be maximal if its order cannot be extended to any proper algebraic extension.

Are these two statements equivalent? I feel they are since I'm still taking an arbitrary order. I think that the problem is with the maximality. If it's wrong, could you help me formulate an equivalent statement for the theorem?

Now, my problem with the proof is as follows:

$\bullet$ In the first part of the proof the author defines a family $\mathcal{F}$ as follows:

$$\mathcal{F}\colon=\{L: L|K \text{ is an alg. ext. and every order $P$ of $K$ s.t. $F\subseteq P$ extends to an order of $L$\}} $$

$\mathcal{F}$ is proven to be nonempty by showing $K\in \mathcal{F}$. The proof continues by taking a chain of fields (ordered by "$\subseteq$") inside $\mathcal{F}$ and the the union of all of these sets. This is the upper bound requested by Zorn's Lemma. Then there exists a maximal element because of Zorn's Lemma.

$\bullet$ My problem is as follows: The author states that by proving that the upper bound is inside $\mathcal{F}$, then the maximal element will also be in $\mathcal{F}$. Is this true?

The doubt in synthesis

When proving that there exists a maximal element for a set with Zorn's Lemma, proving that the upper bound is inside the set (because Zorn's Lemma doesn't guarantee this) implies that the maximal element also inside the set?


There is no meaning to the term "maximal" outside your ordered set. So, yes, the maximal element lies inside $\cal F$, and that is exactly what is guaranteed by Zorn's Lemma.

  • $\begingroup$ I'm not able to understand what you mean. When saying that being maximal has no meaning outside of the ordered set, I think you mean being maximal with respect to the property $L|K$ is an algebraic extension(...). Then why does this imply that maximal element is inside $\mathcal{F}$? I think this is not guaranteed by Zorn's Lemma, since it only guarantees the existence of such element. $\endgroup$ – Ignacio Rojas Nov 28 '17 at 6:55
  • $\begingroup$ Forget the property that you used here. If you have an ordered set, and you want to find a maximal element in that ordered set, then it makes no sense to day that something is maximal and outside of that set. Zorn's Lemma guarantees exactly that such maximal element exists under certain conditions. $\endgroup$ – Asaf Karagila Nov 28 '17 at 6:57
  • $\begingroup$ In other words, when you write "because Zorn's Lemma doesn't guarantee this", you are simply wrong. $\endgroup$ – Asaf Karagila Nov 28 '17 at 7:00
  • $\begingroup$ Oh! I can see it now! Zorn's Lemma states that the upper bound isn't inside the chain necessarily, but it is inside the set. I misread that and thought that the upper bound wasn't inside the set. $\endgroup$ – Ignacio Rojas Nov 28 '17 at 7:05
  • $\begingroup$ Yes, because otherwise, what does it even mean to be an upper bound? $\endgroup$ – Asaf Karagila Nov 28 '17 at 7:06

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