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Does there exist a totally-ordered-without-endpoints proper class $L$ such that every closed interval in $L$ does have the order type of a closed interval in the Conway's surreal numbers, but $L$ as a whole does not have the order type of Conway's surreal numbers?

In case it helps, the thinking that lead up to my posting this question:

I confess proper classes make me a bit uneasy. So for private intuition I use as a crutch set theory below a strongly inaccessible cardinal $\kappa$. If one considers, as a toy surreal numbers, just Conway's surreal numbers with birthdays less than $\kappa$, then it seems to me that one can then imitate the usual long line construction based on $\kappa^+$.

That said, my understanding says one should find the surreal numbers and in particular the ordinals (as individuals) in ZFC but not the totality of surreal numbers. Perhaps that means one can't define any proper class that will play the role of $\kappa^+$ and fulfill my intuition.

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up vote 7 down vote accepted

There are many such structures. For example, simply form L = N x [0,1) in the dictionary order, with the least element deleted, where N is the set of natural numbers including 0, and [0,1) is the semi-open interval of surreals. Since No (i.e. the ordered class of surreals) does not have a cofinal subset and L clearly does, L is not order-isomorphic to No. Moreover, that every closed interval of L is order-isomorphic to a closed interval of No is straightforward.

That fact that No has no cofinal subset follows from Theorem 2 of my paper:“The Absolute Arithmetic Continuum and the Unification of All Numbers Great and Small”, The Bulletin of Symbolic Logic 18 (1) 2012, pp. 1-45. The theorem states that (in NBG with global choice):

No is (up to isomorphism) the unique absolute linear continuum.

An ordered class A is said to be an absolute linear continuum if for all subsets X and Y of A where X < Y (every member of X precedes every member of Y), there is a z in A such that X < {z} < Y.

The absence of a cofinal subset in an absolute linear continuum is established by simply letting Y be the empty set.

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Your $L$ is something of a surreal "short" line, since it is an initial segment of the positive part of No (though not being a set, it need have no least upper bound), right? Am I right to think I'll find every such structure inside No? –  David Feldman Apr 6 '12 at 0:38
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Yes, L has no least upper bound in No. The members of No that are greater than the members of L are positive infinite and there is no least such element. You are also correct that every "surreal long line" in your sense, whether it be short or very long, is contained as a subclass of No. This follows from the fact that in NBG, No is (Up to isomorphism) the unique homogeneous-universal ordered class--it's universal because it contains an isomorphic copy of every ordered class and it's homogeneous because every isomorphism between ordered subsets admits an extension to an automorphism of No. –  Philip Ehrlich Apr 6 '12 at 1:40
    
So am I right to think that convex subclasses of No without endpoints are characterized by a pair a cofinalities, downward and upward, which can range over the infinite regular cardinals, including On, the class of ordinals? And that removing a bounded convex subclass $B$ from No leaves two convex subclasses $l$ and $r$, such that $r$ has downward cofinality equal to the the upward cofinality of $B$ and $l$ has upward cofinality equal to the downward cofinality of $B$? –  David Feldman Apr 6 '12 at 5:39
    
You are correct about the first question but not the second. Consider, for example, B equal to the class of finite and infinitesimal members of No. B has cofinal character omega and coinitial character omega, but your r has cofinal character On and your l has cointial character On. If L, R is a partition of No into convex subclasses, where L < R, then either L has cofinal character On or R has cointial character On. There are partitions where both L and R have character On, but these partitions always contain sub-partitions of the positive members or negative members of an Archimedean class. –  Philip Ehrlich Apr 6 '12 at 12:39
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An Archimedean class of No is a subclass consisting of all members of No that are finite relative to one another. 0 is the sole member of No that is not in an Archimedean class. The class of positive members of an Archimedean class is a convex subclass of No as is the class of negative such members. Let X, Y be a partition of No where X < Y and let X' and Y' be a partition of either the positive or negative members of an Archimedean class where (i) X' < Y' and (ii) X' is cofinal with X and Y' is coinital with Y. There are such cases where X has cofinal and Y has coinitial character On. –  Philip Ehrlich Apr 6 '12 at 22:26
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