3
$\begingroup$

Recently I have been playing around with the idea of well-ordered proper classes, essentially an ordered pair $(P, ≤)$ where P is a proper class and $≤$ is a well-order on $P$.

For well-ordered sets, there is a core equivalence relation known as order-isomorphic - where two well-ordered sets are order isomorphic if there exists a bijection between them such that both the function and its inverse are strictly increasing.

A second core idea is that of a cofinal subset. Given a well-ordered set $A$, then a subset $B \subseteq A$ is called cofinal if for all $a \in A$ there exists $b \in B$ such that $a \leq b$.

Both of these ideas are defined for sets, but there is nothing to stop them from being extended to include proper classes. In doing so, there are many proper classes which are not order-isomorphic to any element of $Ord \cup \{Ord\}$ - where $Ord = \{ x : x$ is an ordinal$\}$. For example the class $Ord \times \{0,1\}$ with the lexicographical order.

However the question I want to ask is this:

Does there exist a well-ordered proper class $P$, such that for all cofinal subclasses $Q$, $Q$ is not order-isomorphic to any element of $Ord \cup \{Ord\}$?

Any example I can think of always has a cofinal subclass order-isomorphic to an element of $Ord \cup \{Ord\}$, which leads me to believe the answer to this question is no - but I don't know how to prove it (or if it even can be proved). Any help would be much appreciated.

  • 0
    Another criterium for your result in NBG would be limitation of size, which yields a negative answer, while its negation probably yields a counterexample. – 2017-02-01
  • 0
    Based on your comment to Eric's answer, it looks like you might not be assuming the axiom of foundation. Is this the case? If so, you should state it explicitly in the question, since foundation is usually taken for granted. – 2017-02-02

1 Answers 1

8

No, there does not. Let $P$ be a well-ordered class, and let us assume that no subset of $P$ is cofinal (if a subset of $P$ is cofinal, that subset is isomorphic to an ordinal and we are done). For each $\alpha\in Ord$, let $f(\alpha)$ be the least element of $P$ that is greater than every element of $V_\alpha\cap P$ (such an element exists since $V_\alpha\cap P$ is not cofinal in $P$). This defines a (nonstrictly) increasing function $f:Ord\to P$ whose image is cofinal. We can easily modify it to make it strictly increasing (recursively define $f'(\alpha)$ to be the $\alpha$th distinct element of the image of $f$). Thus the image is a cofinal subclass which is isomorphic to $Ord$.

  • 0
    The problem I have with this is that I feel it assumes that $P \subseteq V$ which is not necessarily true, especially if $P$ contains elements which are not well-founded. In the case where $P \nsubseteq V$, I feel this method won't produce a cofinal image? – 2017-01-31
  • 0
    Also I assume you mean to say "since $V_\alpha \cap P$ is **not** cofinal in $P$" and the last sentence should say "cofinal subclass" instead of "cofinal subset". Not trying to be overly pedantic, it's just in case anyone comes across this and reads it in future. – 2017-01-31
  • 1
    This argument is working in ZF, so in particular it is assuming Foundation. If you don't assume Foundation then I expect the result is not true, though I don't know how to build a model with a counterexample. – 2017-01-31
  • 0
    The class $W=\{V_{\alpha}:\alpha \in Ord\}$ of sets is defined recursively without Foundation. In the presence of the other axioms of ZF-Infinity, the axiom of Foundation is equivalent to $V=\cup W.$.... Also the recursive def'n of $f $ is well-founded. As far as I can see, your A does not use Foundation. – 2017-02-01
  • 1
    @user254665: The statement that my subclass is cofinal in $P$ uses Foundation, since otherwise $P$ might have an element which is not in $W$ (and hence not in my subclass) but is larger than every element of $P\cap W$. – 2017-02-01
  • 0
    Yes. I missed that point – 2017-02-01