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.
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 $Ord$.