For any well-ordered class $W$ with no maximum element, we can define a successor function $f_W : W \rightarrow W$ by asserting that $f_W(w) = \mathrm{min}\{x > w \mid x \in W\}.$ This allows us to define the notion of a 'limit element of $W$.' In particular, we assert that $\lambda$ is a limit element of $W$ iff $\lambda \in W$ but $\lambda \notin \mathrm{ran}(f).$ So given a well-ordered class $W$ with no maximum element, lets write $L(W)$ for the limit elements of $W$.

For example, if $W$ is the class of ordinal numbers, then the elements of $L(W)$ are precisely the limit ordinals (with the exception that $0$ is an element of $X,$ but $0$ is not a limit ordinal). If $W$ is the class of cardinal numbers, then the elements of $L(W)$ are precisely the weak limit cardinals (similarly exempted). If $L(W)$ is the class of beth numbers, then the elements of $X$ are precisely the strong limit cardinals (similarly exempted).

Question 1. If $W$ is the class of cardinals of the form $2^\kappa$ for some cardinal $\kappa$, what is $L(W)$?

Now if $W$ is a well-ordered, then so too is $L(W).$ Thus, we can iterate finitely many times; so $L^n(W)$ is well-defined for $n \in \mathbb{N}.$ Indeed, I think we can iterate ordinal-many times. Define that

  1. $L^0(W)=W,$
  2. $L^{\alpha+1}(W) = L(L^\alpha(W))$
  3. If $\lambda$ is a limit ordinal, then $L^\lambda(W) = \bigcap_{\alpha \in \lambda}L^\alpha(W).$

Question 2. Are there any issues with the above definition? And if not, can we guarantee that if $W$ is a well-ordered proper class, then $L^\alpha(W)$ is also a (well-ordered) proper class for all ordinals $\alpha$? And, is the proper class of proper classes $\{L^\alpha(W) \mid \alpha \in \mathrm{On}\}$ order-isomorphic to $\mathrm{On}$ when ordered by reverse inclusion?


Clearly $2^\kappa$ is a limit point of $W=\{2^\lambda;\lambda \text{ a cardinal}\}$ iff we have $2^\lambda< 2^{<\kappa}$ for all $\lambda<\kappa$. That is to say, $2^\kappa$ is a limit point iff the continuum function doesn't stabilize below $\kappa$ (and $\kappa$ is a limit cardinal, of course). I don't think there is a nicer description of these cardinals.

There is nothing wrong with your definition (except that you want to have $\alpha+1$ in the successor step, instead of $1+\alpha$). This operation is basically taking the Cantor-Bendixson derivative. The classes $L^\alpha(W)$ will be well-ordered (since they are subclasses of the well-ordered class $W$). Additionally, we can prove by induction that the class $\mathrm{On}$ is embedded in $W$ as an initial segment, which implies that all of the $L^\alpha(W)$ are proper classes (in particular, they are nonempty).

Note that we cannot have $L^\alpha(W)=L^\beta(W)$ for $\alpha<\beta$, since (if nothing else) the least element of $L^\alpha(W)$ is not in $L^\beta(W)$ (and they are both nonempty). Using this, it is straightforward to show that $\alpha\mapsto L^\alpha(W)$ is an order isomorphism.

As a side note, an interesting question would be whether $L^{\mathrm{On}}(W)$ is nonempty. For example, if $W=\mathrm{On}$ then $L^{\mathrm{On}}(W)=\emptyset$. However, it can happen that this class is nonempty, for example if we take $(\mathrm{On}+1)^{\mathrm{On}+1}$ (or something roughly like this).

  • 1
    $\begingroup$ At least one of us is wrong. :-) $\endgroup$ – Asaf Karagila Jul 3 '13 at 9:14
  • $\begingroup$ With regards to the Cantor-Bendixson derivative, thanks for the link! $\endgroup$ – goblin Jul 3 '13 at 9:37
  • $\begingroup$ @AsafKaragila I guess we'll see. En garde! $\endgroup$ – Miha Habič Jul 3 '13 at 10:32

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