[Math] A question about Transfinite Induction

lo.logicset-theory

The Transfinite Induction says: Let $\mathbf{P}(x)$ is a property, assume that, for all ordinal numbers $\alpha $ : If $\mathbf{P}(\beta)$ holds for all $\beta < \alpha$, then $\mathbf{P}(\alpha)$ holds. Then $\mathbf{P}(\alpha)$ holds for all ordinals $\alpha$.

My question is: What is the problem if I replace ordinals with cardinals? I mean could I say that If $\mathbf{P}(\kappa)$ holds for all cardinals $ \kappa< \lambda$, then $\mathbf{P}(\lambda)$ holds. Then $\mathbf{P}(\lambda)$ holds for all cadinals $\lambda$.

Best Answer

If by cardinal you mean an initial ordinal (an ordinal not equinumerous with any smaller ordinal), then your new scheme is merely an instance of the scheme for ordinals. Indeed, you can see it clearly as a special case, if you realize that under AC every cardinal is $\aleph_\alpha$ for some ordinal $\alpha$, and so your proposed cardinal scheme is asserting that if $P(\aleph_\beta)$ for all $\beta\lt\alpha$, then $P(\aleph_\alpha)$ holds. (And actually, the new scheme is equivalent to the old scheme, if you consider replacing $\alpha$ with $\aleph_\alpha$.)

Without AC, however, there is a more general concept of cardinal, by which is meant something like the equinumerosity class of a set. If $Y$ is a set, then a smaller cardinality amounts to a set $X$ such that $X$ injects into $Y$ but not conversely. Without AC, these cardinals are not necessarily well-founded. Thus, the transfinite induction scheme fails for these more general types of cardinals.

For example, it is consistent with ZF that there are infinite sets that are Dedekind finite, so that they are not bijective with any proper subset of themselves. Let $P(X)$ assert that $X$ is not infinite Dedekind finite. If $Y$ is any set and is infinite Dedekind finite, then for any $a\in Y$ the set $X=Y-\{a\}$ is strictly smaller in size (else there is a countable subset of $Y$ by iterating the bijection) and $X$ is also infinite Dedekind finite. In other words, the property $P(X)$ satisfies the induction scheme for cardinalities, but does not hold of all cardinalities (if there are some infinite Dedekind finite sets). So transfinite induction for cardinalities can fail without the Axiom of Choice.

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