Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2007 / B / 7

A cardinal number is an initial ordinal: an ordinal not in bijection with any smaller ordinal. In $ZFC$ every set has the cardinality of a unique such ordinal.

For an infinite cardinal $\kappa$, its cofinality is the least cardinality of an unbounded subset of the ordinal $\kappa$. Equivalently,A subset is unbounded when no ordinal below $\kappa$ bounds all its elements. A cofinal sequence can be taken increasing, and $\mathrm{cf}(\kappa )$ is an infinite regular cardinal.

König theorem for cardinal numbers says that for families of cardinals ${\kappa }_i<{\lambda }_i$ indexed by a set $I$,The sum is the cardinality of the disjoint union of sets $A_i$ of sizes ${\kappa }_i$, and the product is the cardinality of ${\prod }_i{B}_i$ for sets of sizes ${\lambda }_i$.

Consider any function $F:{⨆}_i{A}_i\to {\prod }_i{B}_i$. For each $i$, the set of $i$th coordinates of images of points from the $i$th block has cardinality at most ${\kappa }_i<{\lambda }_i$. Choose $b_i\in B_i$ outside it. Then $b=(b_i{)}_i$ is not the image of any point: a point from block $i$ has the wrong $i$th coordinate. Thus there is no surjection from the sum set onto the product set. Cardinal comparability in $ZFC$ makes the inequality strict in the indicated direction. For the empty index set the statement is $0<1$, also true. This is the diagonal argument underlying the inequality.

First derive the standard consequence of König theorem for cardinal numbers that every infinite cardinal $\rho$ satisfiesLet $\theta =\mathrm{cf}(\rho )$ and take cofinal ordinals ${\alpha }_i<\rho$, $i<\theta$. Put ${\rho }_i=\max (1,|{\alpha }_i|)<\rho$. Their sum is $\rho$: it is at least the size of the union of the cofinal ordinals and at most $\rho \cdot \theta =\rho$. Choose the next larger cardinal ${\rho }_i^{+}\le \rho$. König theorem for cardinal numbers givesThis also covers $\rho ={\aleph }_0$, using finite cardinals and their next larger finite cardinals.

Suppose $\lambda ={\kappa }^{\mu }$ with both $\kappa ,\mu$ infinite. Infinite cardinal multiplication givesIf $\mathrm{cf}(\lambda )\le \mu$, monotonicity of exponentiation would contradict the preceding strict inequality for $\rho =\lambda$. Thus $\mathrm{cf}(\lambda )>\mu \ge {\aleph }_0$. This direction does not require GCH.

Conversely assume GCH and $\mathrm{cf}(\lambda )>{\aleph }_0$. Every function $\omega \to \lambda$ has bounded range. For each infinite cardinal $\nu <\lambda$, GCH givesFinite bounded ranges are also covered by the bound ${\aleph }_0^{{\aleph }_0}\le {\aleph }_1\le \lambda$. Counting functions according to an ordinal bound on their range now gives ${\lambda }^{{\aleph }_0}\le \lambda \cdot \lambda =\lambda$. Constant functions give the reverse inequality. Thereforewhich is the required representation with $\kappa =\lambda$, $\mu ={\aleph }_0$. This proves both directions.