Past exam of the axiomatic set theory course of the University of Oxford / 2011 / 3

Use the usual cardinal arithmetic with the axiom of choice. The König theorem for cardinal numbers states that for any index set $I$ and cardinal numbers ${\kappa }_i<{\lambda }_i$,Represent the cardinal numbers as initial ordinals. Let $A=\{(i,\alpha ):i\in I,\alpha <{\kappa }_i\}$ and $B={\prod }_{i\in I}{\lambda }_i$. There is an injection $A\to B$: send $(i,\alpha )$ to the function whose $i$th coordinate is $\alpha +1$ and whose other coordinates are zero. The unique nonzero coordinate identifies $i$, and its value identifies $\alpha$. All values are allowed because $\alpha +1\le {\kappa }_i<{\lambda }_i$.

For any proposed surjection $F:A\to B$, its $i$th row has at most ${\kappa }_i$ elements, soChoose $b_i$ to be the least omitted ordinal. The function $b=(b_i{)}_{i\in I}$ belongs to $B$ and differs from $F(i,\alpha )$ at coordinate $i$, for every $(i,\alpha )\in A$. Thus $F$ is not surjective. Cardinal comparison now gives the strict inequality. If $I=\varnothing$, the conclusion is $0<1$, also valid.

For an infinite cardinal number $\kappa$, its cofinality isviewing $\kappa$ as an initial ordinal. Equivalently it is the smallest cardinal number which is the domain size of a cofinal map into $\kappa$.

Put $\lambda =2^{\kappa }$ and $\mu =\mathrm{cf}(\lambda )$. Suppose $\mu \le \kappa$, and choose cofinal ordinals ${\alpha }_i<\lambda$ for $i<\mu$. Put $a_i=|{\alpha }_i|<\lambda$. Their union is $\lambda$, so infinite cardinal arithmetic givesApply the König theorem for cardinal numbers to $a_i<a_i^{+}$. Each successor cardinal $a_i^{+}$ is at most $\lambda$, whenceHere the last equalities use the currying law for cardinal exponentiation and square of an infinite cardinal. This contradiction proves the cofinality of a power-set cardinal bound

Put $\mu =\mathrm{cf}(\kappa )$ and choose cofinal ordinals ${\alpha }_i<\kappa$, $i<\mu$. With $a_i=|{\alpha }_i|$, the same union comparison and infinite cardinal arithmetic give ${\sum }_{i<\mu }{a}_i=\kappa$. Since $a_i<\kappa$, their successor cardinals satisfy $a_i^{+}\le \kappa$. The König theorem for cardinal numbers therefore givesThis König bound for the Gimel function applies to regular and singular cardinals alike; choosing cofinal ordinals also covers successor cardinal numbers.

Take the Beth numberCantor's theorem makes this sequence strictly increasing. It is a countable cofinal sequence in $\kappa$, while no finite set is cofinal in the limit ordinal $\kappa$. Hence $\mathrm{cf}(\kappa )={\aleph }_0$. ConsequentlyThus Beth omega is a singular strong limit above the continuum. No Continuum hypothesis is needed for the example.