Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2010 / B / 5

The cardinal sum is the cardinality of a disjoint union, and the cardinal product is the cardinality of the set of choice functions:Here the cardinals are represented by initial ordinals. Tagging by $\alpha$ makes the summands disjoint even when their underlying sets overlap. For an empty index set the sum is zero and the product is one, because there is exactly one empty function. If any factor in a nonempty product is zero, that product is empty.

This is König theorem for cardinal numbers. Let $D={⨆}_{\alpha <\beta }{\kappa }_{\alpha }$ and $P={\prod }_{\alpha <\beta }{\lambda }_{\alpha }$. For any proposed map $g:D\to P$, consider the values of the $\alpha$th coordinates on the $\alpha$th summand:This set has cardinality at most ${\kappa }_{\alpha }<{\lambda }_{\alpha }$. Choose $y_{\alpha }\in {\lambda }_{\alpha }\setminus E_{\alpha }$, using the axiom of choice, and assemble $y=(y_{\alpha })\in P$. For any $(\alpha ,\xi )\in D$, its image disagrees with $y$ at coordinate $\alpha$. Thus $y$ is outside the image of $g$, and no such $g$ is surjective.

Every ${\lambda }_{\alpha }$ is nonzero. The product is nonempty by choice; if $D$ is empty, strict inequality is immediate. Otherwise cardinal comparability principle in ZFC says that $|P|\le |D|$ would yield a surjection from $D$ onto the nonempty $P$, which the diagonal argument excludes. ThereforeThe diagonal coordinate is tied to the summand containing the argument; this is why a sum on the left pairs naturally with a product on the right.

The aleph numbers enumerate the infinite initial ordinals by transfinite recursion:Here ${\kappa }^{+}$ is the least cardinal strictly larger than $\kappa$. The Generalized continuum hypothesis isequivalently $2^{{\aleph }_{\alpha }}={\aleph }_{\alpha +1}$ for every ordinal $\alpha$ in ZFC.

Constant functions give the lower bound ${\aleph }_2\le {\aleph }_2^{{\aleph }_0}$. By the Generalized continuum hypothesis, ${\aleph }_2=2^{{\aleph }_1}$. The currying law for cardinal exponentiation and infinite cardinal arithmetic therefore giveHence $\boxed{{\aleph }_2^{{\aleph }_0}={\aleph }_2}$.

Put $\kappa ={\aleph }_{\omega }$. Apply König theorem for cardinal numbers to the sequences ${\aleph }_n<{\aleph }_{n+1}$:The sum is $\kappa$ because the countable union has cardinality at most $\kappa$ and its summands are cofinal in $\kappa$. Thus the power is strictly larger than $\kappa$, a fact that cannot be replaced by the regular-cardinal calculation of the preceding part.

For an upper bound, use $\kappa \le 2^{\kappa }$ and the currying law for cardinal exponentiation:where the last step is GCH at $\kappa$. There is no cardinal strictly between $\kappa$ and its successor, so

For $\kappa ={\aleph }_{\omega }$, monotonicity of cardinal exponentiation givesThe middle equality uses the currying law for cardinal exponentiation, and the final one uses ${\aleph }_0\cdot \kappa =\kappa$. Hence ${\aleph }_0^{\kappa }=2^{\kappa }$, even before assuming GCH. Applying the Generalized continuum hypothesis yields