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Cardinal exponentiation counts functions from a set of size $\lambda$ to one of size $\kappa$. The currying law for cardinal exponentiation gives $({\kappa }^{\lambda }{)}^{\mu }={\kappa }^{\lambda \cdot \mu }$. These are cardinal operations, distinct from ordinal exponentiation. For example the Generalized continuum hypothesis specifies $2^{\kappa }={\kappa }^{+}$ at infinite cardinals.

Assume the Generalized continuum hypothesis. For necessity, König theorem for cardinal numbers gives ${\rho }^{\mathrm{cf}(\rho )}>\rho$ for every infinite cardinal. But $\lambda ={\kappa }^{\mu }$ implies ${\lambda }^{\mu }=\lambda$ by the currying law for cardinal exponentiation, so $\mathrm{cf}(\lambda )>\mu$. This direction uses only ZFC. Conversely, uncountable cofinality bounds the range of every countable sequence in $\lambda$. Under GCH, each smaller infinite cardinal $\nu$ satisfies ${\nu }^{{\aleph }_0}\le 2^{\nu }={\nu }^{+}\le \lambda$. There are only $\lambda$ ordinal bounds to consider, giving ${\lambda }^{{\aleph }_0}=\lambda$ and the representation with base $\lambda$ and exponent ${\aleph }_0$.

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

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

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 $\boxed{{\aleph }_2^{{\aleph }_0}={\aleph }_2}$.

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

True in ZF. A function from $M$ into the functions $B\to A$ corresponds bijectively to a function from $B\times M$ into $A$:The inverse sends $f$ to the function $m\mapsto (b\mapsto f(b,m))$. These two explicit transformations are inverses, proving the currying law for cardinal exponentiation. Empty domains and codomains cause no exception: the same correspondence still applies, including the unique empty function when its domain is empty. No selection of representatives or choice functions is involved.

False. Take $\kappa =2^{{\aleph }_0}$ and $\lambda ={\aleph }_0$, both infinite. The explicit pairing bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ and the currying law for cardinal exponentiation giveThe power-set cardinal need not be assumed well-orderable. Every bijection used in this calculation is explicitly available in ZF, so the counterexample does not use choice.

Assertion (iii) can be false. Let $\kappa ={\aleph }_0$ and $\lambda =2^{{\aleph }_0}$. Part (ii) and the currying law for cardinal exponentiation giveBy Cantor theorem, $2^{\lambda }>\lambda$, and therefore

Choose sets $K,L,M$ of cardinalities $\kappa ,\lambda ,\mu$. Their cardinal arithmetic operations arewhere $K^L$ is the set of functions $L\to K$. The relation $\kappa \le \lambda$ means that there is an injective function $K\to L$.

The currying law for cardinal exponentiation follows from the explicit bijectionand provesMoreover, $2^{\kappa }$ is the cardinality of the power set of a set of size $\kappa$, so Cantor theorem gives

For completeness, identify each cardinal with its initial ordinal. Suppose that some infinite $\kappa$ violates ${\kappa }^2=\kappa$, and choose the least such cardinal. Well-order the pairs $(\alpha ,\beta )\in {\kappa }^2$ first by $\max \{\alpha ,\beta \}$ and then lexicographically. Every proper initial segment is contained in ${\gamma }^2$ together with finitely many boundary pieces for some $\gamma <\kappa$, and has cardinality below $\kappa$ by minimality. The resulting well-order therefore has cardinality at most $\kappa$. The reverse inequality is immediate from $\alpha \mapsto (\alpha ,0)$, contradicting the choice of $\kappa$. Thus the square of an infinite cardinal satisfies ${\kappa }^2=\kappa$.

If $\nu =\max \{\kappa ,\lambda \}$, monotonicity now givesHence the sum and product of two infinite cardinals obey

Assertion (i) can be false. For any infinite $\lambda$, take $\kappa =(2^{\lambda }{)}^{+}$. Then ${\kappa }^{\lambda }\ge \kappa >2^{\lambda }$, so