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For an infinite well-orderable cardinal number $\kappa$, each strict total order is a subset of ${\kappa }^2$, giving at most $2^{\kappa }$ possibilities by the square of an infinite cardinal. For a lower bound, order $\kappa$ two-element blocks by the ordinal index and independently reverse each block according to a subset of $\kappa$. Transport along $\kappa \times 2\cong \kappa$. Distinct subsets give distinct orders. The usual axiom of choice development supplies these product identities; for well-orderable cardinals they are already provable in ZF.

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

Let $\mathcal{O}$ be the set of strict total orders on $\kappa$. Each is a subset of $\kappa \times \kappa$, so the square of an infinite cardinal givesFor the lower bound, use the set $\kappa \times \{0,1\}$ and order the two-element blocks by their first coordinates, using the well-order of $\kappa$. For every $A\subseteq \kappa$, put $(\alpha ,0)$ before $(\alpha ,1)$ exactly when $\alpha \in A$, reversing them otherwise. This defines a strict total order for each $A$, and different subsets give different orders. Transport these orders to $\kappa$ along a fixed bijection $\kappa \times \{0,1\}\cong \kappa$. Thus $2^{\kappa }\le |\mathcal{O}|$, and the Cantor-Schröder-Bernstein theorem givesIf using the customary axiom of choice development of infinite cardinal arithmetic, its use here is precisely to supply $\kappa \times \kappa \cong \kappa$ and $\kappa \times 2\cong \kappa$. The ordering construction makes no further choices. It proves the cardinality of the family of strict total orders. For $\kappa$ itself an initial ordinal, these particular product identities can also be proved in ZF by well-ordering pairs according to their maximum coordinate, so full axiom of choice is stronger than this instance intrinsically requires.

An aleph number is an infinite initial ordinal. More explicitly,for a limit ordinal $\lambda$. By the well-ordering theorem, every set is equipotent to an ordinal and hence to a unique initial ordinal. If the set is infinite, that initial ordinal occurs in the aleph enumeration. Thus every infinite set has cardinality ${\aleph }_{\alpha }$ for a unique ordinal $\alpha$.

We next prove the square of an infinite cardinal. Suppose otherwise and let $\kappa$ be the least infinite cardinal for which ${\kappa }^2>\kappa$. Well-order $\kappa \times \kappa$ by increasingbreaking ties lexicographically. The predecessors of $(\alpha ,\beta )$ lie inside $\gamma \times \gamma$ for some $\gamma <\kappa$. If $\mu =|\gamma |$, then $\mu <\kappa$; by minimality, ${\mu }^2=\mu$ when $\mu$ is infinite, while the finite case is immediate. Every proper initial segment therefore has cardinality less than $\kappa$.

Recursively assign to each pair the least ordinal below $\kappa$ not already assigned to one of its predecessors. Such an ordinal always exists by the preceding bound, so this constructs an injection $\kappa \times \kappa \to \kappa$. The map $\alpha \mapsto (\alpha ,0)$ gives the reverse injection, and the Cantor-Schröder-Bernstein theorem yields

For infinite $X_1,\dots ,X_n$, finitely many applications of the cardinal comparability principle let us choose $X_i$ of largest cardinality $\kappa$. Thenwhere the final equality follows from infinite cardinal arithmetic. Consequently

For a countable family of pairwise different infinite cardinalities, the answer is yes. Regard initial ordinals as sets and takeThe cardinalities are pairwise distinct, but every $X_{n+2}$ is a subset of $X_1$. Hencewhich is the countable family of distinct infinite cardinalities with a largest member construction.

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

For infinite cardinals $\kappa$ and $\lambda$, the axiom of choice and the square of an infinite cardinal giveThe maximum is a lower bound for both operations, while both are bounded above by respectively $\mu +\mu$ and ${\mu }^2$, where $\mu =\max \{\kappa ,\lambda \}$.