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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.

Assume the axiom of choice. By the well-ordering theorem, every set is bijective with an ordinal, and hence with an initial ordinal. Any two ordinals are comparable, so for any sets $K,L$ either $K$ injects into $L$ or $L$ injects into $K$. Transitivity is immediate, and antisymmetry on cardinalities follows from the Cantor-Schröder-Bernstein theorem. Thus cardinal comparison is a total order.

Conversely, assume the cardinal comparability principle. Given any set $X$, let $h(X)$ be its Hartogs ordinal. By Hartogs theorem, there is no injection $h(X)\to X$. Comparability therefore supplies an injection $X\to h(X)$. Pulling the ordinal order back to $X$ gives a well-order of $X$. Hence every set can be well-ordered, and the well-ordering theorem implies the axiom of choice.

Apply the stated cardinal comparability principle to $x$ and its Hartogs ordinal $h(x)$. The Hartogs theorem rules out an injection $h(x)\to x$, so comparability supplies an injectionOrder $x$ by $u\prec v$ exactly when $j(u)\in j(v)$. This is the restriction of an ordinal well-order and therefore well-orders $x$. Hence every set can be well-ordered. The well-ordering theorem is equivalent to the axiom of choice: for a family of nonempty sets, well-order its union and choose the least member of each set. The comparison statement therefore implies choice.