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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 four principles are:

In the last assertion the upper bound belongs to the partially ordered set; it need not belong to the chain.

Assume the well-ordering theorem. Every set with a chosen well-order has an order isomorphism to a unique ordinal. Any two ordinals are comparable, and the smaller embeds into the larger as an initial segment. Transporting that embedding back to the original sets proves the cardinal comparability principle.

Conversely, assume the cardinal comparability principle and let $X$ be any set. The Hartogs theorem, provable in ZF, supplies an ordinal $h(X)$ for which no injection $h(X)\hookrightarrow X$ exists. Comparison forces an injection $X\hookrightarrow h(X)$. Pulling back the well-order on its image gives a well-order on $X$. ThereforeThe assumed ordinal facts are that every well-ordered set has an ordinal order type and that any two ordinals are comparable; the other assumed theorem is precisely the stated Hartogs theorem.

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.