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The cardinal numbers defined by ${\beth }_0={\aleph }_0$, ${\beth }_{\alpha +1}=2^{{\beth }_{\alpha }}$, and ${\beth }_{\lambda }={\sup }_{\alpha <\lambda }{\beth }_{\alpha }$ at limit ordinals are the beth numbers. In particular ${\beth }_1=|\mathcal{P}(\omega )|$, the cardinality of the continuum. This notation distinguishes iterated power sets from the enumeration by Aleph numbers of well-orderable infinite cardinal numbers.

If $\alpha$ is a nonzero limit ordinal and $({\gamma }_{\xi }{)}_{\xi <\alpha }$ is strictly increasing, then $\mathrm{cf}({\sup }_{\xi <\alpha }{\gamma }_{\xi })=\mathrm{cf}(\alpha )$. A cofinal subsequence of the index order gives one inequality. Conversely, a cofinal family in the supremum determines a cofinal family of indices by choosing an index past each of its values. This also explains the cofinality of limit-indexed aleph numbers.

An infinite cardinal number is a limit cardinal if it is not a successor cardinal; equivalently it is an aleph number with zero or a limit ordinal as index. Uncountable limit cardinals are either singular cardinals or weakly inaccessible cardinals. This concerns succession among cardinals, rather than merely being a limit ordinal.

The least index is $\boxed{\alpha ={\omega }_1}$, withIndeed, for every nonzero limit ordinal $\beta$, continuity of the aleph number enumeration and the cofinality of an increasing ordinal supremum giveIf $\beta <{\omega }_1$ is a nonzero limit ordinal, this cofinality is $\omega$, so ${\aleph }_{\beta }$ is a singular cardinal. The other infinite cardinal numbers below ${\aleph }_{{\omega }_1}$ are successor-indexed or ${\aleph }_0$, and are regular cardinals. For a successor cardinal ${\mu }^{+}$, a cofinal sequence of length at most $\mu$ would express ${\mu }^{+}$ as a union of at most $\mu$ sets of size at most $\mu$, contradicting infinite cardinal arithmetic.

There are ${\aleph }_1$ nonzero countable limit ordinals, and each countable initial segment contains only countably many. In increasing order they therefore have order type ${\omega }_1$. These are precisely the indices of the uncountable singular cardinals below ${\aleph }_{{\omega }_1}$. Hence every ${\sigma }_{\alpha }$ with $\alpha <{\omega }_1$ has countable cofinality, and the next one is ${\aleph }_{{\omega }_1}$, whose cofinality is ${\omega }_1<{\aleph }_{{\omega }_1}$.

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.

An initial ordinal is an ordinal $\kappa$ that is not equipotent to any smaller ordinal. Equivalently, it is the least ordinal having its cardinality. The aleph numbers enumerate the infinite initial ordinals:${\aleph }_{\alpha +1}$ is the least initial ordinal greater than ${\aleph }_{\alpha }$, and at a limit $\lambda$, ${\aleph }_{\lambda }$ is the least initial ordinal greater than every ${\aleph }_{\beta }$ for $\beta <\lambda$.

Each ${\aleph }_{\alpha }$ is infinite and initial by construction. Conversely, let $\kappa$ be an infinite initial ordinal. The infinite initial ordinals below $\kappa$ form a set and therefore have an ordinal order type $\alpha$. The recursive enumeration above lists exactly those predecessors before stage $\alpha$, so its next value is ${\aleph }_{\alpha }=\kappa$.

Read literally, “has cardinality ${\aleph }_{\alpha }$” is too weak: $\omega +1$ has cardinality ${\aleph }_0$ but is not initial. The valid characterization isIf “has cardinality” in the question means “is the cardinal represented by,” this is precisely the requested statement.