The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$

The successor cardinal. For a cardinal $\kappa$ (Cardinal (initial ordinal) and cardinality) write

${\kappa }^{+}:=\mathrm{\aleph }(\kappa ),$

the Hartogs number of $\kappa$ (Hartogs: an ordinal that does not inject into a given set). By For every set $A$ the Hartogs number $\mathrm{\aleph }(A)$ is a cardinal, and for every cardinal $\kappa$ it is the least cardinal strictly above $\kappa$; this is a theorem of ZF this is the least cardinal strictly above $\kappa$, and its existence is a theorem of ZF.

Notation rule, in force on this page and its companion. The superscript $+$ means the successor cardinal only on a cardinal letter $\kappa ,\lambda ,\mu$ or an aleph. On an ordinal letter $\alpha ,\beta ,\gamma ,\xi ,\eta$ the superscript $+$ keeps its published meaning, the ordinal successor ${\alpha }^{+}=\alpha \cup \{\alpha \}$ of Ordinal (von Neumann). The two never agree on an infinite cardinal: $\kappa \cup \{\kappa \}$ is a successor ordinal and therefore not a cardinal at all, while ${\kappa }^{+}$ is much larger. To keep the reader out of the collision, everything below writes $\alpha +1$ for the ordinal successor and reserves ${\kappa }^{+}$ for the cardinal one.

The alephs. By The clauses at $0$, at a successor and at a limit determine exactly one operation $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$, in ZF, and — assuming the Axiom of Choice — exactly one operation $\alpha \mapsto {\mathrm{\beth }}_{\alpha }$; each value is an infinite cardinal, each is strictly increasing and continuous at limits, and $\alpha \le {\mathrm{\aleph }}_{\alpha }$ there is exactly one operation $\alpha \mapsto {\mathrm{\aleph }}_{\alpha }$, defined at every ordinal (Ordinal (von Neumann)), with

${\mathrm{\aleph }}_0=\omega ,{\mathrm{\aleph }}_{\alpha +1}={\mathrm{\aleph }}_{\alpha }^{+},{\mathrm{\aleph }}_{\lambda }=\sup \{{\mathrm{\aleph }}_{\alpha }:\alpha \in \lambda \}(\lambda \text{ a limit ordinal}),$

the limit clause being taken over limit ordinals in the sense of Successor and limit ordinals, and $\sup$ being $\bigcup$ applied to a set of ordinals. Every ${\mathrm{\aleph }}_{\alpha }$ is an infinite cardinal, the operation is strictly increasing, and $\alpha \le {\mathrm{\aleph }}_{\alpha }$; all of this is that corollary, and all of it is ZF.

The beths, assuming the Axiom of Choice, are the parallel operation

${\mathrm{\beth }}_0=\omega ,{\mathrm{\beth }}_{\alpha +1}=2^{{\mathrm{\beth }}_{\alpha }},{\mathrm{\beth }}_{\lambda }=\sup \{{\mathrm{\beth }}_{\alpha }:\alpha \in \lambda \}(\lambda \text{ a limit ordinal}),$

with $2^{\kappa }$ the cardinal power of Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations.

Successor and limit cardinals. An infinite cardinal $\kappa$ is a successor cardinal when $\kappa ={\lambda }^{+}$ for some infinite cardinal $\lambda$, and a limit cardinal otherwise. So ${\mathrm{\aleph }}_{\alpha +1}$ is a successor cardinal for every $\alpha$, and ${\mathrm{\aleph }}_0=\omega$ is a limit cardinal, there being no infinite cardinal below it.

The two identifications.

${\mathrm{\aleph }}_0=\omega ,{\mathrm{\aleph }}_1={\omega }_1\mathrm{.}$

The first is the base clause. The second holds because ${\mathrm{\aleph }}_1={\mathrm{\aleph }}_0^{+}=\mathrm{\aleph }(\omega )$, and $\mathrm{\aleph }(\omega )$ is by definition the first uncountable ordinal ${\omega }_1$ (The first uncountable ordinal ${\omega }_1:=\mathrm{\aleph }(\omega )$); ${\omega }_1$ is uncountable, every ordinal below it is at most countable, it is a cardinal and a limit ordinal, and its existence is a theorem of ZF independently confirms that ${\omega }_1$ is a cardinal, is uncountable (Finite, countably infinite, countable, uncountable), and has every ordinal below it at most countable, which is the same thing said in the language of the ordinal development.

Why the aleph notation is introduced at all, when $\omega$ and ${\omega }_1$ already exist. The subscript is an ordinal index into the cardinals: it makes "the $\alpha$-th infinite cardinal" a term of the language, so that statements such as "${\mathrm{\aleph }}_{\alpha +1}$ is regular" and "${\mathrm{\aleph }}_{\omega }$ is singular" can be quantified over $\alpha$. The published ordinal development deliberately avoids the notation, writing $\omega$ and ${\omega }_1$ throughout, precisely so that an ordinal computation there is never silently read as a cardinal one; that convention is recorded in Ordinal ${\alpha }^{\beta }$ and cardinal ${\kappa }^{\lambda }$ are different operations that share one notation and is not disturbed here. On this page the reverse convention is in force: an aleph is always a cardinal, and an exponential is always the cardinal one.

$\mathrm{\aleph }$ with an argument and $\mathrm{\aleph }$ with a subscript are different things. $\mathrm{\aleph }(A)$ is the Hartogs number of a set $A$, defined for every set in ZF (Hartogs: an ordinal that does not inject into a given set); ${\mathrm{\aleph }}_{\alpha }$ is the $\alpha$-th infinite cardinal. They agree in the one case that matters, $\mathrm{\aleph }({\mathrm{\aleph }}_{\alpha })={\mathrm{\aleph }}_{\alpha +1}$, which is the successor clause, and the same symbol is used because the notation is Hartogs' own.

Where the beths sit. ${\mathrm{\beth }}_0={\mathrm{\aleph }}_0$, by the two base clauses. The two hierarchies then climb by different rules: the aleph step takes the least cardinal strictly above, and the beth step takes the power. Whether they nevertheless agree at every index is the generalised continuum hypothesis, which is not decided by the axioms in use here and is asserted nowhere on this page or its companion.