Every natural number and $\omega$ are cardinals, every infinite cardinal is a limit ordinal, and on the natural numbers the cardinal operations are the published finite counting operations, with $\mid A\mid$ in the finite sense equal to $\mid A\mid$ in the cardinal sense

Work in ZF; no choice principle is used. Let $\mathbb{N}=\omega$ be the von Neumann naturals (The natural numbers $\mathbb{N}$ (von Neumann)), and let ${+}_{\mathbb{N}}$, ${\cdot }_{\mathbb{N}}$ and $m^n$ be the natural-number operations (Exponentiation of natural numbers, $m^n$, and its agreement with the integer power in $\mathbb{R}$ for the power). Then:

(a) Every natural number is a cardinal (Cardinal (initial ordinal) and cardinality), and $\omega$ is a cardinal.

(b) Every infinite cardinal is a limit ordinal (Successor and limit ordinals).

(c) One notation, one meaning. If $A$ is finite (Finite, countably infinite, countable, uncountable) then $A$ is well-orderable and the natural number $\mid A\mid$ of The cardinality $\mid A\mid$ of a finite set is the cardinal $\mid A\mid$ of A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used.

(d) One arithmetic. For $m,n\in \omega$, read as cardinals,

$m\oplus n=m{+}_{\mathbb{N}}n,m\otimes n=m{\cdot }_{\mathbb{N}}n,m^n\text{ as a cardinal }=m^n\text{ as a natural number},$

the natural-number power being that of Exponentiation of natural numbers, $m^n$, and its agreement with the integer power in $\mathbb{R}$ and the cardinal exponential being defined in ZF here because the function set it counts is finite. Moreover $m\oplus n$ and $m\otimes n$ are also the ordinal sum and product of $m$ and $n$ (On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$, $\cdot$ and exponentiation, and for naturals $m,n$ the ordinal $m+n$ and $m\cdot n$ are the natural-number sum and product). No agreement is claimed between the cardinal power and the ordinal power; On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$, $\cdot$ and exponentiation, and for naturals $m,n$ the ordinal $m+n$ and $m\cdot n$ are the natural-number sum and product itself claims none for exponentiation, and none is needed below.

Why this theorem is not optional. Two published items already write $\mid A\mid$: The cardinality $\mid A\mid$ of a finite set, where the value is a natural number and the definition applies to finite sets only, and Cardinal (initial ordinal) and cardinality, where the value is an initial ordinal. On a finite set both apply. Without claim (c) the same symbol would carry two meanings and every finite computation on this page would be ambiguous; with it there is one meaning, and a natural number may be read as a cardinal without comment.

The same holds for $+$ on $\omega$, twice over. Claim (d) closes the second half of a dictionary whose first half is On $\omega$ the ordinal $+$ and $\cdot$ are the Peano operations: $\omega$ is closed under ordinal $+$, $\cdot$ and exponentiation, and for naturals $m,n$ the ordinal $m+n$ and $m\cdot n$ are the natural-number sum and product: the Peano sum, the ordinal sum and the cardinal sum of two natural numbers are one natural number. The three operations diverge immediately above $\omega$, and that divergence is the reason Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations writes $\oplus$ and $\otimes$ rather than $+$ and $\cdot$.

What claim (b) is for. It is used wherever an argument needs to take suprema below an infinite cardinal, or to know that $\max (\xi ,\eta )+1$ stays below $\kappa$ when $\xi ,\eta <\kappa$. The proof is the shift bijection that Cardinal (initial ordinal) and cardinality already describes for ${\omega }^{+}$, carried out at an arbitrary infinite cardinal: prepending or removing one point does not change the size of an infinite well-ordered set, so a successor ordinal above $\omega$ is never an initial ordinal.