Cantor normal form

Remarks

The proof is by induction on $\alpha$. Let $\beta_1$ be the largest ordinal with $\omega^{\beta_1} \le \alpha$. Division with remainder gives $\alpha = \omega^{\beta_1} \cdot c_1 + \rho$ with $0 < c_1 < \omega$ and $\rho < \omega^{\beta_1}$, and the induction hypothesis applies to $\rho$. Example: $\omega^2 + \omega \cdot 3 + 5$ is already in normal form, with exponents $2 > 1 > 0$. The normal form has $\beta_1 \le \alpha$, and equality can occur: for $\varepsilon_0$, the least ordinal with $\omega^{\varepsilon_0} = \varepsilon_0$, the normal form is $\omega^{\varepsilon_0}$. For ordinals below epsilon nought, the exponents are smaller than the ordinal, and repeating the normal form on them gives a finite notation. The same holds with any base $b \ge 2$ in place of $\omega$, with coefficients $0 < c_i < b$.

Going deeper

For a student

The Cantor normal form makes ordinal arithmetic computable on notations. Two ordinals in normal form are compared by their leading terms, as polynomials are compared by degree. Iterating the form on exponents gives the standard notation system for ordinals below $\varepsilon_0$, which appears in Gentzen's consistency proof for Peano arithmetic and in Goodstein's theorem. Beyond $\varepsilon_0$ one needs stronger notations, such as those of the Veblen function.