Ordinal Exponentiation versus Cardinal

Two exponentiations, same notation

The symbol $\alpha^{\beta}$ means completely different things depending on whether $\alpha,\beta$ are read as ordinals (order types, order matters) or as cardinals (sizes, only count matters). On the natural numbers they agree; on infinite arguments they diverge sharply. $\omega$ and $\aleph_0$ name the same underlying set $\{0,1,2,\ldots\}$ but as an ordinal vs. a cardinal respectively, which is exactly where exam traps live.

Ordinal exponentiation by transfinite recursion

For ordinals, $\alpha^{\beta}$ is defined by recursion on the exponent $\beta$:

• Base: $\alpha^{0} = 1$.
• Successor: $\alpha^{\beta+1} = \alpha^{\beta}\cdot\alpha$.
• Limit: $\alpha^{\lambda} = \sup_{\xi<\lambda} \alpha^{\xi}$ for a limit ordinal $\lambda$.

The limit clause is the whole story: at a limit stage you take a supremum of everything below, not a "new" value. This is what makes ordinal powers behave so differently from arithmetic.

The collapse: finite base to the $\omega$ power

Let $a$ be any finite ordinal with $a\ge 2$. Since $\omega$ is a limit ordinal,

$a^{\omega} = \sup_{n<\omega} a^{n} = \sup\{a^{0}, a^{1}, a^{2}, \ldots\}.$

Every $a^{n}$ is again a finite ordinal (a natural number), so this is the supremum of an unbounded set of natural numbers, which is exactly $\omega$:

$2^{\omega} = 3^{\omega} = a^{\omega} = \omega.$

A finite base raised to $\omega$ collapses back to $\omega$. By contrast, raising $\omega$ to a finite power grows: $\omega < \omega^{2} < \omega^{3} < \cdots$, and iterating gives the much larger $\omega^{\omega}$. So the rule of thumb is: finite$^{\omega}=\omega$, but $\omega^{\text{finite}}>\omega$.

The same expression as cardinals

Read $2,3,\aleph_0$ as cardinals and exponentiation counts function spaces, $|B|^{|A|}=|B^{A}|$. Now Cantor's theorem dominates:

$2^{\aleph_0} = 3^{\aleph_0} = \aleph_0^{\aleph_0} = \mathfrak{c} > \aleph_0,$

the continuum, which is strictly bigger than $\aleph_0$. So a finite base to the $\aleph_0$ power explodes to uncountable, the exact opposite of the ordinal collapse.

Note the last row: the ordinal $\omega^{\omega}$ is a countable ordinal (it is still a well-order of a countable set), even though the cardinal $\aleph_0^{\aleph_0}$ is the continuum. Ordinal arithmetic never increases cardinality on countable inputs.

Worked mini-example

Decide whether $\omega < 3^{\omega}$ and whether $\omega < \omega^{3}$ as ordinals. For the first, $3^{\omega}=\sup\{1,3,9,27,\ldots\}=\omega$, so $\omega < 3^{\omega}$ is false (they are equal). For the second, $\omega^{3}=\omega\cdot\omega\cdot\omega$ and already $\omega^{2}>\omega$, so $\omega<\omega^{3}$ is true. Only the second strict inequality holds.

Pitfall: Before evaluating $\alpha^{\beta}$, decide ordinal or cardinal. If the base is finite and the exponent is $\omega$, ordinals give $\omega$ (collapse via supremum) but cardinals give $2^{\aleph_0}=\mathfrak c$ (explosion via Cantor). Mixing the two is the single most common mistake on these problems.