¿Is, for $1\le n<\omega ,\;(\omega+n)^{\omega}=\omega^{\omega}$?

Question

As the title suggests, I am interested in calculating $(\omega+n)^{\omega}$.

My try goes like this: we know that:

$$(\omega+n)^{\omega}=\text{sup}\{(\omega+n)^k|\;k\in\omega\}$$

For instance, I would like to have some formula for $(\omega+n)^k$ which should be more manageable than its current form.

I proceed to prove, by induction on the natural numbers, that:

$$(\omega+n)^k=\omega^k+\dots+\omega^2n+\omega n+n$$

For if $k\in\omega$, and $k\not=0$, then:

$$(\omega+n)^{k+1}=(\omega^k+\dots+\omega^2n+\omega n+n)(\omega+n)=(\omega+n)^k=(\omega^k+\dots+\omega^2n+\omega n+n)\omega+(\omega^k+\dots+\omega^2n+\omega n+n)n$$

And it is not hard to prove, with another induction, that:

$$(\omega^k+\dots+\omega^2n+\omega n+n)n=\omega^kn+\dots+\omega^2n+\omega n+n$$ $$(\omega^k+\dots+\omega^2n+\omega n+n)\omega=\omega^{k+1}+\omega^{k-1}+\dots+\omega^2n+\omega n+n$$

And therefore;

$$(\omega^k+\dots+\omega^2n+\omega n+n)\omega+(\omega^k+\dots+\omega^2n+\omega n+n)n=\;=(\omega^{k+1}+\omega^{k-1}+\dots+\omega^2n+\omega n+n)+(\omega^kn+\dots+\omega^2n+\omega n+n)=\quad=\omega^{k+1}+(\omega^{k-1}+\dots+\omega^2n+\omega n+n+\omega^kn)+\dots+\omega^2n+\omega n+n=\qquad=\omega^{k+1}+\omega^kn+\dots+\omega^2n+\omega n+n$$

And therefore, $k+1$ also satisfies the condition, so the formula actually holds. However, what does:

$$\text{sup}\{\omega^k+\dots+\omega^2n+\omega n+n|\;k\in\omega\}$$

Actually equal to? Is it $\omega^{\omega}$? I am clueless at this point.

Thanks in advance for your time.

Answer 1

Yes, $\sup\{\omega^k+\cdots+\omega^2n+\omega n+n\,:\, k\in\omega\}=\omega^\omega$. If you are concerned about upper bounds, you may use the inequality $\omega^k+\cdots+\omega^2n+\omega n+n\le \omega^{k+1}$. Or directly $\omega^k+\cdots+\omega^2n+\omega n+n\le \omega^\omega$.

In general, for all ordinals $\alpha\ge1$, for all ordinals $\beta> \gamma_k>\gamma_{k-1}>\cdots> \gamma_1>\gamma_0$ and for all ordinals $\delta_{\gamma_i}<\alpha$, $i=0,\cdots,k$, it holds $$\alpha^\beta>\alpha^{\gamma_k}\cdot\delta_{\gamma_{k}}+\alpha^{\gamma_{k-1}}\cdot\delta_{\gamma_{k-1}}+\cdots+ \alpha^{\gamma_1}\cdot\delta_{\gamma_1}+\alpha^{\gamma_0}\cdot\delta_{\gamma_0}$$

A more "algebraic" way for the whole problem:

$$\omega^\omega\le (\omega+n)^\omega\le (\omega\cdot 2)^\omega\le (\omega^2)^\omega=\omega^{2\cdot \omega}=\omega^\omega$$

Of course, here one needs to prove/remember that the two rules that hold for ordinal exponentiation are $\alpha^\beta\alpha^\gamma=\alpha^{\beta+\gamma}$ and $(\alpha^\beta)^\gamma=\alpha^{\beta\cdot \gamma}$, whereas in general $(\alpha\cdot\beta)^\gamma\ne \alpha^\gamma\cdot\beta^\gamma$.

Answer 2

Clearly $\omega^\omega\le(\omega+n)^\omega$. Also, $(\omega+n)^\omega\le (\omega^2)^\omega=\omega^{(2\cdot \omega)}=\omega^\omega$, and the equality follows.

If you do not feel comfortable with the move from $(\omega^2)^\omega$ to $\omega^{(2\cdot \omega)}$, simply note the left-hand side is $\omega\cdot\omega\cdot\omega\cdot\dots$, where there are $2\cdot\omega=\omega$ factors, all equal to $\omega$.