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

My try goes like this: we know that:


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.

  • 1
    $\begingroup$ If $x, y$ are ordinals and $y=\cup y>0$ then $x^y=\sup_{k<y}x^k.$ So $(\omega +n)^{\omega}=$ $\sup_{k<\omega}(\omega +n)^k\le$ $ \sup_{k\in \omega}(\omega^2)^k=$ $=\sup_{k<\omega}\omega ^{2k}=$ $\omega^{\omega}.$ $\endgroup$ – DanielWainfleet May 2 at 9:32

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 $(\alpha\cdot\beta)^\gamma\ne \alpha^\gamma\cdot\beta^\gamma$.

  • $\begingroup$ Yes, I exactly thought about proving that $\omega^k+...+\omega^2n+\omega n+n\le\omega^{k+1}$ just when I posted the answer... your shorter answer was pretty enlightening though. $\endgroup$ – Akerbeltz May 1 at 22:44
  • $\begingroup$ Oh, I see you did the same thing I suggested. :-) $\endgroup$ – Andrés E. Caicedo May 1 at 23:26

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 $\omega$.

  • $\begingroup$ I know the basic properties of the exponentiation of ordinals. Thanks for your answer. $\endgroup$ – Akerbeltz May 1 at 23:23
  • 3
    $\begingroup$ These kind of move tends to simplify many messy computations: rather than trying to proceed directly with the given quantities, replace them with more manageable ones. Even if you do not end up with an equality, it may result in useful bounds. $\endgroup$ – Andrés E. Caicedo May 1 at 23:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.