# What ordinal is bigger ? $\omega^{\omega+1}, (\omega+1)^{\omega}$

elementary-set-theory

What ordinal is bigger ?

$$\omega^{\omega+1}, (\omega+1)^{\omega}$$

I think $$\omega^{\omega+1}>(\omega+1)^{\omega}$$

$$\omega^{\omega+1}=\bigcup_{\alpha <\omega+1} \omega^\alpha >\bigcup_{\alpha <\omega} \omega^\alpha = \omega^{\omega}$$

$$(\omega+1)^{\omega}=\bigcup_{\alpha <\omega} (\omega+1)^\alpha$$

Is it correct $$(\omega)^{\omega}=(\omega+1)^{\omega}$$?

I have no idea how to prove this problem.

I'd be grateful for your help!

Note that for all ordinals $$\alpha$$, we have $$\alpha^\omega = \sup\limits_{n < \omega} \alpha^n$$.

We wish to prove that for all $$n \in \mathbb{N}$$, $$(\omega + 1)^n < \omega^n \cdot 2$$.

Let us prove this by induction. The base case is immediate.

We now proceed with the inductive step. Suppose $$(\omega + 1)^j < \omega^j \cdot 2$$. Then

$$$$\begin{split} (\omega + 1)^{j + 1} &= (\omega + 1)^j \cdot (\omega + 1) \\ &\leq (\omega^j \cdot 2) \cdot (\omega + 1) \\ &= (\omega^j \cdot 2) \cdot \omega + (\omega^j \cdot 2) \cdot 1 \\ &= \omega^j \cdot (2 \cdot \omega) + \omega^j \cdot 2 \\ &= \omega^j \cdot \omega + \omega^j \cdot 2 \\ &= \omega^{j + 1} + \omega^j \cdot 2 \\ &< \omega^{j + 1} + \omega^j \cdot 2 + \omega^j \cdot \omega \\ &= \omega^{j + 1} + \omega^j \cdot (2 + \omega) \\ &= \omega^{j + 1} + \omega^j \cdot \omega \\ &= \omega^{j + 1} \cdot 2 \end{split}$$$$

Therefore, we see that for all $$n \in \mathbb{N}$$, we have $$\omega^n \leq (\omega + 1)^n < \omega^n \cdot 2 < \omega^n \cdot \omega = \omega^{n + 1}$$.

From here, it follows that $$(\omega + 1)^\omega = \omega^\omega$$.

Now the question is how to compare $$\omega^\omega$$ with $$\omega^{\omega + 1} = \omega^\omega \cdot \omega$$. This is quite easy using the distributive property, since we see that $$\omega^\omega \cdot \omega = \omega^\omega \cdot (1 + \omega) = \omega^\omega + \omega^\omega \cdot \omega$$. Therefore, we see that $$\omega^\omega < \omega^{\omega + 1}$$.

Thus, we see that the larger of the two ordinals is in fact $$\omega^{\omega + 1}$$.