Ordinal $\alpha$ such that $\alpha = \omega_{\alpha}$?

Question

I am asked whether or not there exists an ordinal $\alpha$ such that $\omega_{\alpha}$ where we define:

1). $\omega_{0} = \omega$

2). $\omega_{\alpha+1} = \gamma(\omega_{\alpha})$

3). $\omega_{\lambda} = sup\{\omega_{\alpha} \mid \alpha < \lambda\}$ for a non-zero limit ordinal $\lambda$

Where, if I remember correctly, $\gamma(\alpha)$ is the least ordinal that cannot be injected into $\alpha$

Given these definitions, and the question of finding such an ordinal $\alpha$, the natural thought is something along the lines of $\alpha = \omega_{\omega_{\omega_{\ddots}}}$, but I'm not even sure this object is a well defined ordinal.

Trying to define it in terms of the recursive relationship, I just end up with it inside of its own definition.

Given this, does there then exist such an ordinal $\alpha$? I feel as if there can't be anything else, but experience has told me that maths can have some bizarre counter examples, so I don't really know what to think anymore.

Answer 1

The map $$\alpha\mapsto\omega_\alpha$$ has two important basic properties: it's increasing $(\alpha<\beta\implies\omega_\alpha<\omega_\beta$) and continuous at limit ordinals (for limit $\lambda$ we have $\omega_\lambda=\sup\{\omega_\beta:\beta<\lambda\}$). Your question is then answered by the following fundamental result about infinitary combinatorics:

Suppose $F$ is increasing and continuous. Then $F$ has a fixed point - that is, there is some $\gamma$ such that $F(\gamma)=\gamma$.

Proof sketch: First, we show by induction that $F(x)\ge x$ for all $x$. Then, consider the supremum of the sequence $$0,F(0), F(F(0)), F(F(F(0))), ...$$

Answer 2

The answer is yes. The mapping function $\alpha\mapsto\omega_\alpha$ is both continuous and increasing, which means it must have a fixed point. We can find this fixed point by taking the supremum - $\sup\{0,\omega,\omega_\omega,\omega_{\omega_\omega},...\}$ and in fact if we start with any two ordinals $\alpha_{1,2}$ which aren't fixed points of the function and we know that there exists an ordinal $\alpha_1<\gamma<\alpha_2$ which is a fixed point of the function, then if we take the two supremums $\sup\{\alpha_1,\omega_{\alpha_1},\omega_{\omega_{\alpha_1}},...\}$ and $\sup\{\alpha_2,\omega_{\alpha_2},\omega_{\omega_{\alpha_2}},...\}$ we will in fact get different fixed points. If we take a function $F(\alpha)$ which maps $\alpha$ to the least $\omega$-fixed point above it, then we take the set $\{F(\alpha)|\alpha\in\text{Ord}\}$ and since we already know that for different vaues of $\alpha$ with fixed points inbetween we get different fixed points, then it follows that there are as many ordinals satisfying $\alpha=\omega_\alpha$ as there are ordinals.