Prove that if: $g: \omega_1\rightarrow\omega_1$ then for some $\alpha<\omega_1$: $\forall \beta<\alpha: g(\beta)<\alpha$ Prove that if $g: \omega_1\rightarrow\omega_1$ then for some $0<\alpha<\omega_1$,  $\forall \beta<\alpha: g(\beta)< \alpha$
How should I prove this theorem. I don't even know how to start
 A: We generally need to work with the things we have. So what do we have? we have two things:
The first thing we are given are well-orders. Ordinals are well-ordered. And well-orders invite recursion. The second thing, is that we are given $\omega_1$. Why not $\omega$? Why not $\omega+\omega$ or $\omega^3$?
Well, the function $g(n)=n+1$ is a function from $\omega$ to itself, but no $n<\omega$ satisfies this closure property. I leave it to you to find out why $\omega+\omega$, or $\omega^3$, or any countable ordinal, are not going to work either. There will be some function that avoids this condition.
So what is so different about $\omega_1$? Well. Two things, it is uncountable, and its cofinality is uncountable (which in this case means that it is regular). One of these properties should be used, therefore.

So let's try the recursion. Start with $\alpha_0=0$, and we should just try to approximate the closure point $\alpha$. What does that mean? Well, it means that we will construct a sequence, such that each ordinal is more and more closed under $g$.
So $\alpha_1$ will be the least ordinal such that $g(\alpha_0)=g(0)<\alpha_1$. This way, we will ensure that $g(0)<\alpha$. 
Next, we need to make sure that $g(\beta)<\alpha$ for all $\beta\leq\alpha_1$. So $\alpha_2=\sup\{g(\beta)\mid\beta\leq\alpha_1\}$.
From here, you should be able to continue and define the sequence $\alpha_n$. Finally, take $\alpha=\sup\{\alpha_n\mid n<\omega\}$, and prove that it is a closure point as wanted.

So what did we use about $\omega_1$? Not that it is uncountable, as much as it is closed under countable limits. Namely, that its cofinality is uncountable. As an exercise, find a function $g\colon\omega_\omega\to\omega_\omega$ which has no closure point.
A: The negation of the statement you want to prove is (think about it, it's just standard logic):
$$\forall 0 < \alpha < \omega_1: \exists \beta < \alpha: g(\beta) \ge \alpha$$
So assume the above holds.
For definiteness we can define (by well-orderedness)  $g(\alpha) = \min \{\beta < \alpha: g(\beta) \ge \alpha\}$.
So $g: \omega_1\setminus\{0\} \to \omega_1$ is a regressive function. 
Now apply the pressing down lemma (or Fodor's lemma), to derive a contradiction. 
