I see that this question might be trivial but I can't seem to figure it out myself:
Suppose that $F:ON\to ON$ is a class function: that is, for every ordinal $\alpha$ there is unique ordinal $F(\alpha)$ corresponding. Suppose that $F$ satisfies following conditions:
- $\alpha < \beta \implies F(\alpha)<F(\beta)$ i.e. $F$ is monotonic
- $F(\lambda)=\bigcup\{F(\gamma):\gamma<\lambda\}$ for limit ordinals $\lambda$
Show that, then there exists $\alpha \in ON$ such that $F(\alpha)=\alpha$.
The hint suggested to prove $\alpha\le F(\alpha)$ for every ordinals, which can be done by induction on $\alpha$. Also suggested by hint, I constructed a "candidate" $\alpha$: By recursion, there is $f:\omega \to f[\omega]$ such that $f(0)=0$ and $f(n+1)=F(f(n))$, and then set $$\alpha=\bigcup f[\omega]$$
The problem is I don't see how this might help me. It is clearly suggested that above $\alpha$ works, but I have no information regarding $F(\alpha)$ at this point that proving $F(\alpha)\subset \alpha$ seems impossible.
It does seem like people don't want this thread to be closed. So let me provide an answer, that is a variation of the answer I previously provided here:
Let's say that a function $F \colon \operatorname{On} \to \operatorname{On}$ is normal iff it is strictly increasing (i.e. $\alpha < \beta$ implies $F(\alpha) < F(\beta)$) and continuous (i.e. $F(\lambda) = \sup_{\alpha < \lambda} F(\alpha)$ for all limit ordinals $\lambda$. Note that $\sup_{\alpha < \lambda} F(\alpha) = \bigcup_{\alpha < \lambda} F(\alpha) = \bigcup F[\lambda]$, by definition).
The 'Normal Function Lemma' states: Let $F \colon \operatorname{On} \to \operatorname{On}$ be a normal function. Then the class $F' := \{ \alpha \in \operatorname{On} \mid F(\alpha) = \alpha \}$ is closed (i.e. if $(\alpha_{\xi} \mid \xi < \theta)$ is a strictly increasing sequence of $F$-fixed points, then $\alpha := \sup_{\xi < \theta} \alpha_{\xi}$ is an $F$-fixed point) and unbounded (i.e. for all $\alpha \in \operatorname{On}$ there is some ordinal $\alpha < \beta$ such that $F(\beta) = \beta$). Moreover, for any regular ordinal (recall that regular ordinals are cardinals) $\omega \le \lambda$ there are arbitrarily large fixed points of $F$ of cofinality $\lambda$.
We will actually prove that for any sequence $(\alpha_{\xi} \mid \xi \le \lambda)$ such that $\lambda$ is a limit ordinal, $\alpha_{0}$ is an arbitrary ordinal, $\alpha_{\xi+1} := F(\alpha_{\xi})$ for all $\xi < \lambda$ and finally $\alpha_{\theta} := \sup_{\alpha < \theta}$ for all limit ordinals $\theta \le \lambda$, the ordinal $\alpha_{\lambda}$ is a fixed point of $F$ (whose cofinality is the cofinality of $\lambda$). In particular, this applies to your situation if we let $\alpha_{0} = 0$ and $\lambda = \omega$.
Proof. If $(\alpha_{\xi} \mid \xi < \theta)$ is a strictly increasing sequence in $F'$, then $$F(\sup_{\xi < \theta} \alpha_{\xi}) = \sup_{\xi < \theta} F(\alpha_{\xi}) = \sup_{\xi < \theta} \alpha_{\xi}.$$
It thus suffices to prove that for any regular cardinal $\lambda \geq \omega$ and any $\alpha_{0} \in \operatorname{On}$ there is some $\alpha_{0} < \alpha$ such that $F(\alpha) = \alpha$ and $\alpha$ has cofinality $\lambda$: Recursively construct a sequence $(\alpha_{\xi} \mid \xi < \lambda)$ as follows. $\alpha_{0}$ is as above and given $\alpha_{\xi}$, we let $\alpha_{\xi+1} := F(\alpha_{\xi})$. If $\theta < \lambda$ is a limit ordinal and $(\alpha_{\xi} \mid \xi < \theta)$ has already been constructed, let $\alpha_{\theta} := \sup_{\xi < \theta} \alpha_{\xi}$. Finally, let $\alpha := \sup_{\xi < \lambda} \alpha_{\xi}$. Since $(\alpha_{\xi} \mid \xi < \lambda)$ is strictly increasing, we have that $\operatorname{cf}(\alpha) = \operatorname{cf}(\lambda) = \lambda$ and furthermore, by the construction of $(\alpha_{\xi} \mid \xi < \lambda)$, $$ \begin{eqnarray*} F(\alpha) &= \sup_{\beta < \alpha} F(\beta) \\ &= \sup_{\xi < \lambda} F(\alpha_{\xi}) \\ &= \sup_{\xi < \lambda} \alpha_{\xi +1} \\ &= \alpha. \end{eqnarray*} $$ Since $\alpha_{0} < \alpha_{1} < \alpha$, the claim follows. QED