Is the class of all ordinals producible through ordinal collapsing functions countable? I was recently learning about ordinal collapsing functions, which are a systematic way to produce very large countable ordinals.  Before I talk about collapsing functions, I will briefly explain some ordinals:
$$\epsilon_0=\omega^{\omega^{\omega^{\dots}}}\bigg\}\omega\text{ powers going up}$$
$$\epsilon_1=\epsilon_0^{\epsilon_0^{\epsilon_0^{\dots}}}\bigg\}\omega\text{ powers going up}$$
$$\epsilon_2=\epsilon_1^{\epsilon_1^{\epsilon_1^{\dots}}}\bigg\}\omega\text{ powers going up}\\\vdots$$
$\epsilon_\omega$ is thus the limit of this sequence, and from there, we get another ordinal:
$$\zeta_0=\epsilon_{\epsilon_{\epsilon_{\ddots}}}$$
This generalizes into the Veblen function.

We can now tackle the ordinal collapsing function:
$$C(\alpha)=C(\alpha)_0\cup C(\alpha)_1\cup C(\alpha)_2\cup\dots$$
where
$$C(\alpha )_{{n+1}}=C(\alpha )_{n}\cup \{\beta _{1}+\beta _{2},\beta _{1}\beta _{2},{\beta _{1}}^{{\beta _{2}}}:\beta _{1},\beta _{2}\in C(\alpha )_{n}\}\cup \{\psi (\beta ):\beta \in C(\alpha )_{n}\land \beta <\alpha \}$$
$$C(\alpha)_0=\{0,1,\omega,\Omega\}$$
where $\Omega$ is larger than any ordinal we are going to produce (sometimes we set it to be $\omega_1$)
$\psi(\alpha)$ is then the smallest ordinal not in $C(0)$.  $\psi(0)$ is seen above, and it is
$$\psi(0)=\epsilon_0$$
Then, $\psi(1)$ is clearly $\epsilon_1$.
We can continue this, and we find that for any $\alpha<\zeta_0$, we have
$$\psi(\alpha)=\epsilon_\alpha$$
But the interesting thing is that we can never surpass $\zeta_0$ due to the forced restriction of "finite combination of..." which is simply beyond the power of $\epsilon_\alpha$, that is
$$\psi(\alpha)=\zeta_0\forall\zeta_0\le\alpha<\Omega$$ and so this is where we pull in the magic $\Omega$:
$$\psi(\Omega)=\zeta_0$$
This now let's us pocket $\zeta_0$ into our $C(\Omega)$ and we can continue on.  We get stuck once again at $\zeta_1$, which we bail out using
$$\psi(\Omega\cdot2)=\zeta_1$$
and so on.  Eventually, we reach $\psi\left(\Omega^{\Omega^\Omega}\right)$ which is the supremum of Veblen ordinals.
And after that, we reach the supremum of the 'first' ordinal collapsing function, since we can never reach $\psi\left(\Omega^{\Omega^{\Omega^{\Omega^{\dots}}}}\right)$.
We thus introduce $\psi_1(\alpha)$, which is the same as the previous with the only addition that $C_1(0)=\{0,1,\omega,\Omega,\Omega_1\}$, which by using $\Omega_1$, we may now surpass $\psi\left(\Omega^{\Omega^{\Omega^{\Omega^{\dots}}}}\right)$.  Likewise, we'll hit a limit, and start all over with $\psi_2(\alpha)$.
And we can go on and on, and perhaps even do some crazy things like
$$\psi_{\psi_\omega(0)}(0)$$

Anyways, I was wondering if the class of all ordinals producible through ordinal collapsing functions is countable or uncountable.  I imagine that its countable, though a friend of mine believes it may be uncountable.  Does anyone know?

 A: The set of ordinals expressible in a notation system using an ordinal collapsing function will be countable, because every ordinal will be expressed using a finite set of starting ordinals and a finite set of operations some finite number of times.  For example, using the ordinal notation system you give above, we can let 
$$C_0=\{0,1,\omega,\Omega\}$$
and
$$C_{{n+1}}=C_{n}\cup \{\beta _{1}+\beta _{2},\beta _{1}\beta _{2},{\beta _{1}}^{{\beta _{2}}}, \psi(\beta):\beta _{1},\beta _{2}\in C_{n}\}$$
Then
$$C = \bigcup_n C_n$$
is the set of all ordinals expressible in this notation system.  Each $C_n$ is finite, so $C$ will be countably infinite.  The same reasoning goes for any other ordinal collapsing function, except sometimes the starting set or set of operations could be countably infinite; this still leads to a countably infinite set of ordinals.
I feel I must mention that the definition of $\psi_1$ you give is nonstandard.  A typical ordinal notation for higher cardinals will be something like:
$$C_0 (\nu, \alpha) = \Omega_\nu \cup \lbrace 0 \rbrace$$
$$C_{n+1} (\nu, \alpha) = \lbrace \beta + \gamma, \varphi(\beta, \gamma), \psi_\mu(\delta) | \beta, \gamma, \delta \in C_n (\nu, \alpha); \delta < \alpha; \nu \le \mu  \rbrace $$
$$C (\nu, \alpha) = \bigcup_{n = 1}^{\infty} C_n (\nu,\alpha) $$
$$\psi_\nu (\alpha) = \min \lbrace \beta | \beta \notin C(\nu,\alpha) \rbrace $$
Note in particular that $C_0(1,\alpha)$ contains $\Omega_1$ as a subset, and therefore $\psi_1(\alpha)$ is always uncountable (of cardinality $\Omega_1$).  So $\psi_1$ collapses ordinals of large cardinality down to ordinals of cardinality $\Omega_1$, and those large uncountable ordinals help $\psi_0$ produce really large countable ordinals.  This is much more powerful than the system you describe; for example, $\psi_{\psi_\omega(0)}(0)$ in your system is going to be much less than $\psi_0(\Omega_3)$ in the standard system.
In this system, the $C_n(\nu,\alpha)$'s will be uncountable for positive $\nu$, so if this is what you meant by "produces" then you can get uncountable sets from ordinal collpsing functions.  But, you cannot actually express all the ordinals in $C_n(\nu,\alpha)$; the set of expressible ordinals will be countable, from the previous argument.
