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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?

Simply Beautiful Art
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    It's absolutely countable. You can describe all of these ordinals using a finite string of symbols in a finite language (specifically, English). There are only countably many such strings. – Reese Johnston Mar 07 '17 at 23:08
  • @Reese Could you prove that to me? Its not as obvious as I'd like, and I didn't restrict this to a finite amount of function iterations by the way. – Simply Beautiful Art Mar 07 '17 at 23:11
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    Every finite string of symbols in a finite language can be coded as a finite integer: represent each symbol by a positive natural, and take the product of $p_i^{a_i}$, where $p_i$ is the $i$th prime and $a_i$ is the $i$th symbol. As for the iterations: where do you stop? If you allow any ordinal number of iterations, then every ordinal is reachable this way. If you require that the number of iterations itself be describable, we're back to countable. – Reese Johnston Mar 07 '17 at 23:25
  • I lost the thread at $\psi(\Omega)$. Up until that point it looked like $\psi$ was a strictly increasing ordinal function, which implies $\psi(\alpha)\ge \alpha$. But then how can $\psi(\Omega)$ possibly be countable if you can set $\Omega=\omega_1$? Did you mean $\Omega=\omega_1^{CK}$? – hmakholm left over Monica Mar 07 '17 at 23:34
  • @HenningMakholm This part and a bit above it may help you. Indeed, note that the definition involving finite combinations restricts the "strictly increasing nature" in some odd ways. – Simply Beautiful Art Mar 07 '17 at 23:40
  • @SimplyBeautifulArt: No, that doesn't deconfuse me. If each ordinal adds a new number to $C(\alpha+1)$ that wasn't in $C(\alpha)$, then by the time we reach $C(\omega_1)$ there will be uncountably many different ordinals in $C$, which means that their supremum is at least $\omega_1$ -- and every ordinal that is smaller than one of those can, by definition, be made by a finite combination of some of the uncountably many elements of $C$. – hmakholm left over Monica Mar 07 '17 at 23:46
  • @HenningMakholm No. How are you going to construct $\zeta_0$ using a finite amount of addition, multiplication, and exponentiation of the previous elements? By definition, if we can't make it, we can't add it in. Thus, $\zeta(\alpha)=\zeta_0$ occurs first at $\alpha=\Omega$ by a sort of definition. (or something like that, sorry, I just started trying to wrap my head around this since yesterday) – Simply Beautiful Art Mar 07 '17 at 23:50
  • @SimplyBeautifulArt: By definition whatever we add is something we cannot make. Sooner or later $\zeta_0$ is going to be added, simply by virtue of all smaller ordinals already having been made or added, such that $\zeta_0$ is the smallest ordinal not yet made. So $\zeta_0$ must become $\psi(\alpha)$ for some $\alpha$ not larger than $\psi$. And from that point onwards, of course it is easy to "make" $\zeta_0$ simply because it is already in $C$. – hmakholm left over Monica Mar 07 '17 at 23:55
  • Actually: it seems that your definition "$\psi(1)$ is the first ordinal not representable from $C(1)$" is something different from the Wikipedia article you link to describes. It says that $\psi(\alpha)$ must be made from $C(0)$ (rather than $C(\alpha)$, as your description suggests) using ordinal operations and a restricted $\psi$ function. – hmakholm left over Monica Mar 07 '17 at 23:59
  • @HenningMakholm What do you mean? I see this: Then $\psi (\alpha )$ is defined as the smallest ordinal not belonging to $ C(\alpha )$. – Simply Beautiful Art Mar 08 '17 at 00:01
  • @SimplyBeautifulArt: What you quoted just there is different from $\psi(a)$ your description in the question where you said that $\psi(\alpha)$ is the smallest ordinal not representable from $C(\alpha)$. You're also defining $C(\cdots)$ completely differently from the Wikipedia article -- what Wikipedia describes is that each time you want to find a $\phi(\alpha)$ you start anew from ${0,1,\omega,\Omega}$ -- it doesn't build $C(\alpha+1)$ simply by appending the single element $\psi(\alpha)$ to $C(\alpha)$ like you specify. – hmakholm left over Monica Mar 08 '17 at 00:06
  • And Wikipedia's $C(0)$ is not just ${0,1,\omega,\Omega}$ but everything that can be built from those four. – hmakholm left over Monica Mar 08 '17 at 00:07
  • Ah, I see what you mean. Well, does that make it more clear why it stops at $\zeta_0$ @HenningMakholm ? And thanks, I think I'll fix that. – Simply Beautiful Art Mar 08 '17 at 00:09
  • @SimplyBeautifulArt: Yes, from Wikipedia's description I understand it. – hmakholm left over Monica Mar 08 '17 at 00:13
  • Is your question whether $C(\alpha)$ is countable for every countable $\alpha$? – martin.koeberl Mar 08 '17 at 02:17
  • @martin.koeberl No. My question is whether or not the set of outputs of a countably infinite combination of $\psi_\alpha(\beta)$ for $\alpha,\beta<\Omega$ is countable or not. – Simply Beautiful Art Mar 08 '17 at 02:20
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    @SimplyBeautifulArt Are you still unsure? Reese's answer seems very good. – Deedlit Mar 08 '17 at 07:08
  • @Deedlit I suppose I am waiting for a full answer to appear below. – Simply Beautiful Art Mar 08 '17 at 11:54
  • I'm surprise no one has mentioned here that the ordinals make up a proper class rather than a set. – Michael Hardy Aug 27 '17 at 22:46
  • Oh, you are absolutely right @MichaelHardy I'm not very well-versed in set theory. – Simply Beautiful Art Aug 27 '17 at 22:48

1 Answers1

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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.

Deedlit
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