T Sequences - Explicit Enumeration of $\epsilon_0$

Aug 2012
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I think this is what he WANTS to do since what he wrote makes little sense. But he writes things very differently and I think he makes several mistakes in his post.
It's pretty cool that you worked this out. Do you think this gives an enumeration of $\varepsilon_0$? I thought a bit about how you would enumerate $\varepsilon_0$. You could use the usual snaking diagonals proof of the countability of the rationals, applied over and over to enumerate our way up to $\varepsilon_0$.

$\varepsilon_0 = \bigcup_{\mathbb N} {^{n}} \omega$ where $^{n} \omega$ is an $n$-fold exponential tower of $\omega$'s.

It suffices to enumerate each of the $^{n} \omega$, then enumerate the enumerations, and apply the snaking diagonals.

It's obvious how to enumerate $\omega$. Now $\omega^2$ is just $\omega$ copies of $\omega$, and you can stack them vertically and snake the diagonals.

Likewise $\omega^3 = \omega^2 \omega$, which is $\omega$ copies of $\omega^2$ and you can "stack and snake" those. And likewise for $\omega^n$. Stacking and snaking the union we have enumerated our way to $\omega^\omega = \ ^{2} \omega$. We can then convince ourselves this will work for any finite power tower, and we attain an enumeration $\varepsilon_0$.

That's the best I can conceptualize enumerating $\varepsilon_0$. It would be a lot of work to come up with a formula for it, but in principle it's along the lines of the Cantor pairing function.
 
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Oct 2009
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It's pretty cool that you worked this out. Do you think this gives an enumeration of $\varepsilon_0$? I thought a bit about how you would enumerate $\varepsilon_0$. You could use the usual snaking diagonals proof of the countability of the rationals, applied over and over to enumerate our way up to $\varepsilon_0$.

$\varepsilon_0 = \bigcup_{\mathbb N} {^{n}} \omega$ where $^{n} \omega$ is an $n$-fold exponential tower of $\omega$'s.

It suffices to enumerate each of the $^{n} \omega$, then enumerate the enumerations, and apply the snaking diagonals.

It's obvious how to enumerate $\omega$. Now $\omega^2$ is just $\omega$ copies of $\omega$, and you can stack them vertically and snake the diagonals.

Likewise $\omega^3 = \omega^2 \omega$, which is $\omega$ copies of $\omega^2$ and you can "stack and snake" those. And likewise for $\omega^n$. Stacking and snaking the union we have enumerated our way to $\omega^\omega = \ ^{2} \omega$. We can then convince ourselves this will work for any finite power tower, and we attain an enumeration $\varepsilon_0$.

That's the best I can conceptualize enumerating $\varepsilon_0$. It would be a lot of work to come up with a formula for it, but in principle it's along the lines of the Cantor pairing function.
A formal proof is a bit annoying, but I'm fairly sure that this enumerates $\varepsilon_0$ if you insert enough extra rules. You can go further and enumerate $\varepsilon_n$ this way and beyond. But there seems to be a limit to what you can do with this method: https://en.wikipedia.org/wiki/Church–Kleene_ordinal So you won't get to $\omega_1$.
 
Jun 2014
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Ok Micrm@ss, I read my text on notation (good suggestion and thank you, but the experiment must go on!). I am ready to continue working off the intro to $T$ sequences you proposed where you compute $A_n$ for each $n \in \mathbb{N}$, and $n$ represents the number of iterations if trying to relate to my OP.

Let $E$ be the class of all ordinals $\alpha$ such that $\alpha$ is equal to the product of two ordinals:
$$\alpha = a \cdot b, \text{ where } a,b \in Ord$$
Now, to write rule 4, we must clarify that we are looking to find $\mu, \beta, \gamma \in (A_n \cap E)$ such that:
$$\mu = a \cdot b, \beta = a \cdot (b + 1), \text{ and } \gamma = a \cdot (b + 2)$$
To try and write these the way Micrm@ss is, I would then say:

4) $x \in E_n$ if and only if there is some $\mu, \beta, \gamma \in A_n$ such that $\mu, \beta, \gamma \in E$ and $\mu = a \cdot b, \beta = a \cdot (b + 1), \text{ and } \gamma = a \cdot (b + 2)$, where $a$ and $b$ are any ordinals, and $x = a \cdot (b + \omega)$.

Personally, I think that is a crappy way to write these out and I like my way much better. My way is cleaner, easier, and clearer to me. It may not be that way to you and I’m ok with that, but remember:


This is fun. ... At the end of the day this is just an enumeration of $\omega^2$, so I can't emphasize enough that it isn't too complicated. We can do this. With all the crazy cranks that visit this forum who think what they are saying is obviously correct, and all the locals who repeatedly bash them down, I'd say this has got to be a worthwhile experiment if nothing else...
As it sits, I'm siding with the cranks right now. I might appreciate that book Micrm@ss on notation, but even you decided to snub me. That's terrible. If I was the typical crank trying to break math, disprove Cantor, etc., then I can't even imagine how hopeless it would be for me to get some real help here. I'm afraid everyone would just be mean. :ninja:
 
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Jun 2014
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Hey, I worked it out, lol... :p ;) :D

It's pretty cool that you worked this out. Do you think this gives an enumeration of $\varepsilon_0$?
There are a number of explicit enumerations of $\epsilon_0$. The point here is mainly that, even if there was some mistake in my calcs, they could be fixed easily. I only did the calcs once and didn't put a ton of effort into it, so mine may be wrong and that's ok. It's stated as such.

Here are some more common ways to enumerate $\epsilon_0$:

https://math.stackexchange.com/questions/2016997/is-an-algorithm-known-for-ordering-natural-numbers-in-the-epsilon-0-ordinal

I still want to go about this...:

For any ordinal $\alpha$, there are $\alpha^3$ triplets that can be made from the elements of $\alpha$ (if allowing triplets where two or more elements of the triplet can be the same ordinal). This gives us an exact number of triplets to choose from for each $\alpha$ if comprising a $T$ sequence that inserts one and only one ordinal $\alpha$ into the sequence per rule in a fashion where each rule is of the form: “if the triplet (a,b,c) can be formed from some initial segment of a $T$ sequence defined using all rules less than rule $\alpha$, where a,b,c < $\alpha$, then $\alpha$ gets inserted into the sequence.”

How about that, still with me?
...where I keep $|f((\mu,\beta,\gamma))| \leq \aleph_0$ as discussed here:

Concluding Remarks:

Assuming the above calculations are correct, the goal is to discuss a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ compiled by adding rules to the above (or some similar) $T$ sequences. Each $T^{\alpha}$ will be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably. Also, let $T^0 = 1,2,3$.

Each additional rule must meet four specific criteria:

1) Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence (except $T^0$). To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

2) It must hold true that $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

3) A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } \mu = a, \beta = b, \gamma = c \implies \{ \bigcup_{\delta < \alpha} \{x : x \in T^{\delta} \} \}, \text{ where } a,b,c < \bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$$

4) No new rule should should result in $|f((\mu, \beta, \gamma))| \geq \aleph_0$ for any particular triplet $(\mu, \beta, \gamma)$ in the class of all triplets.

Of particular interest is what $T^{\omega_1}$ would equate to after $\omega_1$ rules are added in a fashion that meets the above criteria and assuming no modifications are made to the basic $T$ sequence model to accommodate a transfinite $T$ (note that we could still start with $T = 1,2,3$ and let the iterative process run, so $T^{\omega_1}$ must equate to something despite making no alterations to the $T$ sequence model to accommodate a transfinite $T$, and it may be that $\{x : x \in T^{\omega_1} \} = \omega_1$ ??? :eek: ).
 
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Aug 2012
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There are a number of explicit enumerations of $\epsilon_0$.
I was not aware of that prior to your SE link and the checked answer was a hell of a ride. Cantor normal form, that makes sense. Note to self learn more about that.

I should mention something that might help you to understand some of what goes on in this thread. From where you sit, you apparently think that Micrm@ss and I are both Olympian gods of math. That is not true. Micrm@ss is an Olympian god of math. I've been reading his stuff for years on "that other site." I'm a putzer with a couple of years of grad school and an Internet connection. Micrm@ss knows far far far more about ordinals than I do. Perhaps this helps you understand the thread better.


The point here is mainly that, even if there was some mistake in my calcs, they could be fixed easily.
Um. No. the point is whether your $T$ thing enumerates $\varepsilon_0$. You've offered no proof nor even a plausibility argument.

I only did the calcs once and didn't put a ton of effort into it,
so why should I?

so mine may be wrong and that's ok. It's stated as such.
Let us know when you post something you're willing to stand by.

That's a rockin' thread man, thanks for the info.


...where I keep $|f((\mu,\beta,\gamma))| \leq \aleph_0$ as discussed here:
You must think that means something to me. Ooh am I being mean again?
 
Jun 2014
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A formal proof is a bit annoying, but I'm fairly sure that this enumerates $\varepsilon_0$ if you insert enough extra rules. You can go further and enumerate $\varepsilon_n$ this way and beyond. But there seems to be a limit to what you can do with this method: https://en.wikipedia.org/wiki/Church–Kleene_ordinal So you won't get to $\omega_1$.
Let's take a gander at some large countable ordinals, including $\omega_1^{CK}$:

https://en.wikipedia.org/wiki/Large_countable_ordinal

$\omega_1^{CK}$ is the first ordinal that is not the "order type of any recursive well-ordering of the integers."

I believe that $\omega_1^{CK}$ is a little itty bitty baby ordinal, quite literally infinitesimally small comparatively speaking, to other ordinals that $T$ sequences can enumerate. The reason is simple. Even if we took every triplet that could be made from the elements of $\omega_1^{CK}$ (there are $(\omega_1^{CK})^3$ triplets) and said that each triplet implied $\omega_1^{CK}$ elements could be added to the sequence, we still wouldn't exhaust the capabilities of my unmodified $T$ sequence model while simultaneously exhausting all the elements of $\omega_1^{CK}$.

It takes uncountably many rules or, if using generalized rules like I do in my OP, it takes individual rules that add uncountably many ordinals to $T$ to exhaust my model. I was considering modifying the model to accommodate uncountably many rules via a transfinite $T$, but I can't modify it yet for $T^{\omega_1}$ because I'm having trouble coming up with what needs to be modified (if anything) because we could still run a $T$ sequence with that many rules if compiled as I described above.
 
Jun 2014
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Now that we've agreed upon some basic notation, I've put in quotes below the basic collection of $T$ sequences enumerating $\omega + 1$ that I had described previously in narrative form. Please take a moment to make sense of it if you are new to this thread.

For those familiar with the $T$ sequence model, the goal now is to discuss a particular transfinite sequence of $T$ sequences: $(T^{\alpha})_{\alpha \in Ord}$. Each $T^{\alpha}$ will be a $T$ sequence generated using $\alpha$ rules, where $\alpha$ is some ordinal. An uncountable $\alpha$ would require some clarification and for $T^{\alpha}$ to be a transfinite sequence, presumably. Let $T^0 = 1,2,3$.

For any ordinal $\alpha$, there are $\alpha^3$ triplets that can be made from the elements of $\alpha$ (if allowing triplets where two or more elements of the triplet can be the same ordinal). This gives us an exact number of triplets to choose from for each $\alpha$ if comprising a $T$ sequence that inserts one and only one ordinal $\alpha$ into the sequence per rule in a fashion where each rule is of the form: “if the triplet (a,b,c) can be formed from some initial segment of a $T$ sequence defined using all rules less than rule $\alpha$, where a,b,c < $\alpha$, then $\alpha$ gets inserted into the sequence.”

Each additional rule must meet four specific criteria:

1) Each $T$ sequence may contain only ordinals, but not every $T$ sequence is an enumeration of an ordinal. Equivalently, $\{ x : x \in T \}$ may or may not be an ordinal based on the rules used to generate $T$. I wish to create a transfinite sequence $(T^{\alpha})_{\alpha \in Ord}$ where $\{ x : x \in T^{\alpha} \}$ is an ordinal for each element of the sequence (except $T^0$). To do this, it must be ensured that each rule, when added to the previous rules, results in a new $T$ sequence that again enumerates some ordinal.

2) It must hold true that $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \} = \{x : x \in T^{\alpha - 1} \}$ whenever $\alpha$ is not a limit ordinal.

3) A new rule $\alpha$ should imply an ordinal that is equal to $\bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$ so as to extend the sequence:

$$\text{Rule } \alpha \text{ : } \mu = a, \beta = b, \gamma = c \implies \{ \bigcup_{\delta < \alpha} \{x : x \in T^{\delta} \} \}, \text{ where } a,b,c < \bigcup_{\delta < \alpha} \{ x : x \in T^{\delta} \}$$

4) No new rule should should result in $|f((\mu, \beta, \gamma))| \geq \aleph_0$ for any particular triplet $(\mu, \beta, \gamma)$ in the class of all triplets.

Of particular interest is what $T^{\omega_1}$ would equate to after $\omega_1$ rules are added in a fashion that meets the above criteria and assuming no modifications are made to the basic $T$ sequence model to accommodate a transfinite $T$ (note that we could still start with $T = 1,2,3$ and let the iterative process run, so $T^{\omega_1}$ must equate to something despite making no alterations to the $T$ sequence model to accommodate a transfinite $T$, and it may be that $\{x : x \in T^{\omega_1} \} = \omega_1$ ??? :eek: ).





Description of This $T$ Sequence: The following is a model for a $T$ sequence using $\omega + 1$ rules that in turn enumerates $\omega + 1$. It is a basic $T$ sequence in that no rule adds more than a single triplet to the sequence (whereas other $T$ sequences discussed in this thread used more generalized rules that added infinitely many ordinals to the sequence and were therefore more confusing to some).

Rules: The following rules are used to define function $f$:

$$\text{Rule 1 : }\mu = 3, \beta = 2, \gamma = 1 \implies \{ 0 \}$$
$$\text{Rule } z \text{ : }\mu = 0, \beta = 1, \gamma = z - 1 \implies \{z\}, \text{ where } 2 \leq z < \omega$$
$$\vdots$$
$$\text{Rule } \omega \text{ : }\mu = 1, \beta = 2, \gamma = 3 \implies \{\omega\}$$
Define function $f$:

$$f((\mu,\beta,\gamma)) = \bigcup \{ x : \mu,\beta,\gamma \implies x \}, \text{ where } \mu,\beta, \text{ and } \gamma \text{ are ordinals and } x \text{ is a set defined by the above rules}$$

Define function $g$:

For any set of ordinals, $A$, let $g(A)$ be the set of all ordered triplets that can be made from the elements of $A$:
$$g(A) = \{ (a,b,c) : a,b,c \in A \}$$


Define the sequence $T$:

Define a sequence $T = t_1, t_2, t_3, \dots$ via iterations where:

Step 1) $t_1 = 1, t_2 = 2,$ and $t_3 = 3$.

Step 2) Each $t_n$, where $n \geq 4$, is defined by the previous elements of the sequence. Starting with $n = 4$:

a) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{1, 2, 3 \}$ on the first iteration.

b) Let $B = \{ f((\mu,\beta,\gamma)) : (\mu,\beta,\gamma) \in g(A) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new sets of ordinals implied by letting function $f$ range over $g(A)$.

c) Let $C = (\bigcup B) \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \in \mathbb{N}$. This step removes any redundant elements from $\bigcup B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \in \mathbb{N}$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots, t_{n+j-1} = c_j$.

e) If $|C| = |\mathbb{N}|$ (not applicable for this particular $T$ sequence), then let $T’ = t’_1, t’_2, t’_3, \dots$, be a subsequence of the remaining undefined elements of $T$ and set $t’_1 = c_1, t’_3 = c_2, t’_5 = c_3, \dots$.

Step 3) Proceed to the next undefined index $j$ in $T$, set $n = j$, and repeat step 2.
 
Jun 2014
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I suppose I've been poking this enumeration of $\omega_1$ in both this forum and the Physics forum for a while now without anyone batting an eye, so I'm sort of starting to wonder. Usually you guys are relentless in debunking statements like that, so I suppose I'll have to go full blown crank here:

I assert that $T^{\omega_1}$, if compiled as described above, is an enumeration of $\omega_1$. I further assert that this contradicts the definition of $\omega_1$, and that there are no uncountable ordinals as a result, so as to provide a solution to the continuum hypothesis. I'll further note that the axiom of choice, which implies that all sets can be well ordered in ZFC, would imply that all sets in ZFC are countable.

Math is broken! Why can't you see!!! You guys are all [insert some sort of forum-friendly insult here]!!!!! Bwwwaaaaahhhhaaaaahahaaaaha!!!!!!! :D
 
Oct 2009
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I suppose I've been poking this enumeration of $\omega_1$ in both this forum and the Physics forum for a while now without anyone batting an eye, so I'm sort of starting to wonder. Usually you guys are relentless in debunking statements like that, so I suppose I'll have to go full blown crank here:

I assert that $T^{\omega_1}$, if compiled as described above, is an enumeration of $\omega_1$. I further assert that this contradicts the definition of $\omega_1$, and that there are no uncountable ordinals as a result, so as to provide a solution to the continuum hypothesis. I'll further note that the axiom of choice, which implies that all sets can be well ordered in ZFC, would imply that all sets in ZFC are countable.

Math is broken! Why can't you see!!! You guys are all [insert some sort of forum-friendly insult here]!!!!! Bwwwaaaaahhhhaaaaahahaaaaha!!!!!!! :D
All sets in ZFC being countable isn't a wrong statement. It follows from the Lowenheim-Skolem-Tarski theorems and is known as Skolem's paradox.

In any case, if you think you have a good proof of your assertions, go publish them in a journal.