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October 27th, 2019, 12:32 AM   #1
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Having Sum Fun Counting Ordinals

This is a draft. Knowing me, it is probably nonsensical and filled with errors. I'm trying to enumerate some very large countable ordinal assuming of course that math isn't broken and I can't enumerate $\omega_1$ itself. ... And no, jic, I don't think math is broken. Why? Do you?

I'll take a moment to step back from this and at some point return to see what corrections, comments, etc., I care to make. Feel free to look at it funny in the meantime.


Define $t(\alpha)$ for any ordinal $\alpha$:

For $\alpha \geq 2$, let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$.


Define $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$:

Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be almost regressive such that:

1) $\phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective,

2) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and

3) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{-1}(b)\} < min\{ \phi_{\alpha}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{-1}(b)\}$.


Define function $f$:

$$f(x) = \phi_{t^{-1}(x)}^{-1}(x)$$


Define $k(\alpha)$ for any ordinal $\alpha$:

$$k(\alpha) = \{ x : f(x) = \alpha \}$$


Define $h(\alpha)$ for any ordinal $\alpha$:

$$h(\alpha) = min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \}$$


Define function $g$:

For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$:
$$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$


Define the sequence $T$:

Define a transfinite sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where:

Step 1) Let $t_1 = 0, t_2 = 1, t_3 = 2$, and iteration counter $m = 1$.

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., $A = \{0, 1, 2 \}$ on the first iteration.

b) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{-1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.

c) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$.

d) If $|C| \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$.

e) If $|C| = \aleph_0$, 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) Let $j$ be the first ordinal such that $t_j$ is undefined. Set $n = j$, increase the iteration counter by letting $m = m + 1$, and then repeat step 2.
Thanks from idontknow

Last edited by AplanisTophet; October 27th, 2019 at 12:46 AM.
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October 27th, 2019, 02:03 PM   #2
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Quote:
Originally Posted by AplanisTophet View Post

Define $t(\alpha)$ for any ordinal $\alpha$:

For $\alpha \geq 2$, let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$.
If $\alpha$ is an arbitrary ordinal, you have to account for "tuples" of order type $\omega + 1$ or even $\omega_{\omega_\omega}$ or worse.

You idea would generalize better and be more clear if you regard each tuple as a function from a given ordinal to some set. Indeed that's exactly what a high-school tuple is. A point $(x_0, y_0)$ in the Euclidean plane is a particular function $\{x, y \} \to \mathbb R$, namely the one that sends $x$ to $x_0$ and $y$ to $y_0$; and the coordinate plane is the set of all such functions.

So now we have no trouble making sense of an "ordinal-indexed tuple," namely that it's a function from some ordinal to some set.

Now you have not told us what the range of this function is. What are the elements of the tuples? Are they ordinals too? I'll assume they are.

So an "ordinal-indexed tuple" is a function $f : On \to On$, where $On$ is the class of ordinals. I confess that I still don't quite believe in functions defined on proper classes. Where do they live? But set theorists have some legit way to make sense of it so no matter.

Does that seem like what you have in mind?

Last edited by Maschke; October 27th, 2019 at 02:39 PM.
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October 28th, 2019, 12:19 PM   #3
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Originally Posted by Maschke View Post
If $\alpha$ is an arbitrary ordinal, you have to account for "tuples" of order type $\omega + 1$ or even $\omega_{\omega_\omega}$ or worse.
I'll gladly take the "$(\omega + 1)$-lets" of variables, as that is the range of $\phi_{\omega+1}$ across $\omega_1 \setminus \{0\}$. I don't consider $\phi_{\alpha}$ for any $\alpha \geq \omega_1$ in the OP though, so the "worse" you note is not actually "worse" as far as my OP is concerned and the $(\omega+1)$-lets you think of as accidental issues are completely intentional.

Don't forget about $\omega$-lets though either. I want that function normal.

I'm not sure what to say about the rest of your post as a result.

Last edited by AplanisTophet; October 28th, 2019 at 12:22 PM.
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October 28th, 2019, 12:43 PM   #4
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Quote:
Originally Posted by AplanisTophet View Post
I'll gladly take the "$(\omega + 1)$-lets" of variables, as that is the range of $\phi_{\omega+1}$ across $\omega_1 \setminus \{0\}$. I don't consider $\phi_{\alpha}$ for any $\alpha \geq \omega_1$ in the OP though, so the "worse" you note is not actually "worse" as far as my OP is concerned and the $(\omega+1)$-lets you think of as accidental issues are completely intentional.

Don't forget about $\omega$-lets though either. I want that function normal.

I'm not sure what to say about the rest of your post as a result.
You said $\alpha$ was an arbitrary ordinal but now you seem to be backtracking. I'm trying to help you clarify your exposition but you seem to prefer obfuscation, as seen in your quoted text here.
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October 28th, 2019, 02:41 PM   #5
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These are some potential questions and clarifications that I've noted as being possibly necessary. There may be more of course:

1) Typo - My definition of $t(\alpha)$ headlines “for any ordinal $\alpha$” and then starts by restricting to $\alpha \geq 2$. This particular work only makes use of $2 \leq \alpha < \omega_1$ as a domain for function $t$, so I should change the headline for my definition.

2) For the definition of $t$, I don’t need to use doublets, triplets, quadruplets, and so on, but it’s easier because $a=b \implies \{a,b\}=\{b,a\} = \{a\} = \{b\}$ whereas $a=b$ does not $\implies (a,b)=(b,a)=(a)=(b)$.
3) I could define $C$ to make sure there are never any duplicate elements in $T$ perhaps... Right now there are never duplicate elements if the cardinality of $C \neq \aleph_0$ on each iteration.


I think that if anyone questions anything it will be the following things:

1) When $\alpha \geq \omega$, I think someone may question whether there must be an element $\kappa$ such that $\kappa \in \phi_{\alpha}(\kappa)$. Fodor’s lemma requires that there must be such an element I believe (Fodor’s lemma would have $\kappa$ map to a constant if $\phi$ were ‘fully regressive’ instead of just ‘almost regressive’), so I don’t think that is an issue.

2) The following line is the one that should cause some stir and that I expect people to question for accuracy. As of right now I still stand by it, but why wouldn’t I (I’m a self-professed crank)?:
“The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.”


Quote:
Originally Posted by Maschke View Post
You said $\alpha$ was an arbitrary ordinal but now you seem to be backtracking. I'm trying to help you clarify your exposition but you seem to prefer obfuscation, as seen in your quoted text here.
If you don't fully understand Fodor's lemma, club sets, normal functions, fixed points, etc., then you probably won't be able to comment and help me like you normally are able to.

That said, your only complaint about my work is the same in every thread lately and it's always something to do with my notation. It's not notation this time absent any typo so don't start that. I will assist you if you like in figuring out my work, but that is a courtesy on my behalf because I don't think you can assist me. Feel free to prove me otherwise of course. I would prefer to do that over PM, but I can't stop you from posting in my threads like you typically do.

Last edited by AplanisTophet; October 28th, 2019 at 02:46 PM.
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October 28th, 2019, 03:27 PM   #6
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I can't stop you from posting in my threads like you typically do.
Fuck you.

Last edited by Maschke; October 28th, 2019 at 03:30 PM.
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October 29th, 2019, 11:13 AM   #7
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They might be able to edit that out once this becomes the most famous math forum thread ever.
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October 29th, 2019, 01:12 PM   #8
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Quote:
Originally Posted by AplanisTophet View Post
These are some potential questions and clarifications that I've noted as being possibly necessary. There may be more of course:

1) Typo - My definition of $t(\alpha)$ headlines “for any ordinal $\alpha$” and then starts by restricting to $\alpha \geq 2$. This particular work only makes use of $2 \leq \alpha < \omega_1$ as a domain for function $t$, so I should change the headline for my definition.

2) For the definition of $t$, I don’t need to use doublets, triplets, quadruplets, and so on, but it’s easier because $a=b \implies \{a,b\}=\{b,a\} = \{a\} = \{b\}$ whereas $a=b$ does not $\implies (a,b)=(b,a)=(a)=(b)$.
3) I could define $C$ to make sure there are never any duplicate elements in $T$ perhaps... Right now there are never duplicate elements if the cardinality of $C \neq \aleph_0$ on each iteration.


I think that if anyone questions anything it will be the following things:

1) When $\alpha \geq \omega$, I think someone may question whether there must be an element $\kappa$ such that $\kappa \in \phi_{\alpha}(\kappa)$. Fodor’s lemma requires that there must be such an element I believe (Fodor’s lemma would have $\kappa$ map to a constant if $\phi$ were ‘fully regressive’ instead of just ‘almost regressive’), so I don’t think that is an issue.

2) The following line is the one that should cause some stir and that I expect people to question for accuracy. As of right now I still stand by it, but why wouldn’t I (I’m a self-professed crank)?:
“The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.”

If you don't fully understand Fodor's lemma, club sets, normal functions, fixed points, etc., then you probably won't be able to comment and help me like you normally are able to.
It's all definitions for the most part. There is this claim, which I believe must be false:

“The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.”

If $B$ is countable for each iteration then $T$ remains a regular sequence of only order type $\omega$ after $\omega_1$ iterations. Each iteration adds elements to $T$, so there are at least $\omega_1$ ordinals in $T$. That is a contradiction because it implies $T$ enumerates an ordinal with cardinality greater than or equal to $\omega_1$.

At the same time, I think $B$ must be countable for each iteration, so I am confused. Hopefully this helps others help. Sorry I don't understand myself.

Last edited by ISP; October 29th, 2019 at 01:14 PM.
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October 29th, 2019, 08:50 PM   #9
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I hope this is the last draft after having stepped back for a while to see what I could fix and clarify. Please help now and thank you!!!!

ISP pinpoints the issues:


Quote:
Originally Posted by ISP View Post
If $B$ is countable for each iteration then $T$ remains a regular sequence of only order type $\omega$ after $\omega_1$ iterations...

At the same time, I think $B$ must be countable for each iteration...
Yes, but only if I don't forget to make use of the iteration counter! Thank you ISP!!

The following is the issue that needs addressing:

Quote:
Originally Posted by AplanisTophet View Post
The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.

...


Define $t(\alpha)$ for any ordinal $\alpha \geq 2$:

Let $t(\alpha)$ equal a doublet of variables $(a,b)$ if $\alpha = 2$, a triplet of variables $(a,b,c)$ if $\alpha = 3$, a quadrulplet of variables $(a,b,c,d)$ if $\alpha = 4$, and so on, for any ordinal $\alpha$. Note that $a=b$ does not imply $(a,b) = (b,a) = (a) = (b)$, whereas it does imply $\{a,b\} = \{b,a\} = \{a\} = \{b\}$ in general.


Define $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$:

Let each element of $(\phi_{\alpha})_{2 \leq \alpha < \omega_1}$ be almost regressive such that:

1) $\phi_{\alpha} : \omega_1 \setminus \{0\} \rightarrow \{ t(\alpha) : a,b,c,\dots \in t(\alpha) \implies a,b,c,\dots < \omega_1 \}$ is bijective,

2) $a,b,c,\dots \leq \kappa$ for each $a,b,c,\dots \in \phi_{\alpha}(\kappa)$, and

3) $\zeta < \alpha \implies min\{ \phi_{\zeta}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\zeta}^{-1}(b)\} < min\{ \phi_{\alpha}^{-1}(b) : \exists k \in b \text{ where } k \geq \phi_{\alpha}^{-1}(b)\}$.


Define function $f$:

$$f(x) = \phi_{t^{-1}(x)}^{-1}(x)$$


Define $k(\alpha)$ for any ordinal $\alpha$:

$$k(\alpha) = \{ x : f(x) = \alpha \}$$


Define $h(\alpha)$ for any ordinal $\alpha$:

$$h(\alpha) = min\{ t^{-1}(x) : x \in k(\alpha) \text{ and } \forall y \in x(y < \alpha) \}$$


Define function $g$:

For any set of ordinals, $A$, let $g(A)$ be the set of all ordered doublets, triplets, quadruplets, and so on, that can be comprised from the elements of $\omega_1 \cap A$:
$$g(A) = \{ t(\alpha) : t(\alpha) \setminus (A \cap \omega_1) = \emptyset \text{ and } 2 \leq \alpha < \omega_1 \}$$


Define the sequence $T$:

Define a (potentially transfinite) sequence $T = t_1, t_2, t_3, \dots$ over $\omega_1$ iterations where:

Step 1) Let $t_1 = 0, t_2 = 1, t_3 = 2$, and iteration counter $m = 1$.

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

a) If $m$ is a countable limit ordinal and $T$ is of order type $\omega$, free up space in $T$ by first letting $(s_n)_{n \in \mathbb{N}}$ be defined for odd indexes and undefined for even indexes: $s_{n \cdot 2 - 1} = t_{n}$. Then, set $t_1 = s_1, t_2 = s_2, t_3 = s_3, \dots$.

b) Let $A = \{ t_i \in T : i < n \}$. E.g., $A = \{0, 1, 2 \}$ on the first iteration.

c) Let $B = \{ f(x) : x \in (g(A) \setminus \{ x \in g(A) : t^{-1}(x) > h(f(x))\} ) \}$. Using the previous elements of the sequence $A$, this step creates a set $B$ of all the new ordinals implied by letting function $f$ range over $g(A)$. The restriction of $g(A)$ to $A \cap \omega_1$ ensures that $B$ remains countable for this particular $T$ sequence.

d) Let $C = B \setminus A$ and let $c_1, c_2, c_3, \dots$ be an enumeration of $C$ that is also well ordered if $|C| \neq \aleph_0$. This step removes any redundant elements from $B$ before potentially well ordering them so that we can add them to $T$.

e) If $|C| \neq \aleph_0$, then set $t_n = c_1, t_{n+1} = c_2, t_{n+2} = c_3, \dots$.

f) If $|C| = \aleph_0$, 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) Let $j$ be the first ordinal such that $t_j$ is undefined. Set $n = j$, increase the iteration counter by letting $m = m + 1$, and then repeat step 2.

Last edited by AplanisTophet; October 29th, 2019 at 09:49 PM. Reason: fix iteration counter
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October 29th, 2019, 09:59 PM   #10
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Quote:
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Step 3) Let $j$ be the first ordinal such that $t_j$ is undefined. If $j>n$, set $n = j$. Increase the iteration counter by letting $m = m + 1$ and then repeat step 2.
Fixed. This should do it. Sorry for taking my sweet time working the bugs out. We're down to just that one assertion in the immediately preceding post. Thanks again for any help!
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