Limits of Two Ordinal Sequences

[AplanisTophet]

TL;DR Summary

Examining the limits of ordinal sequences

Let $\omega_1$ be the first uncountable ordinal such that $x$ is an element of $\omega_1$ if and only if it is either a finite ordinal or there exists a bijection from $x$ onto $\omega$.
I want to define a matrix such that the matrix contains each element of $\omega_1$ only once.
To get the first column, I might start by listing upwards from the bottom row in the standard order all the elements of $\omega_1$ before removing the limit ordinals. The remaining (non-limit) ordinals will comprise the first column of the matrix.
For the second column, I might start by listing the (limit) ordinals not contained in the first column before removing the limit ordinals of that list. The remaining ordinals will comprise the second column.
For the third column, I might again start by listing the ordinals not contained in the previous columns before removing the limit ordinals of that list.
Continue ad infinitum, such that the matrix contains each element of $\omega_1$ only once. Here is the matrix:
$$

\begin{matrix}

\vdots & \vdots & \vdots & & \vdots \\

{\omega + 1}_{(0, \omega)} & {\omega^2 + \omega}_{(1, \omega)} & {\omega^3 + \omega^2}_{(2, \omega)} & \dots & {\omega^{\omega+1}+\omega^{\omega}}_{(\omega,\omega)} & \dots \\

\vdots & \vdots & \vdots & & \vdots \\

3_{(0,2)} & {\omega*3}_{(1,2)} & {\omega^2*3}_{(2,2)} & \dots & {\omega^{\omega}*3}_{(\omega,2)} & \dots\\

2_{(0,1)} & {\omega*2}_{(1,1)} & {\omega^2*2}_{(2,1)} & \dots & {\omega^{\omega}*2}_{(\omega,1)} & \dots \\

1_{(0,0)} & {\omega}_{(1,0)} & {\omega^2}_{(2,0)} & \dots & {\omega^{\omega}}_{(\omega,0)} & \dots \\

\end{matrix}

$$
Note that $a_{(x,y)} + b_{(x,0)} = c_{(x,y+1)}$ for any elements of the matrix $a,b,$ and $c$ located in the same column (column $x$), where $a$ is in row $y \geq 0$, $b$ is in the bottom row, and $c$ is in row $y+1$.
**Proof: The matrix has a last column**
The matrix above only contains elements of $\omega_1$, but consider a second matrix created the same way as the above one using all the ordinals instead. We know that $\omega_1$ would appear at position $(\omega_1,0)$ of this second matrix, $\omega_1*\omega+\omega_1$ would appear at position $(\omega_1,\omega)$, and $\omega_1 + \beta$ would appear at $\text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x,\omega)}$. That is, there must be a final column of the first matrix above with order type $\omega$ because it can be the only column of that order type where, if the order type was any greater, the column would contain elements that are not in $\omega_1$. This contradicts the standard notion that the first matrix above can have a last column or that the order type of the last column must be $\omega_1$ instead of $\omega$, but the matrix does have a last column of order type $\omega$ because it must.
Let $\alpha = \text{the ordinal at matrix position } (\text{last column},1)$.
Let $\beta = \text{the ordinal at matrix position } (\text{last column},0)$.
**Questions:**
Is it true that $\alpha = \beta + \beta \implies \beta < \alpha \implies \beta < \omega_1 \implies$ there exists a bijective function from $\beta$ onto $\omega$?
If so, then could we go on to show that $\zeta = limit_{x \rightarrow \omega_1}^{\text{matrix position } x,\beta}$ is also in $\omega_1$. And since $\zeta \in \omega_1$, can't we now show that $\theta = limit_{x \rightarrow \omega_1}^{\text{matrix position } x,\zeta}$ is also in $\omega_1$, ad inifinitum, so as to start enumerating the matrix and contradict the assumption that $\omega_1$ is uncountable?
If the limits of the rows are countable and we put them into a sequence such that they are the range of a normal function $f$ (ie, $f(1) = \beta, f(2) = \zeta, f(3) = \theta$, and so on), then does this normal function contradict the fixed-point lemma because it doesn't have any fixed points nor is the set of all fixed points unbounded in $\omega_1$?
**My Proposed Conclusion**
There is no first uncountable ordinal because there cannot be. Assuming the existence of $\omega_1$ leads to the above countable matrix containing all of its elements, contradicting the assumption that $\omega_1$ is uncountable. I note that the above matrix takes the shape of a 'quarter circle'. I also note that there are uncountable cardinals and I believe Cantor found the first one, but nevertheless, the continuum hypothesis rests on the existence of an uncountable ordinal and there are none so it is solved.
Thank you for reading!
B.J.K. 10-21-21

 

[AplanisTophet]

$$
\begin{matrix}
\vdots & \vdots & \vdots & & \vdots &| \text{outside of } \omega_1 \text{ matrix}\rightarrow & \vdots & & \vdots \\
{\omega + 1}_{(0, \omega)} & {\omega^2 + \omega}_{(1, \omega)} & {\omega^3 + \omega^2}_{(2, \omega)} & \dots & {\omega^{\omega+1}+\omega^{\omega}}_{(\omega,\omega)} & | & {\omega_1 + \beta}_{(\text{last column}, \omega) \text{ } \cong \text{ } \text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x,\omega)}} > \omega_1 & & {\omega_1 * \omega + \omega_1}_{(\omega_1, \omega)} \\
\vdots & \vdots & \vdots & & \vdots & | & ------------ \vdots ------------&-- & \vdots \\
3_{(0,2)} & {\omega*3}_{(1,2)} & {\omega^2*3}_{(2,2)} & \dots & {\omega^{\omega}*3}_{(\omega,2)} & \dots & {\beta * 3}_{(\text{last column}, 2)} & | & {\omega_1 * 3}_{(\omega_1, 2)} \\
2_{(0,1)} & {\omega*2}_{(1,1)} & {\omega^2*2}_{(2,1)} & \dots & {\omega^{\omega}*2}_{(\omega,1)} & \dots & {\beta * 2}_{(\text{last column}, 1)} & | & {\omega_1 * 2}_{(\omega_1, 1)} \\
1_{(0,0)} & {\omega}_{(1,0)} & {\omega^2}_{(2,0)} & \dots & {\omega^{\omega}}_{(\omega,0)} & \dots & {\beta}_{(\text{last column}, 0)} & | & {\omega_1}_{(\omega_1, 0)} \\
\end{matrix}
$$

 

[AplanisTophet]

The axiom of choice implies the well ordering theorem in ZF set theory, which in turn implies that uncountable cardinals can be well ordered and thus the existence of uncountable ordinals. If the original post's proposed conclusion is correct, then there are no uncountable ordinals and the axiom of choice can be abandoned. ?

 

[SSequence]

There was quite a bit of explanation on your question on MSE (in particular, by an expert on the topic). I don't think I can add too much to it. Nevertheless, I will add a few points that I think might be relevant.

As a preliminary, I would just add that you don't necessarily have to agree with set theory philosophically or consider it satisfactory or anything like that. However, the question here is that whether your argument follow "formally" from "strictly set-theoretic" perspective (to be precise, from the axioms of ZFC and the usual rules of (classic) FOL).

The axiom of choice implies the well ordering theorem in ZF set theory, which in turn implies that uncountable cardinals can be well ordered and thus the existence of uncountable ordinals. If the original post's proposed conclusion is correct, then there are no uncountable ordinals and the axiom of choice can be abandoned. ?

Existence of uncountable ordinals should also be a theorem of ZF (without choice) I think. Upon a simple search on MSE, seems to be confirmed by the following thread/question (for example, see the first comment below the question):
https://math.stackexchange.com/ques...-do-i-need-to-believe-in-uncountable-ordinals
The reasoning involved def. seems to be a bit complicated though.

Also, ordinals (uncountable or not) by definition are well-ordered by the $\in$ relation.

=================================================

Now proceeding to OP, what you have described in the OP is a bijective function $F:\omega_1 \times \omega_1 \rightarrow \omega_1-\{0\}$ (the $\omega_1 \times \omega_1$ "matrix").

However, the line of reasoning that you give in the main post, I don't know how you are proceeding with it (in other words, I can't see how it follows formally). If the matrix is $\omega_1 \times \omega_1$ then by definition there is no "last column".

 

[AplanisTophet]

The expert on MSE never saw this paper as far as I know. The comments are from before it was fully posted so Noah has never heard any "last column" talk. That said, you are mistaken in that I've had no feedback from anyone yet.

I agree with set theory philosophy, whatever that means. No problem with Cantor's Theorem for example. Generally would never have a reason to doubt the existence of $\omega_1$, etc.

You need choice to prove a well ordering exists for uncountable sets in ZF.

Telling me basics about the standard well ordering of the ordinals is not helpful. I am well aware of their structure, limits, Fodor's lemma, etc.

I would assume, like most anyone else, that the matrix in my OP is an $\omega_1 \text{ x } \omega_1$ matrix absent proof to the contrary.

Infinite sequences don't have last elements. An infinite sequence of columns doesn't have a last column. Or, at least without proof. I agree. See the section of my paper entitled "Proof: The matrix has a last column."

Thank you for taking the time.

 

[AplanisTophet]

You need choice to prove a well ordering exists for uncountable sets in ZF.

Please allow me to rephrase. You need choice to prove a well ordering exists for uncountable [cardinals] in ZF. I say this, even though there may not be a difference between the two statements.

 

[SSequence]

Regarding post#6, eh nevermind about that. You are right about that:

The axiom of choice implies the well ordering theorem in ZF set theory, which in turn implies that uncountable cardinals can be well ordered ...

I misread the part in bold as "uncountable ordinals".

Nevertheless, existence of $\omega_1$ is a theorem of ZF and can be shown without choice (mentioned in the link in post#4).

==============================

Also, if you have a modified version, then I suggest you also make these changes to your MSE version of question. Now regarding post#5, quoting from the OP:

**Proof: The matrix has a last column**
The matrix above only contains elements of $\omega_1$, but consider a second matrix created the same way as the above one using all the ordinals instead. We know that $\omega_1$ would appear at position $(\omega_1,0)$ of this second matrix, $\omega_1*\omega+\omega_1$ would appear at position $(\omega_1,\omega)$, .....

Beyond that part I quoted, I can't follow what you wrote.Anyway, one thing I can comment on is what you wrote later on:

If the limits of the rows are countable and we put them into a sequence such that they are the range of a normal function $f$ (ie, $f(1) = \beta, f(2) = \zeta, f(3) = \theta$, and so on), then does this normal function contradict the fixed-point lemma because it doesn't have any fixed points nor is the set of all fixed points unbounded in $\omega_1$?

The very first row of the $\omega_1 \times \omega_1$ matrix in OP is indeed a normal function. However, if you look at the second row or third row for example, they fail the "continuity condition" very early on, so these rows are not normal functions.

 

[AplanisTophet]

The existence of a set containing all the countable ordinals is a theory in ZF, much like the set of all sets that do not contain themselves once was a theory of 'set theory', but that doesn't mean $\omega_1$ is a countable set and there is no proof that $\omega_1$ is uncountable in ZF (or else we'd have solution to CH already, I believe). Assume there is a surjective function onto a set's powerset and you'll find the assumption faulty. Form $\omega_1$ in ZF and then apply my matrix and you'll find that the uncountable assumption is faulty too (not that the axiom schema of specification cannot be applied to create the set as it can, which is merely what you're saying).

I kept editing the MSE version. You can look at the edit history if you like. The versions posted now at MSE, here, and MyMathForum are all the same. Again, no feedback from Noah on these, or anyone else for that matter.

If you can't follow beyond what you quoted, then that's where we stop. The rest of your comments cannot be meaningful at that point. The partial sentence and following full sentence you are struggling with are these:

"... and $\omega_1 + \beta$ would appear at $\text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x, \omega)}$. That is, there must be a final column of the first matrix above with order type $\omega$ because it can be the only column of that order type where, if the order type was any greater, the column would contain elements that are not in $\omega_1$."

If that is the case, then let me ask rhetorically, what is $\text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x, \omega)}$? We know it's not equal to the ordinal at matrix position $(\omega_1, \omega)$, which is equal to $\omega_1*\omega + \omega_1$. We know it's also greater than $\omega_1$ because matrix position $(\omega_1,0) = \omega_1$, so the preceding column must have an element greater than $\omega_1$ and less than $\omega_1*\omega + \omega_1$ at vertical position $(\text{last column},\omega) \cong \text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x, \omega)}$. It is an infinite "last column matrix." (I like the sound of that) The problem is that we've never been able to say something like "last column" in a case like this because we've never been able to prove, until now, that there must be a column of order type $\omega$ in the $\omega_1$ matrix. The $\omega_1$ matrix would contain an element greater than $\omega_1$ if there wasn't a last column of order type $\omega$, however, so there must be a last column. We're not speculating that there isn't anymore, as would be the standard assumption until now.

 

[SSequence]

If that is the case, then let me ask rhetorically, what is $\text{limit}_{x \rightarrow \omega_1}^{\text{matrix position }(x, \omega)}$?

When you write $\mathrm{matrixPosition}(x,\omega)$ do you mean:
(i) $x$-th row and $\omega$-th column
or
(ii) $x$-th column and $\omega$-th row
?

At any rate, in both cases we have:
$\mathrm{limit}_{x \rightarrow \omega_1} [\mathrm{matrixPosition}(x,\omega)]=\sup\{\mathrm{matrixPosition}(x,\omega) \,\,| \,\, x \in \omega_1\}=\omega_1$.==============================One point here that I will repeat (also mentioned towards end of post#7) is that if one has a (class) function such as $f:\mathrm{Ord} \rightarrow \mathrm{Ord}$ then, generally speaking, we have no guarantee that for any arbitrary limit ordinal $l$ we will have [ofc it is very easy to construct functions that fail this property]:
$f(l)=\sup\{f(i)\,|\, i \in l\}$
If a function doesn't satisfy this property for every limit ordinal $l$ then it is not normal. Hence it is no longer guaranteed to have any fixed points (let alone unbounded number of them).

 

[AplanisTophet]

When you write $\mathrm{matrixPosition}(x,\omega)$ do you mean:
(i) $x$-th row and $\omega$-th column
or
(ii) $x$-th column and $\omega$-th row
?

We'll have to address this before anything else you write makes sense. The second post of this thread contains the matrix if you need help visualizing it. Matrix position (x,y) implies x is the column and y is the row.

I think it's fun, given the nature of the OP, to envision the column number as the x-axis and the row number as the y-axis sort of like a cartesian plane. That is a little bit of a stretch perhaps, just the way I envision it.

 

[AplanisTophet]

Big "if" here, so just ignore this post for now perhaps. I just want to make a note of what I'm thinking so I can revisit it later if, under some miraculous set of circumstances, my original post is 100% accurate.

If there are no uncountable ordinals, then perhaps the class of all ordinals is still uncountable. That cannot be true either, however, because if it were, then the class of all ordinals could be considered a set that is in fact the next ordinal, making it the first uncountable ordinal from which to again start building more ordinals off of and bring us back to where we started. From there, my matrix could again be applied to show it is not uncountable, forcing a contradiction. Therefore, the class of ordinals is countable and seemingly expands into a $2^{\aleph_0}$ space that can never be filled, much like our universe, is my speculation. Lol, and by Universe I mean a sequence of black holes, each of which is an ordinal so to speak.

 

[AplanisTophet]

I should mention that $g(x) = \beta * x$ would be a normal function and that $\omega_1$ could be partitioned using the sequence $[0,\beta), [\beta, \beta*2), [\beta*2,\beta*3), \dots$, which is why $\omega_1$ is countable given $\beta < \omega_1$ (again, assuming by some miracle that the OP is accurate).

That said, I put the function $f$ in my original list of questions for a reason. Consider making a normal function out of the 'limits' of the rows, some of which would be in the first column I speculate, so as to climb up the matrix that way. I'm guessing it would work out just fine...

The first column can be represented by the function $z(x) = x + 1$, which is not a normal function under the standard definition.

 

[AplanisTophet]

I should mention that $g(x) = \beta * x$ would be a normal function and that $\omega_1$ could be partitioned using the sequence $[0,\beta), [\beta, \beta*2), [\beta*2,\beta*3), \dots$, which is why $\omega_1$ is countable given $\beta < \omega_1$ (again, assuming by some miracle that the OP is accurate).

I don't want to keep posting with "if" the OP is correct type ideas, but I'll point out that some choice is generally required to show that a countable collection of countable sets is countable so the above mentioned partition isn't necessarily countable.

Rhetorically, maybe we can keep the axiom of choice if the well ordering theorem wasn't made to imply uncountable ordinals in a slightly altered form of ZF that could also accommodate the class of (countable) ordinals...

Anyways, I may post some philosophic stuff eventually, but I think that I'll give the math in this thread a rest now. Again, thanks for reading and I hope you enjoyed it!

 

[AplanisTophet]

Anyways, I may post some philosophic stuff eventually, but I think that I'll give the math in this thread a rest now. Again, thanks for reading and I hope you enjoyed it!

I have written a philosophic, scientific, and religious statement to accompany my work. I will likely post it in the other forum under my true name (Brendon) when I feel so inclined. Just working out a few kinks as to the timing of things. I will post a link here in this thread if and when I post my statement. It probably won't be until December.

 

[berkeman]

Thread closed temporarily for Moderation...

Update -- the OP has left us, so the thread will stay closed.