Have I gone wrong in putting irrationals in one to one correspondence with natural numbers?

[Phar2wild]

Hi,

New member here and have been dabbling with some aspects of George Cantor's work.

I think I have found a way to put the irrationals in one to one correspondence with natural numbers
but thousands of mathematicians over the years might disagree. Is there a subtle error ( or even a
blatant one ) in my reasoning?

My attempt is here and it's short and simple:

phar2wild.ca

Comments here or replies to phar2wild@gmail.com would be appreciated.

Thanks,

Robert Hall

 

[Greg]

Hi Robert and welcome to MHB. :D

How do you conclude that " ... All decimal expansions will be on the list. ..."?

 

[Phar2wild]

I'm starting with an irrational number and cycling through all digits so that the first ten on the list will each start with a different digit and everything pass the first digit will be identical for the first ten entries on the list. It should not matter which number I start with since every possible decimal expansion will be created at some time.
For the first 100 on the list the entries will start with 00,01,02 through 99 but not in that order. Each of these will be followed by 15926535897932384626433 ( and so on ) this string of digits being the sequence of the digits of pi that starts at the third decimal place of pi.

My first attempt was a sort of bifurcation ( or should I say a "ten"furcation ) but this only gave me rational numbers of an ever increasing number of decimal places. A kindly math professor pointed out that any decimal expansion that I created with my first method could not produce an irrational number. I see that clearly now.

It is my contention that you cannot name a decimal expansion that would not at some point be created by this method.
Therefore all irrationals would eventually be churned out.

Robert

 

[Deveno]

Apply Cantor's second diagonal method to your list-what is your conclusion?

How will you verify that $e$ (for example) is on your list? Surely you can find a series of numbers on your list that grow ever closer to $e$, but that is not quite the same thing. For $e$ to be on your list, the digits have to match "all the way down".

Starting with $\pi - 3$ just confuses the issue, your first number is just $\pi - 3$, your second is $\pi - 2.9$, your third is $\pi - 2.8$..."and so on", creating the following correspondence:

$(\pi - 3) \leftrightarrow 0$
$(\pi - 3) + 0.1 \leftrightarrow 0.1$
$(\pi - 3) + 0.2 \leftrightarrow 0.2$
$\ \ \vdots$
$(\pi - 3) + 0.9 \leftrightarrow 0.9$
$(\pi - 3) + 0.01 \leftrightarrow 0.01$
$(\pi - 3) + 0.11 \leftrightarrow 0.11$
$(\pi - 3) + 0.21\leftrightarrow 0.21$

so we're only going to get as many "readings" on your odometer as we have *finite* decimals between $0$ and $1$ (there's no special advantage to starting with $\pi - 3$).

There's a world (literally) of difference between "infinitely close" and "infinitely (absolutely) equal".

The ancients thought of the real line as a *continuum*. With ANY kind of decimal approximation (no matter what random reading we start our crazy odometer with), we only get 10 choices ("stops") at digit $x$. The chance that a particular irrational number will be "hit" by one of those 10 choices is statistically 0. It DOES NOT MATTER how "far down we go" in the digits, we're still going to "come up short" (or long). Exactitude is a sort of intractable problem (no one has ever written down $\pi$ in decimal form, and no one ever will).

Base 10 ("decimals") is also a distraction, here-you may as well limit the digits to $0$ and $1$, you don't actually get "fewer" readings (because we have an infinite number of rotors). You can think of these as "base two" decimals, or numbers inscribed on a perfect ruler. Again, the fact that we can get "as close as we like" to any conceivable irrational isn't "good enough", it has to be "turtles all the way down".

Your list will always be chasing numbers it doesn't quite match, and adding the ones you *know* you missed doesn't help, either.

 

[Phar2wild]

Deveno,

Thank you for replying and taking the time; your answer is interesting and I will have to think about it.

Please clarify which Cantor proof you are referring to:

I assume the second is the proof that you can always create another irrational number not on the list
versus the first which I believe creates all the fractions and not leaving any out.
Robert Hall

 

[Deveno]

Deveno,

Thank you for replying and taking the time; your answer is interesting and I will have to think about it.

Please clarify which Cantor proof you are referring to:

I assume the second is the proof that you can always create another irrational number not on the list
versus the first which I believe creates all the fractions and not leaving any out.
Robert Hall

Yes, that's the one. For definiteness-

Suppose your odometer readings are:

$a_1 = d_{11}d_{12}d_{13}\dots$
$a_2 = d_{21}d_{22}d_{23}\dots$
$a_3 = d_{31}d_{32}d_{33}\dots$
$\ \ \vdots$

So that, in particular, for any natural number $k$, we have:

$a_k = d_{k1}d_{k2}d_{k3}\dots$, where $d_{kj}$ is the digit in the $j$-th decimal place.

We define a real number $b$ with $0 < b < 1$, by:

$b = 0.c_1c_2c_3\dots$ where $c_j$ is a digit chosen in the following way:

If $d_{jj} = 1$ then $c_j = 2$, otherwise $c_j = 1$.

Note that $b \neq 0$, since it has a non-zero digit. Note that since no digit is 9, that $b < 1$.

Now $b \neq a_k$ for any $k$, since it differs from the decimal expansion for $a_k$ in the $k$-th place, at the very least.

So the $a_k$ cannot represent a complete list of all real numbers between 0 and 1, since $b$ does not occur on it.

 

[Phar2wild]

Deveno,

In reply to your posts: ( In no particular order )

Item 1

Quote: "Apply Cantor's second diagonal method to your list-what is your conclusion?" Unquote.

There is a conflict between what I claim and Cantor's second proof and it is not to be lightly dismissed. If there is a list of all possible numbers can there be a number that is not on the list? Cantor's idea was original and brilliant but to claim that these are numbers that can't be on a list of all possible numbers is illogical. This seems to have the makings of a true paradox but I believe that the statement "You can't have a number that is not on a list of all possible numbers" trumps Cantor's proof. This should be accepted as ( for want of a better word ) some kind of axiom; it is simple, understandable and ironclad.

Item 2

My choice of pi ( I don't know how to get my computer to print the pi symbol! ) is entirely arbitrary and is of no significance ). Any irrational number will do as a start point.Item 3

Yes, I understand the concept of Cantor's second proof. I have seen it in many math books mostly for the layman and I see that yours is more precise and defined than most. I just wanted to be sure we were on the same page when I asked which proof.Item 4 ( This is the big one ):

Here I show quotes from your post ( from various parts of your reply though it is not my intention to take anything out of context. )

Quote:

the digits have to match "all the way down".

It DOES NOT MATTER how "far down we go" in the digits, we're still going to "come up short" (or long).

the fact that we can get "as close as we like" to any conceivable irrational isn't "good enough", it has to be "turtles all the way down".

Unquote

I submit that I have shown a way to create a list of all possible decimal expansions and thus all the irrationals. I have taken an existing irrational and changed the digits one at a time to create a new number. This is the same process that Cantor used with his second diagonal proof; he created a new irrational one decimal place at a time. Somehow he got "all the way down." Of course we have no hope of ever writing a single irrational sequentially but the process is possible. My method is tied to Cantor's; If his proof is valid so is my process. If mine is not then his proof must fall.

Sincerely

Robert H

 

[HallsofIvy]

Early on you say "Imagine if you will an odometer; the kind that is in your car and keeps track of the miles or kilometers that you have driven.
This odometer starts at zero and rotates to 1, then 2 and so on.
After a while it racks up to 150,678 or whatever number you can envision within the limitations
of how it was designed."

Your odometer shows only integers and only a finite number of integers at that.

 

[Phar2wild]

To HallsofIvy,

I assume you have my claim at phar2wild.ca

It goes on to say:

Quote:

Now imagine a very special odometer; different in two ways from the one in your car:

First, instead of operating from right to left, this one operates from left to right
so that the first rotor to complete a revolution is on the left side.

Second, this odometer has an infinite number of rotors.

Unquote.

The integers follow a decimal point so that there is an irrational number created that is between 0 & 1.

Robert

 

[HallsofIvy]

If you mean that there are an infinite number of rotors to the right of the decimal point, then where does your "list" begin?

 

[Phar2wild]

It begins with an arbitrary number. I choose pi because it just popped into my head.
( Actually pi-3 )
This odometer ( actually just a metaphor for the process like the Infinite Hotel is used to illustrate some properties of infinite sets ) will work its way through all permutations. It will generate all the irrationals.

Robert

 

[Deveno]

It begins with an arbitrary number. I choose pi because it just popped into my head.
( Actually pi-3 )
This odometer ( actually just a metaphor for the process like the Infinite Hotel is used to illustrate some properties of infinite sets ) will work its way through all permutations. It will generate all the irrationals.

Robert

The initial "seed value" of the odometer (its first setting) doesn't matter. If it truly goes through all the permutations, we may as well imagine it starts at $0000\dots$.

But then it can be seen that it merely counts, just as the natural numbers do.

The odometer *will* reach every permutation of some finite *initial* string. But this is not *every* permutation of every infinite string.

To match ONE given irrational number, it can take an infinite number of turns, which will be a countable infinity. For example, if we are trying to match $2 - e$, then we may have to wait up to $9$ "turns" until the first digit is $7$ (certainly no more than that). We may have to wait an additional $90$ (but perhaps less) turns to have the first two digits be $71$, and then perhaps $900$ more until we have the initial string $718$.

So after a very long (infinite) time, we may have a "perfect match" (note how we keep having to wait an increasingly long, but finite, interval to get the "next digit").

And that's just *one* irrational. We have uncountably more to get to. You are badly underestimating how difficult it is to match ALL the infinite strings.

I'll try to illustrate with an analogy:

An infinite string is like a power series, while a finite string is like a polynomial. This power series:

$e^{x} = 1 + x + \dfrac{x^2}{2} + \dfrac{x^3}{3!} +\cdots + \dfrac{x^k}{k!} + \cdots$

can be approximately by a polynomial to any "error" desired, but approximation is not EQUALITY. There is, in fact, NO polynomial, no matter how large a degree, which approximates it EXACTLY, or even particularly WELL outside of a certain interval (this interval will get quite large, but it will never include EVERY integer).

What I think you are trying to get at, is a kind of theorem called the "Downward Skolem Lowenheim Theorem" which states that if a (first-order) set theory admits a model, it admits a countable model. In more ordinary terms, this amounts to saying that that we can declare uncountable collections "too big to be called sets".

The trouble with applying this to the real numbers is that the completeness axiom (or equivalently, the Cauchy convergence theorem, or the "least-upper bound property") is not a first order logical statement: it quantifies not over individual real numbers, but over predicates defined on real numbers (these predicates define bounded sets of real numbers). In short, the real numbers form a subset of the POWER SET of the rationals, and things get a bit more complicated when you talk about infinite subsets of an infinite set.

There's too many possibilities for your odometer to reach. It seems like a reasonable extension of inductive reasoning to suppose that "infinite combinations of a finite number of things" are as many as "finite combinations of an infinite number of things", but it just ain't so.

 

[Phar2wild]

First to clarify: I am not a mathematician, so while I find your explanation interesting I would have to struggle through some of the things that you mentioned. And I will have to look up the "Downward Skolem Lowenheim Theorem" ( First I heard of it! )

However I have been contemplating Cantor's basic work for some time and I try to be a rational ( pun intended ) being!

So I will skip, for now, some of what you have taken the time to write because I just want to make the following points in response to your post.

I did not imagine the odometer set to .00000000... to begin. To do so I would have started at zero and then your argument would make perfect sense and you would be absolutely correct.

I started with an irrational number. ( to create one would take too long! :-) ) So each turn of the odometer a new irrational is created. I suppose it will also create the rationals along the way.

I will stick to my assertion that Cantor did what I did; he created an irrational one digit at a time. And what's more he was prepared to create more than one! I request that the same opportunity be afforded to me to use this process.

If I fall, we ( Cantor and I ) fall together!

Again, many thanks for your time.
Robert H

P.S. Re: "infinite combinations of a finite number of things". Isn't this just impossible? Example?

 

[Deveno]

First to clarify: I am not a mathematician, so while I find your explanation interesting I would have to struggle through some of the things that you mentioned. And I will have to look up the "Downward Skolem Lowenheim Theorem" ( First I heard of it! )

However I have been contemplating Cantor's basic work for some time and I try to be a rational ( pun intended ) being!

So I will skip, for now, some of what you have taken the time to write because I just want to make the following points in response to your post.

I did not imagine the odometer set to .00000000... to begin. To do so I would have started at zero and then your argument would make perfect sense and you would be absolutely correct.

I started with an irrational number. ( to create one would take too long! :-) ) So each turn of the odometer a new irrational is created. I suppose it will also create the rationals along the way.

I will stick to my assertion that Cantor did what I did; he created an irrational one digit at a time. And what's more he was prepared to create more than one! I request that the same opportunity be afforded to me to use this process.

If I fall, we ( Cantor and I ) fall together!

Again, many thanks for your time.
Robert H

P.S. Re: "infinite combinations of a finite number of things". Isn't this just impossible? Example?

Your odometer is an example-we have a finite number of digits (0-9) and we are arranging them in an infinite sequence.

The fallacy lies in you imagining that you will create *every* irrational number. In the language of cardinal arithmetic, if we call the cardinality of the natural numbers $N$, we have:

$10^N = 2^N > N$.

By the $>$ sign, I mean that there is no 1-1 correspondence between a set with $2^N$ elements and a set with $N$ elements.

You'll have to take my word for this (for now, you can research it if you like), but the number of elements of the power set $\mathcal{P}(A)$ (the set of all subsets of a set $A$), is $2^{|A|}$. This is easy to verify (via induction) for finite sets.

The reason $2^A$ is also used for $\mathcal{P}(A)$, is because there is a 1-1 correspondence between $\mathcal{P}(A)$ and *functions* $f:A \to \{0,1\}$.

To see this, let $S \subseteq A$, and define the function:

$\chi_S: A \to \{0,1\}$ (called the *characteristic function* of $S$) by:

$\chi_S(a) = 1$ if $a \in S$
$\chi_S(a) = 0$ if $a \not\in S$.

Cantor showed there is no 1-1 correspondence between $|A|$ and $|\mathcal{P}(A)| = 2^{|A|}$ for ANY set $A$. His most general argument runs as follows:

Suppose there were such a 1-1 correpsondence $f:A \to \mathcal{P}(A)$. Such a function would be onto $\mathcal{P}(A)$, and we will show this cannot happen.

Let $B = \{x \in A: x \not\in f(x)\}$

Now either $B$ has something in it, or it doesn't. Let's suppose there isn't any such $x$'s in $B$. This means that for EVERY $x \in A$, $x \in f(x)$. So every element of $A$ maps to some non-empty subset of $A$. So no element of $A$ maps to the empty set (which is a bona-fide element of $\mathcal{P}(A)$) ,and $f$ is thus not onto, a contradiction.

So $B$ must have something in it, then. Therefore, there is some $x \in A$ with $x \not\in f(x)$. Since $f$ is assumed onto, we must have for some $y \in A$, that $f(y) = B$.

Now if $y \in B$, then (since $f(y) = B$), then $y \in f(y)$, and thus $y \not\in B$.

However, if $y \not\in B$, then $y \in A$ (since $B \subseteq A$) and $y \not\in f(y) = B$, and thus $y \in B$.

Both of these are self-contradictions, so no such non-empty set can exist.

Since $B$ can neither be empty, nor non-empty, and is definitely *some* subset of $A$, $f$ cannot be onto, there cannot be any $y \in A$ such that $f(y) = B$.

In a bit more detail for the natural numbers case:

Let's form some subsets of the natural numbers:

$\{1\}, \{2,3\}, \{2,4,6,\dots\}, \{1,3,5,\dots\}, \{2,4,8,16,32,\dots\}, \{3,5,7\}$ and so on. Note some of these subsets are infinite (the third one is meant to be the even natural numbers, the fourth the odd natural numbers, the fifth the powers of 2 starting with 2, and the other sets are finite).

Suppose we assume we can pair these (completely) in some fashion with natural numbers:

$1 \leftrightarrow \{1\}$
$2 \leftrightarrow \{2,3\}$
$3 \leftrightarrow \{2,4,6,\dots\}$
$4 \leftrightarrow \{1,3,5,\dots\}$
$5 \leftrightarrow \{2,4,8,16,32,\dots\}$
$6 \leftrightarrow \{3,5,7\}$

Note that 1 is paired with a set that contains 1 (and so is 2), but that 3,4,5,6 are not.

Call numbers that are paired with a set that contains them "greedy", and numbers that are paired with a set that does not contain them "pure".

Consider the subset $S \subseteq \Bbb N$ consisting of all "pure" numbers. This is a subset of the natural numbers, and so is paired with some natural number $k$:

$k \leftrightarrow S$.

Is $k$ "greedy"? If so, then $k \in S$, but everything in $S$ is "pure".

Is $k$ "pure"? If so, then $k \in S$, and is thus "greedy".

Now it could be that *all* numbers are "greedy", so that $S$ is empty. In that case, no natural number maps to the empty set, and our correspondence is not complete.

So there is no natural number $k$ that can possibly be paired with $S$, violating our original assessment that such a 1-1 complete pairing could be made.

Thus the set of all subsets of natural numbers is *bigger* that the set of natural numbers, not all infinities are created equal.

 

[Phar2wild]

Reply to Deveno:

Sorry, when you said "infinite combinations of a finite number of things" I was assuming something along the lines of combinatorics which I think to mean would give me the power set of the set of digits from 0 to 9. I did not consider that you were indicating that a digit could be used more than once in the combination.

Yes I am imagining that I will create every irrational number. I start with pi-3 ( Been trying to get the pi symbol to show up properly as I type but I haven`t had much luck ) for the first item on my list. Second item is (pi-3)+.1 Third is (pi-3)+.2 and so on.
After first rotor has made 1 complete rotation ( ten clicks ) it time to change the second place after the decimal. After a thousand clicks I have a thousand different irrationals. So far the only changes have been to the first three decimal places. I keep going to the tenth decimal place and even the 345,835,444,968th decimal place. I could continue without end and ask myself is there a limit, any number of places after the decimal where my magical imaginary machine must stop because it has gone too far?
The key point that I am making here is that when Cantor devised his second diagonal proof he asked us to imagine creating a new irrational by sequential steps of assembling the number based on the properties of a different decimal place of each irrational on an existing list.

Cantor generated numbers with a infinite number of decimal places and he could generate an infinite number of them.I generate numbers with an infinite number of decimal places and I can generate an infinite number of them.Thanks for listening,

Robert H

 

[HallsofIvy]

Cantor generated numbers with a infinite number of decimal places and he could generate an infinite number of them.

No, he didn't. He assumed the existence of the real numbers, he did not "generate them". In fact, one could interpret his result as saying that it is impossible to "generate" the real numbers in the way you mean.

 

[Phar2wild]

No, he didn't. He assumed the existence of the real numbers, he did not "generate them". In fact, one could interpret his result as saying that it is impossible to "generate" the real numbers in the way you mean.

No, I'm pretty sure he "generated" the first digit after the decimal place of his "new" number by taking the first digit after the decimal place in the first number on the supposed list of all irrationals and changing it. Same for the second digit. He may have "assumed" that he could continue indefinitely and create a new irrational this way. Cantor showed a "method" to create a new irrational. I show a method to create a list of all the irrationals.

 

[Deveno]

No, I'm pretty sure he "generated" the first digit after the decimal place of his "new" number by taking the first digit after the decimal place in the first number on the supposed list of all irrationals and changing it. Same for the second digit. He may have "assumed" that he could continue indefinitely and create a new irrational this way. Cantor showed a "method" to create a new irrational. I show a method to create a list of all the irrationals.

You're using the word "all" far too loosely. Ask yourself: how do you know any irrational number, such as say, $\sqrt{2}-1$ is reachable on your odometer?

Yes, you've created a way of generating a countable number of real numbers, and in fact, I believe it can be proven that all of them are irrational. What you have not done, is prove your list is exhaustive.

With exact equality (of every digit), a miss is as good as a mile. Matching most, or the first several million, digits doesn't count. The real number 0 has only one decimal expansion:

$0 = 0.00000\dots$

and two real numbers $a,b$ are equal if and only if $a - b = 0$.

(There is one caveat: we need to identify pairs that end in an infinite string of 0's and an infinite string of 9's in a certain way, using as our guideline:

$0.99999999\dots = 1.00000000\dots$).

I also believe you are badly mis-interpreting Cantor's proof: it is a proof by contradiction, that assumes such a complete list of reals (including all rationals AND irrationals) between 0 and 1 *could* be made. His "generated" number, which is not on the list, means the assumption that a complete list could be made, is erroneous.

Applying his proof to your odometer, we'd now need a SECOND odometer, starting with our new "unreachable" number. We can interweave both odometer readings into a new list, like so:

$a_1,b_1,a_2,b_2,a_3,b_3,\dots...$

where $a_k$ and $b_k$ are the $k$-th readings from odometer 1 and odometer 2, respectively.

Applying Cantor's diagonal method again, we'll get a new number not reached by EITHER odometer 1, and odometer 2. So, now, we need a third...

In fact, even an INFINITE number of odometers won't be enough (although the proof of this is a bit involved, and uses the axiom of choice, which some mathematicians prefer to eschew). And if that's not enough, I'm out of ideas...your move, sir.

 

[Phar2wild]

You're using the word "all" far too loosely. Ask yourself: how do you know any irrational number, such as say, $\sqrt{2}-1$ is reachable on your odometer?

Yes, you've created a way of generating a countable number of real numbers, and in fact, I believe it can be proven that all of them are irrational. What you have not done, is prove your list is exhaustive.

With exact equality (of every digit), a miss is as good as a mile. Matching most, or the first several million, digits doesn't count. The real number 0 has only one decimal expansion:

$0 = 0.00000\dots$

and two real numbers $a,b$ are equal if and only if $a - b = 0$.

(There is one caveat: we need to identify pairs that end in an infinite string of 0's and an infinite string of 9's in a certain way, using as our guideline:

$0.99999999\dots = 1.00000000\dots$).

I also believe you are badly mis-interpreting Cantor's proof: it is a proof by contradiction, that assumes such a complete list of reals (including all rationals AND irrationals) between 0 and 1 *could* be made. His "generated" number, which is not on the list, means the assumption that a complete list could be made, is erroneous.

Applying his proof to your odometer, we'd now need a SECOND odometer, starting with our new "unreachable" number. We can interweave both odometer readings into a new list, like so:

$a_1,b_1,a_2,b_2,a_3,b_3,\dots...$

where $a_k$ and $b_k$ are the $k$-th readings from odometer 1 and odometer 2, respectively.

Applying Cantor's diagonal method again, we'll get a new number not reached by EITHER odometer 1, and odometer 2. So, now, we need a third...

In fact, even an INFINITE number of odometers won't be enough (although the proof of this is a bit involved, and uses the axiom of choice, which some mathematicians prefer to eschew). And if that's not enough, I'm out of ideas...your move, sir.

I suspect that there are many things in math the are subtle so just to make sure we are on the same page, is this what Cantor was saying: ( My perception of it anyway and this is as simple as I can make it ) ?

"Given a list of all possible irrationals I can show that there are irrational numbers that are not on the list."

Thanks.

Robert

 

[HallsofIvy]

Well, what Cantor proved was that if there were a list that purported to include all real numbers, then he could find a real number that was not on the list. That is sufficient to prove that the real numbers are not "countable". Since it is easy to show that the set of rational numbers is countable, it follows that the set of irrational numbers is not countable.

 

[Phar2wild]

Reply to Deveno:

First, let me again thank you and HallsofIvy for your replies and input. My purpose in these posts is to get my idea out there and defend it.
Getting back to your quoteQuote:

the digits have to match "all the way down".

It DOES NOT MATTER how "far down we go" in the digits, we're still going to "come up short" (or long).

the fact that we can get "as close as we like" to any conceivable irrational isn't "good enough", it has to be "turtles all the way down".

UnquoteI would like to bring your attention to two mathematical objects ( I would hazard a guess that you are already know of them ):

1. Cantor dust.

Description ( Quoted directly from http://jwilson.coe.uga.edu/EMAT6680Fa11/Frailey/FractalEssay.htm )

"It is formed by removing the middle one third of a line segment, and repeating the process for the resulting lines an infinite number of times."

2. Koch snowflake

Description ( Quoted directly from Koch Snowflake -- from Wolfram MathWorld )

"It is built by starting with an equilateral triangle, removing the inner third of each side, building another equilateral triangle at the location where the side was removed, and then repeating the process indefinitely"

I'm hoping that it is OK to use these quotes here as I only want to use this to add weight to my argument and not to plagiarize anything.

The point that I want to make is that these objects ( Which seem to be accepted by the math community ) are generated by a process that is described and must be repeated an infinite number of times. The process used sequential steps.

If we deny that we can get "all the way down" then we must deny that Cantor Dust and Kock Snowflakes can be created as mathematical objects. They are incomplete and we have no business describing their properties which would not be valid anyway since they could not be "created".

My "Infinite odometer" is no different. Yes it is imaginary; it requires no oil and never wears out. :-) It uses a process that must be repeated an infinite number of times.If my infinite odometer cannot do the job and go "all the way down" then it must come up short. Let us number the rotors from the left. The first is "1", the second is "2" and so on.
It is easy to imagine that rotor "1" makes a revolution. Likewise rotor # 2. If the odometer cannot make it "all the way down", then there there must be a first rotor of an infinite number of rotors that will never rotate.

Question: What is the ordinal number of this rotor?

Thanks,

Robert Hall