Are there smaller orders (cardinalities) of infinity?

I am using this source as a basis for the language to ask this question.

Considering the topic of degrees of infinity, are there smaller degrees than ℵ0 (aleph null, also called ω)?

Following the logic of the article, ω can be constructed, essentially of natural numbers, and the article states,

This assumption is called the Axiom of Infinity. The smallest such I is ω. It is the set of natural numbers.

Consider the following set construction.

• Consider ω, the set of natural numbers.

• Take proposed ℵ-1 to be the prime numbers, as an example.

Proof of Hypothesis

[Please pardon if there is simpler or alternative notation to the following]

Demonstration that this should equate to a smaller infinity.

Foundation

1. Define P(n) as the nth prime number.

P(1)=2, P(2)=3, P(3)=5, P(100)=541, P(200)=1223, and so on.

• Postulate: P(n) is an infinite set.
• We will refer to the set of prime numbers as Pprimes.
2. Define NP(n) as the count of primes up to that number (n = ω - 1 in all cases, although this is irrelevant. You could likewise define NP(1) to be == 0).

NP(1)=1 , NP(2) [2,3], NP(3)=2 [2,3]
NP(4)=3 [2,3,5], NP(5)=3, NP(6)=4 [2,3,5,7]
NP(100)=25, NP(541)=100
NP(200)=46, NP(1223)=200

3. Define RP(n) to be the percentage of prime numbers, beginning at 2 (n = ω - 1).

Define RP(n) = NP(n) / (n - 1)

This produces a set something like this.

• RP(1) = 1
• RP(10) = 0.444
• RP(100) = 0.2525252525
• RP(200) = 0.2311557789
• RP(500) = 0.1903807615
• RP(1000) = 0.1681681682
• RP(1250) = 0.1633306645
1. A trendline of this function demonstrates that this approximates to 1 / ln(n). Although I wouldn't know how to find the source, I remember that this is an already established fact, so it is treated as a postulate..

As shown:
RP(n) ~ 1 / ln(n)

Demonstration

Now, consider the following in terms of degrees of infinity.

1. RP(n) approximates to 1 / ln(n) as n → ∞.
2. The limit of ln(n) as n → ∞ is ∞, and so the limit of 1 / ln(n) as n → ∞ = 0.
• Consideration: Do we have to consider the order of the infinity here?
3. The limit of RP(n) as n → ∞ is 0 (zero).
• This indicates that the total percentage of prime numbers in ω is exactly 0%.
• Consideration: For any n in ω, there is always infinitely (of at least the order of ℵ0) more non-primes.
• e.g., simply to avoid overcounting, we can use, n2, n3, n4, etc. Failure to factor in the numbers where primes are multiplied with each other only would only further lower the percentage, which is already trending towards 0 (zero).

Apparent Conclusion:

1. Since there are infinite members (numbers) in ω,
2. And there are infinite prime numbers in ω,
3. Yet there are infinitely more non-primes (ℵ0) in ω than there are primes (total percentage is 0%, theoretically),
4. Ergo, the set of prime numbers (Proposed ℵ-1) is a lower degree of infinity than ℵ0, despite them both being indexed by n, which is of the order of ω.
5. This is because there is no one-to-one correlation between ℕ and Pprimes, as n → ∞, because...
6. ℕ is infinitely larger than Pprimes (aka, ℵ-1), wherein as n → ∞, there are always infinite primes (of the order of ℵ0) in ℕ as compared to every member in Pprimes.
7. Further, since Pprimes can be indexed by n, one could then theoretically index the set Pprimes(n) with the set Pprimes(n) itself, hence Pprimes(P(n)). This would then result in a proposed set, ℵ-2, infinitely smaller than ℵ-1, followed by ℵ-3, ℵ-4, ℵ-5, ... and so forth.
• That is to say, if you took a subset of Pprimes as only the prime numbers that are represented by prime indicies in the original Pprimes (we would take the 1st prime number, the 2nd, 3rd, not the 4th, the 5th, 7th, 11th, and so on), not only would it be infite itself and be infinitely smaller than ℕ, but it would also be infinitely smaller than Pprimes itself, or ℵ-1, thus ℵ-2.
• This is because, following the same example, following the same analysis, there will always be infinitely less items in ℵ-n than in its master, or parent-set, ℵ(-(n+1)).
• -2 then is Pprimes(P(n)), ℵ-3 = Pprimes(PprimesP(n)), and so on. Each being infinitely less than the previous infinity.

Analysis

The basis for this sub-set, ℵ-n, is not based on prime numbers, per-se, but is based on the creating an infinite sub-set of ℕ, hence ω or ℵ0, whose representation in ℕ is theoretically 0 (zero) as n → ∞.

Whenever such a sub-set occurs, if such a sub-set can be indexed by n, if it is then indexed by itself, it will always, therefore, produce an infinite set, which is represented by an absolute 0% representation in its parent set in the whole.

Because the parent set is always of an infinite magnitude greater for every n, and because there exists no one-to-one correlation between any item, it demonstrates itself to be of an order of magnitude less than ℵ0, hence, ℵ-n.

Each is progressively 'less infinite' than its predecessor, and yet is still infinite.

Projections

Given that the above holds true, this could then raise the possibility of sparse-ness of infinitude.

As there appears to be no limit to the infiniteness of the level of infinity (ℵ), there would then hold to be no limit to the sparseness of infinity, while still being infinity (ℵ-∞).

Each progressive level is, of course, infinite itself, but each is infinitely less infinate than the infinity before.

Edit 1

After review, I had thought that bijection covered it, but I have considered some more. After having reviewed other similar questions, now with the similar questions and more specific nomenclature (cardinality, bijection, etc), I have further inquiry.

While I appreciate the often in-depth answers that have been given to similar threads, there remains certain questions as to the details of the proof of these things. I will explain.

I particularly favored the Hilbert Hotel scenario of an infinitely large hotel accomodating an infinitely large arrival, of varying orders of infinity. This and more deeply studying Cantor's theorem illustrated further that the problem is indeed in degrees of infinities.

After studying these two examples (from various sources, not just the above), what I see in the degrees of infinitude is that it is much like dimensionality. In the same way that you cannot represent a 2-dimensional reality within 1-dimensional space, the move from, say ℕ to ℝ is an increase in complexity.

I had postulated before that, given a finite number of decimals, one could easily represent 3-D space by the interleaving method, as it was called in the wikipedia article above. Simply let every third digit represent a value from the X, Y, an Z plane, repsectively, but, as the Hilbert Hotel paradox represents, as you can allow up to infinite coaches with up to infinite teammembers, and even have infinite carriers arrive, each with infinite coaches, each with infinite team members, and still supply rooms in an infinite hotel, the article indicates you cannot do this "infinitely", even if the members themselves are finite.

The complexity of infinities appears quite similar to dimensionality. Add a dimension, add a layer of infinity. Yet, in the case of the transition from the original set of Cantor's Theorem, ℕ to ℝ, we do not see an increase in dimentionality, but we see a "filling in the gaps" between items. We see, between each point on the countability index a linear infinity (ℕ). While ℕ consists of 1, 2, 3 .. ∞, in the realm of ℝ, we now have ℕ between 1 and 2, and ℕ between 2 and 3, and so on, all the way up to ∞. This is shown by simple observation--The numbers after the decimal point are simply written in reverse order, with leading and trailing zeros likewise omitted on either end.

What this means is that ℝ really consists of a 2-dimensional space (A, B), where A & B are both in the set of ℕ. Our particular notation is nonessential. Additionally, for a typical x-y axis, one could conceive that it is really four dimensions of infinity, two for each x and y term. Since each r can be written as (A,B), an x-y point could be represented as (xA, xB, yA, yB). Were one inclined, one could add a second "decimal point" and add a third infinity to each of them (eg., 123.456.789), although what use it would be, and what it would mean is not clearly represented by our natural world. Another way to think of it would be that the decimal point added a "depth" between each number in the linear infinity of ℕ. If a third were added, it could be called "width", and then "height", etc. (e.g., 123.456.789.3892.3892). Each simply adds a complexity. But, is there a lessening degree to this? Let's look at what I see.

We consider the prime numbers, or possibly other infinite sets wherein the percentage of the whole of ℕ tends towards 0% as n → ∞..

The argument against it is that because it can be written as f(n), where n ∈ ℕ? Is it true that it is a bijection, simply because it can be written as f(n)?

My thoughts are no. This is not a valid axiom. Why? Because of Cantor's own demonstration. By using the diagonal logic, Cantor developed a system of attempting to number the members of ℕ by placing the numerators on the vertical axis and the denominators on the horizontal axis. Then, by employing a diagonal approach, he established an ordered, one-to-one correlation from the elements of ℝ to ℕ. His proof was then to show that while every number from ℕ was duly represented, every number from ℝ was not (in essence).

In this way, whether realized or not, Cantor has established that for every function of the form f( ℕ ), it could be re-written as as a function on ℝ. f( ℕ ) becomes f( g( r ) ), where r ∈ ℝ and g(r) is the diagonal ordering from Cantor's grid. Thus, a new function Fr(r) is born ( f(g(r)) for short).

What is interesting, therefore, is that since ∣ ℝ ∣ > ∣ ℕ ∣, for every element of ℕ, there is a corresponding value in ℝ, but the reverse is NOT true, and you cannot get a complete inverse of the function generating any value in ℝ from ℕ, for the simple reason of above (∣ ℝ ∣ > ∣ ℕ ∣).

As already demonstrated this is simply due to the fact of two infinities. ℕ consists of one infinity, the one to the left of the decimal only. ℝ consists of two, one before and after.

But, going back to the single number line then, we see that ℕ was represented linearly along its entire length. Then, between each point, an singular linear infinity of points was added between each member. This produced an extra infinity, producing an added cardinality, and is what Cantor proved.

But, consider the mathematics of the Primes. I will refer to this set as ℙ, since I found its unicode value, and it's simpler than my previous notation.

As described above, as n → ∞, RP(n) goes to 0 (1 / ln(n)). At infinity, this means that, effectively, on average, there are an infinite number of spaces between each point. What this has done is create a set that has at, at average, a linear infinity between them (in the scope of ℕ, since this is P(n)). At first glance, this appears to be exactly where Cantor proved the increase of cardinality of ℕ and ℝ, but lets consider other avenues.

In the case of the Hilbert Hotel, which greatly modeled this situation, we appear to have a situation wherein we can counter this. The issue, however, is that we see that for each prime added, we are in fact, adding not only an infinite number of elements to our complexity (infinity in depth), such as an unlimited number of football players (per the wiki article), which could be represented as P(n)2, P(n)3, P(n)4, ... , but we are indeed actually adding an entire powerset (by definition, oddly enough), every time we add a prime--each combination of every already present, which are infinite, in every combination, in every count! In the finite this works, but it does not work in the infinite, as Cantor demonstrated. Again, this is exactly the case for the increase in Cardinality previously noted. The fact that each additional prime adds a true powerset (plus some, because each element can be included infinitely), we see an indication that this is indeed a cardinality issue.

That is, for every prime P(n) added, as n → ∞, we represent in realm of ℕ a multiplicity of infinitude, to the degree of the "ultra-powerset".

And, finally, the basic argument, which states that since P(n) can be written as a function on ℕ that it is therefore bijection, we refer again to Cantor's own arguments that all numbers in the domain of ℕ can be represented by f( g(r) ), where r ∈ ℝ, because ∣ ℝ ∣ > ∣ ℕ ∣, but that the reverse is not true, because every number in ℝ cannot be represented by a number in ℕ.

In the same way, therefore, it seems to be apparent, and this is my question, while it is true that one could write P(n) where n ∈ ℕ, you could also write P(r), where r ∈ ℝ. Because of Cantor's ordering via the diagonal grid, he established a well-ordering, and hence, established that such a construct, P(r) is plausible. However, because we already know that ∣ ℝ ∣ > ∣ ℕ ∣, we cannot say that P(r) is bijective of P, because that would make ∣ ℝ ∣ = ∣ ℕ ∣, which is false. Regardless of index, following this argument, it appears we cannot say that any more than we can say that P(n) is bijective on P, because, as it seems, ∣ ℝ ∣ > ∣ ℕ ∣ > ∣ ℙ ∣. Since the one is greater than the other, the simple fact that we can index P by the elements of ℕ is not proof of bijectivity.

It would seem, perhaps with caveats then, that a given function F(n) is only bijective on ℕ when there exists an inverse function from F(n) to N(f), where every number in ℕ is capable of being generated, in a one-to-one fashion, from the elements of F. In relation to ℙ, simply because ℙ can be written as P(n), it does not make it bijective any more than P(r) is.

What would be needed to show that P(n) is bijection, then, would seem be to be to show in another way that every element from ℕ can be generated from P(n), so that you could construct a function N(p) which would include all of ℕ. As it seems on cursory analysis (without a thorough study), since 1 / ln(n) → ∞, this is impossible, P(n) is then not bijective. Or, in the language of the Hilbert Hotel, an infinite layers of possibly-finite sets are being introduced with each arrival, or, in the Cantor's language, each numeral n is producing an additional powerset of its own, hence, increasing its cardinality with each arrival.

I still ask, based on the dialogue above, whether it is then true that ∣ ℕ ∣ > ∣ ℙ ∣.

I simply as you to attempt to answer my logic, and point me to the logical inconsistency in the above discourse (as opposed to simply reverting back to Cantor's arguments, bijection, etc.). That is, at this point, I need a little bit more, as these themselves are now drawn into question.

Thank-you.

• My appologies on the length of the question. Please, if you plan to down-vote, consider the topic of the conversation. Many questions have addressed a cardinality less than ℵ0, however, none have addressed the specific claims of bijection and the theorems involved. My original question had come at the issue from a different direction, and when I realized the similarity to the other questions (despite it being unique), I realized I needed to answer them. The original question was much more concise. The second, in response. Please, consider reading my entire argument when reviewing. – user188464 Oct 30 '14 at 13:06
• Benjamin, this is an insanely long and hard to read question. This is not a forum where you present an idea and open it to discussion. It's a Q&A site. I can't see anywhere that there is a well-defined question, and maybe it's because I haven't read that huge block of text in front of me, but that's because I looked at the website you linked in the first line, and it didn't make me want to continue reading either. Not to mention that you start a question with "I have an idea for $\aleph_{-1}$" it immediately gives of a crankish vibe. Since that is easily refutable by studying basic set theory. – Asaf Karagila Oct 30 '14 at 13:14
• Okay, I agree. However, the question is in the title. If ℵ-1 is easily defeated (I had conceeded that point, and dismissed the question at first, but after after review, added my edit), then the answer is in the answer to the comment below, which was: Because of Cantor's diagonal argument itself, one could similarly represent the primes as P(r), using his grid. This would no more prove that P is of the same cardinality of N than it would prove that P is of the same cardinality of R. Should I simply reopen a new question an focus exclusively on this aspect of it, since its the focus? – user188464 Oct 30 '14 at 13:17
• (My too-long edit was simply attempting to come at this one point, the bijection claim, from three different avenues at the same time, but, effectively, the one above of P(r) should be enough to move forward) – user188464 Oct 30 '14 at 13:19
• (1) You're not using the diagonal argument right, and we have a huge list of questions that already don't use the diagonal argument right. (2) If $A\subseteq\Bbb N$ and $A$ is infinite, then the function $a\mapsto|\{a'\in A\mid a'<a\}|$ is a bijection. There, a bijection between the primes and the natural numbers. What might bother you is that this might not be recursive, or primitive recursive, or we didn't give a formula for the function. But all that is not a contradiction to Cantor and set theory. It shows that you have a misconception as to how cardinals and set theory works out. – Asaf Karagila Oct 30 '14 at 13:21

You haven't defined "order of infinity" in any of this interesting effort, so that in the end it says very little. If "order of infinity" means cardinality, then, no, there are just as many primes as natural numbers. There is a one-to-one correspondence between primes and naturals: map $1$ to the first prime $2$ to the second prime, and so on. As you're probably aware, there are infinitely many primes, so this is a bijection, even though we eventually have to map $n$ to a prime very much larger than $n$. If "order of infinity" means a notion of density, then, indeed, as you show there are far fewer primes than natural numbers. This is a perfectly valid concept, but it shouldn't be called $\aleph_{-1}$, because $\aleph$ is generally a notation reserved for cardinalities.

• You say "there are just as many primes as natural numbers". Could you please address the discussion above, also in terms of the one-to-one correspondence? Partic with regards to the limit of the Percentage of Primes tending towards zero, hence, there being infinitely more normal numbers than primes, although they are both infinite? This was my point, showing that there are not as many primes as natural numbers, and simply saying there are does nothing to defeat the logic above. You say Map 1 to the 1st prime, 2 to the 2nd, and so on. This was exactly my point as developed, that it doesn't. – user188464 Oct 30 '14 at 6:45
• That is to say... They cannot map one-to-one, because while it works in the finite realm to say such, as the limit goes to infinity, adding one more prime number adds an infinite number of natural numbers, before progressing forward. At infinity, which is what we're talking about, it fails, even as the limit of 1 / ln(n) goes to 0. The total population of prime numbers amidst the natural numbers is effectively 0 (zero) at infinity, which contradicts your statement, and was my point. – user188464 Oct 30 '14 at 6:53
• Appears I was wrong, and bijection was it after all. Had to study more precisely what Cantor was getting at. Thanks. – user188464 Oct 30 '14 at 10:16

I should clarify a few things you got wrong in the original post. Particularly the edited part.

First, you appear to only be using only finite decimal expansions on all your real numbers.
But this will not even get you out of the rational numbers as the finite decimal $a_n...a_1a_0.d_1d_2...d_m$ can be written as $a_n...a_1a_0d_1d_2...d_m/10^m$.

This means that for any number with an infinite decimal expansion you will be unable to use the decimal part as an integer as in your 'dimension'-argument.

Onto Cantors diagonal argument.
His first diagonal argument was not between $\mathbb{N}$ and $\mathbb{R}$, but between $\mathbb{N}$ and $\mathbb{Q}$, (or more precisely between $\mathbb{N}$ and $\mathbb{N}^2$).
It is in his second diagonal argument it is between $\mathbb{N}$ and $\mathbb{R}$, but what he proves there is the exact opposite, that there could not be any bijection between them and so $\mathbb{N}$ has a strictly less cardinality than $\mathbb{R}$.