more numbers.

[Keiji]

I am going to say that the real number line is a "squarely infinite" number line.

Why?

1. If you take any two numbers on the real number line there are infinite numbers between them.
2. If you count how many of these "blocks" of numbers there are in the real number line there are infinitely many of these "blocks".
Thus we have infinite blocks * infinite numbers in a block = infinity squared, or squarely infinite.

This is why a floating point number always needs two "integer dimensions": the number to a certain number of SF, and the power of ten it is raised to. Both of these numbers are INTEGERS, and the integer number line is "linearly infinite".

You can also represent a decimal (terminating) number as two integers: the part before the decimal point, and the part after the decimal point.

And finally, you can represent any real rational number as a/b where a and b are integers.

Problem 1. This is inconsistent! I stated above and showed why the real number line is squarely infinite. However, if the real number line was squarely infinite, it would not contain irrational numbers.

Now the set of integers can be folded into a set of integers from 0 to infinity. This makes the set of integers a "linearly infinite" series.

The set of real numbers cannot be folded. However, if you represent a rational number as a/b, then pair a and b, you receive an integer value.

Problem 2. Integers are linearly infinite, but we just obtained an exactly equivalent number in a linearly infinite set from a squarely infinite set. This should not be possible.

Now, I'm sure many of you have read on this forum about my "number circle" - with infinity at the top, -1 at the left, 1 at the right, 0 at the bottom, and so forth. (If you haven't, go search for it, it's on here). In a number circle, you can divide it up into a finite number of blocks and each block has infinite numbers in it.

Problem 3. This makes it linearly infinite. However, the number circle contains all real numbers, so it should be squarely infinite.

That's all I can think of for now. I'd appreciate it if anyone could solve these problems.

[pat]

The more standard terminology is that the integers are countably infinite. They have the infinity Aleph-null (ℵ₀).

The real numbers cannot be put into one-to-one correspondence with the integers. Thus, they have a larger infinity. It is Aleph-one (ℵ₁) (assuming the continuum hypothesis).

1. If you take any two numbers on the real number line there are infinite numbers between them.
2. If you count how many of these "blocks" of numbers there are in the real number line there are infinitely many of these "blocks".
Thus we have infinite blocks * infinite numbers in a block = infinity squared, or squarely infinite.

There are ℵ₁ numbers between any two numbers on the real number line. There are ℵ₀ blocks of those numbers. You've got apples and oranges, you're putting them together and saying it makes a huge pile of oranges.

In fact, ℵ₀ squared is still ℵ₀. The case in point is that there are the same number of positive rational numbers as positive integers. You need at least 2^ℵ₀ (the set of all subsets of positive integers) to get to a larger infinity.

Problem 1. I believe you mis-stated. Maybe I'm just not catching antecedents well. You say "I stated above and showed why the real number line is squarely infinite". I agree that it is. You stated that. And, except for the apples/oranges problem mentioned earlier, you showed it. Then, you say that "if the real number line was squarely infinite, it would not contain irrational numbers". I suppose that depends... if you're not really talking about the real numbers but are talking instead about the rational numbers or the finite decimals, then you're not actually talking about the real number line. In that case, all you showed was that the real number line was as big or bigger than "squarely infinite". And, it is.... so that's no big shock.

Your "squarely infinite" is less than ℵ₁ but not less than ℵ₀, so it must be ℵ₀.

Problem 2. By virtue of the pairing we know that "squarely infinite" is "linearly infinite". And, by recursion, "n-foldly infinite" is "linearly infinite" for any finite n (where "n-foldly" means representable by an n-tuple of integers).

Problem 3. When you break the number circle up into a finite number of blocks, each of those blocks has ℵ₁ numbers in it. Thus, the real number circle is of order at least ℵ₁. If, again, you want to restrict yourself to finite decimals or rationals on the number circle, then you've got a finite number of "linearly infinite" segments and you're linearly infinite. But, as we saw in problem 2, "squarely infinite" is "linearly infinite".

ps. The number circle is an inverse stereographic projection of the real number line. Take the real number line. Put a circle of radius 1/2 tangent to the number line at 0. Now, draw a line from any point on the number line to the point on the circle diametrically opposed to zero. This line will intersect the circle at one point (other than the point diametrically opposed to zero). Identify that intersection point with the point on the number line.

[Keiji]

I see. So my big mistake was talking about the real numbers instead of the real rational numbers.

If ℵ₀ = ∞, what is ℵ₁?

And thanks for the ps. That would explain it perfectly. Of course, infinity is at the top because a tangent at exactly the top of the circle would be parallel to the number line and so never reach it.

[pat]

Cantor's point with ℵ₀ and ℵ₁ was that there is more than one infinity. The symbol ∞ is inadequate. Usually, one uses a lower-case omega ω to represent the cardinality of the integers and a lower-case c to represent the cardinality of the real numbers.

The omega has the implication that it's after all of the other things. The 'c' is for continuum.

[Keiji]

I know that... but what IS ℵ₁?

[pat]

It's the infinity of the real number line... as opposed to the infinity of the integers.... I'm not sure what you're asking.....

[Keiji]

I mean, define it.

I could define ℵ₀ like this: ∀x∈ℝ, ℵ₀ > x.

So how do you define ℵ₁?

[pat]

The only way that I can think of is this... | ℘(ℤ) | = ℵ₁

But, I believe it is also ℵ₁ = ℵ₀^ℵ₀.

[houserichichi]

ℵ₁ and |℘(ℤ)| are two different entities depending on the axioms that one is dealing with which is a result of the continuum hypothesis being a statement independent of the ZF axioms of set theory. The continuum hypothesis essentially says that the next transfinite cardinal number after ℵ₀ is ℵ₁ and that the cardinality of the reals (the continuum, which is defined as |℘(ℤ)|) is equal to ℵ₁.

ℵ₁ as far as I have ever understood it is the cardinality of the set of countable ordinals, however (un)enlightening that may be.

[Keiji]

The only way that I can think of is this... | ℘(ℤ) | = ℵ₁

And what does ℘ mean? (I assume it's a function)

But, I believe it is also ℵ₁ = ℵ₀^ℵ₀.

Impossible. ℵ₀^ℵ₀ is ℵ₀ hyper-3 ℵ₀. ℵ₀*ℵ₀ is ℵ₀ hyper-2 ℵ₀ and ℵ₀+ℵ₀ is ℵ₀ hyper-1 ℵ₀. So why should hyper-1 and hyper-2 give ℵ₀ and hyper-3 give ℵ₁?

I imagine that if ℵ₁ could be defined in terms of ℵ₀ using only an operator, it would be ℵ₀ hyper-ℵ₀ ℵ₀.

[houserichichi]

℘ is the power set and pat was saying that ℵ₁ is the cardinality of the power set of the integers.

I'm going to look the latter up and get back to you on it too, but I also think I may have read the same thing. The reason your hyper-functions don't work the same is because alephs are not numbers.

In certain set theories ℵ₁ = c = 2^ℵ₀ while in others this is not the case. This is why trying to "get" ℵ₁ from ℵ₀ is impossible because it depends entirely on the set theory that one is working in - it's an undecidable statement in ZF set theory.

[Keiji]

On second thought, I'm not so sure that n^ℵ₀ = ℵ₀.

Let a and b be constants such that a∈ℝ and a≠0, and the same for b.

a/0 = ℵ₀ and b/0 = ℵ₀.
a/0 + b = (a+0b)/0 = a/0 = ℵ₀.
So ℵ₀ + a = ℵ₀.

a/0 = ℵ₀
a/0 = a(1/0), so ℵ₀ = aℵ₀.
So ℵ₀ = aℵ₀.

Similarly, (1/0)^a = 1^a/0^a = 1/0, so ℵ₀^a = ℵ₀.

But I can't think of a way to prove that a^ℵ₀ = ℵ₀, since doing so would require simplifying a^1/0.

[PWrong]

Am I supposed to see an Aleph symbol? All I see is a little square.

℘ is the power set and pat was saying that ℵ1 is the cardinality of the power set of the integers.

Is that because a real number can be written as an infinite list of integers?

[houserichichi]

I'll avoid the aleph symbol, I wasn't aware people couldn't see them too. I'll use A to represent Aleph

At any rate, A₀^A₀ is certainly not A₀ by the simple fact that 2^A₀ is equal to the continuum which is defined to be bigger than A₀ (and it can be shown, in case you want to argue the fact, that the set of integers Z cannot be mapped bijectively into its power set P(Z), thus

A₀ := |Z| < |P(Z)| = 2^A₀ := |R| = c

Since 2 < A₀ one would imagine that 2^A₀ < A₀^A₀

[Keiji]

If A₀^A₀ > 2^A₀ = A₁, then what is n for A_n = A₀^A₀?

Would A₀^A₀ be A₂, or perhaps A_A₀?

By the way, I'm moving this to Genius Discussions.

[houserichichi]

Found my answer in a set of notes I took back at school.

If c is the continuum, that is c := 2^A₀ then

c^A₀ = (2^A₀)^A₀ = 2^A₀^A₀ = 2^A₀ = c

(I won't bother showing that A₀A₀ = A₀ but just believe me that it does and results from finding |N x N|)

Now, if 2 ≤ z ≤ c then

c = 2^A₀ ≤ z^A₀ ≤ c^A₀ = c

So to answer your question, since 2 < A₀ < c by definition we must have that A₀^A₀ = c and I was incorrect in my assumption in my previous post of the strict limitations on the ordering (< rather than ≤...I got a little hasty).

Also, 2^A₀ = c ≠ A₁...careful about the assumption.

(This took far too long to type out...frig)

Edit by Rob: <= to ≤ and != to ≠

[Keiji]

Found my answer in a set of notes I took back at school.

If c is the continuum, that is c := 2^A₀ then

c^A₀ = (2^A₀)^A₀ = 2^A₀^A₀ = 2^A₀ = c

(I won't bother showing that A₀A₀ = A₀ but just believe me that it does and results from finding |N x N|)

No problem; I proved that in my last post anyway (since A₀A₀ = A₀²).

Now, if 2 ≤ z ≤ c then

c = 2^A₀ ≤ z^A₀ ≤ c^A₀ = c

So you are saying that for any value of n greater than one, n^A₀ is the same?

Also, 2^A₀ = c ≠ A₁...careful about the assumption.

You said in your last post that A₁ may or may not be equal to c. What else could it equal?

[houserichichi]

So you are saying that for any value of n greater than one, n^A₀ is the same?

Any number greater than 1 but less than the continuum (2^A₀ = c). There are transfinite cardinals bigger than the continuum, for instance 2^c since a set (of any size, but for instance of size c) cannot be mapped bijectively into its power set (in this case of size 2^c) which actually turns out to be the cardinality of the set of real valued functions on [0,1].

You said in your last post that A₁ may or may not be equal to c. What else could it equal?

A₁ is the cardinality of the set of countable ordinals. The continuum hypothesis says that there is no cardinal number between A₀ and c but, as mentioned previously, it has been shown to be independent of ZF and ZFC axiomatic set theory so it's both true and not true. If the continuum hypothesis is true (depending on choice of axioms) then A₁ = c. Otherwise it's not. If CH is true then the cardinality of the set of countable ordinals is then equal to the power set of the integers which is equal to the continuum which is the cardinality of the reals. That's only if CH is true.

[quickfur]

Basically, the upshot of all this is that "infinity" is not a single entity, but a concept that encompasses a multitude of different things. The common terms "infinity" or "infinitely large" aren't precise enough to pin things down.

First of all, the "infinity" used in calculus is not the same thing as the Cantorian infinities (cardinals/ordinals), and it is this infinity that's frequently associated with the infinity symbol. Some complex number analysis use the set C U {infinity} to make the reasoning cleaner, but this infinity has nothing to do with the Cantorian infinities.

The Cantorian infinities, as others have said, come in two varieties: the ordinals, measuring sequence, and the cardinals, measuring size. In the finite realm, these two concepts are one and the same; however, once you go beyond the finite realm, these two concepts give rise to two very different set of infinities. Hence, when speaking of infinite things, one should not confuse the two.

The omegas are all ordinals, and consist of all well-ordered sets. (Well-ordered meaning that they have a less-than relation defined between their members, such that this relation is linear, and every possible subset of the set has a least element.) Note that multiple ordinals may correspond with the same cardinal (i.e., multiple ordinals have the same "number of elements"), simply by rearranging elements. For example, if you take the set N of natural numbers, and rename 0 to be omega, and define omega to be the greatest element in the set, then you end up with the set that is isomorphic to what Cantor names Omega+1. However, omega+1 has exactly the same number of elements as N itself, in terms of size. They are just arranged funny.

Cardinals completely disregard all orderings, and measure sets in terms of pure size. Two sets are defined to be of the same cardinality (the "same size") if there exists at least one bijection between them (a 1-to-1 and onto function between their elements). The idea, of course, is that we can't count up to an infinite quantity in order to compare the number of elements in two infinite sets, but if we can establish a 1-to-1 mapping between them with no elements left over, then they must be of the same size. (E.g., at a birthday party, if you have N hats and M people, and after passing out all the hats there are no hats left and no one with no hat, then you conclude N=M. The advantage of this approach is that you don't have to know what N and M are, which we can't know when N and M are infinite.)

Many infinite sets have the same cardinality: for example, the set of positive integers, the set of integers, the set of all rational numbers, the set of two-dimensional integer coordinates, the set of N-dimensional integer coordinates (where N is a finite integer, of course). The proof of all this is better left to another post. The point is that even sets that we "intuitively" consider as "larger", such as the set of rational numbers vs. the set of whole numbers, are of the same size. In this particular case, the size is Aleph₀.

Given so many sets that are the same size even though we imagine them to be different, we may be tempted to think that there's only one infinite cardinal, Aleph₀. However, Cantor proved that if you take a set S of cardinality Aleph₀ and build a set P(S) consisting of all possible subsets of S, then P(S) is strictly larger than S. In other words, card(P(S)) > card(S). Or, using another notation, 2^S > S. (CAVEAT: do not confuse this notation with integer/real number exponentiation, because it is NOT. It is simply a notation generalized from the fact that the set of all subsets, or power set, of a finite set X has 2^X elements. This ONLY applies to finite sets.) In fact, it can be proved that for sets of any cardinality, 2^S > S.

It is well-known that the set of all subsets of natural numbers has the same cardinality as the set of real numbers. That is to say, card(R) = 2^Aleph₀. This is strictly larger than Aleph₀, so there are "infinitely more" real numbers than there are natural numbers. The size of R is often denoted as c, so c = 2^Aleph₀.

However, what is not so straightforward is whether Aleph₁ = c. Aleph₁, by the way, is the cardinality of set of all well-ordered sets of cardinality Aleph₀. In general, Aleph_n is the cardinality of the set of all well-ordered sets of cardinality Aleph_n-1. It can be proved that Aleph_n+1 > Aleph_n for all n. That is, the Alephs are strictly larger than preceding Alephs in terms of cardinality.

But the problem is, we don't know exactly how the Alephs correspond with the power sets. We know that 2^Aleph₀ = c, and c > Aleph₀, and Aleph₁ > Aleph₀. But we don't know whether 2^Aleph₀ is equal to, greater than, or less than Aleph₁. That is, we don't know if there's anything "between" Aleph₀ and c. The so-called "Continuum Hypothesis" is that 2^Aleph₀ = Aleph₁.

Sadly, the Continuum Hypothesis (CH) is an undecidable statement, as predicted by Godel's Incompleteness theorems. We cannot prove whether it's true or false from the axioms of set theory (ZFC). It is not possible to prove it.

So what do we do? We cannot prove CH, but it has profound implications for several important theorems in mathematics, so what mathematicians have done is to explore either possibility, in the hopes that eventually, after we learned enough about the consequences of choosing CH either way, we can pick the more "preferred" option and make it a new axiom.

Anyway, the point of all this is that if you want to compare the Alephs to the power sets, then you need to state whether or not you're assuming CH, because the conclusions you draw will be drastically different depending on your choice.

Now coming back to the topic, we need to understand that infinity is a very complex concept, and so when we use the term "infinity" we need to be clear what exactly we're referring to.

If you're talking about orderings and sequences, then "infinity" refers to transfinite ordinals -- and these have well-defined meanings for addition, multiplication, and exponentiation. (However, subtraction, division, or roots are not well-defined on these things.) And it is generally true that given two ordinals V and W, V+V = 2*V, V+V+V+...(W times) = V*W, and V*V*V*...(W times) = V^W.

If you're talking about the size of sets, then you're talking about cardinals, and if you have two infinite cardinals A and B, then A+A = A*A = Aⁿ = A (where n is a finite number), and A+B = A*B = max(A,B). However, 2^A > A, and A^B is greater than either A or B.

If you're talking about "infinity" as used in calculus, e.g., in sets like C union {infinity}, then infinity+infinity, infinity*infinity, and infinity^infinity are all nonsense and have no meaning, because these operations are not well-defined.

[houserichichi]

Very thorough. Golf clapping!

[quickfur]

Very thorough. Golf clapping!

Ack. I noticed I made some mistakes.

1) For two cardinalities A and B, A^B is actually equal to A if B < A. Therefore, c^Aleph₀ = c. However, if B > A, then A^B > A, and also A^B > B. So, c² = c, but 2^c > c.

2) Aleph_n+1 > Aleph_n is true only if n is finite. If n is transfinite, this may or may not be true. For example, Aleph_n + 1 = Aleph_n, so Aleph_{Aleph_n} = Aleph_{Aleph_n+1}, but Aleph_{Aleph_n+1} > Aleph_{Aleph_n}. Also, there is a huge member of the Aleph series sometimes denoted by Epsilon, which has the (possibly astounding) property that Aleph_Epsilon = Epsilon itself. One could think of it as the limit of the sequence Aleph₁, Aleph_Aleph₀, Aleph_{Aleph_Aleph₀}, ... Aleph_{Aleph_{Aleph_{Aleph_{... (Aleph₀ times)}}}} = Epsilon.

3) About the Continuum Hypothesis, I believe it is actually proven that Aleph₁ cannot be greater than c, because all possible cardinalities have an associated Aleph, and so there can't be anything between Aleph₀ and Aleph₁. The problem is that we don't know (and can't know) which Aleph equals which member of the powerset series of cardinals.