Is Cantor's Diagonal Argument Flawed in Its Application to Positive Integers?

[CRGreathouse]

Any non-finite set that you can inject into the naturals will also have cardinality aleph-null. However you can't do this with the real numbers which is why they have aleph-one cardinality.

Beth-one cardinality, not aleph-1. Unless you have the CH all you can say is that the reals have at least aleph-one cardinality.

Are you sure about that? The entry in Wikipedia (!) on p-adics (http://en.wikipedia.org/wiki/P-adic_number) indicates otherwise (see I've been doing my homework ). It seems to me you are referring to 10-adic integers.

I've never read that article, but yes I'm talking about the 10-adic integers. I imagine that in both ...111 = -1/9, though.

 

[HallsofIvy]

What you mean is cardinality. It isn't really a size of a set (by the way a set does not have to be one of integers), I would say its more closer to the number of elements in a set, but I don't like this wording either (I don't really have a satisfactory definition :shy:).

Aleph-null is the first cardinal, which is used to describe the set of natural numbers. Any non-finite set that you can inject into the naturals will also have cardinality aleph-null. However you can't do this with the real numbers which is why they have aleph-one cardinality. This you prove by using cantors diagonal argument via a proof by contradiction. Also it is worth noting that 2^{\aleph_0}=\aleph_1 (I think you need the continuum hypothesis for this). Interestingly it is the transcendental numbers (i.e numbers that aren't a root of a polynomial with rational coefficients) like pi and e.

Sorry for the long post, hope this helps.

No, you don't need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1. You need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1= c[/itex] where c is the cardinality of the real numbers.<br /> <br /> But what does &quot;it&quot; in &quot;Interestingly it is the transcendental numbers (i.e numbers that aren&#039;t a root of a polynomial with rational coefficients) like pi and e&quot; refer to? Not, presumably \aleph_1= c(assuming continuum hypothesis) because that includes all real numbers, not just transcendental numbers.

 

[CRGreathouse]

No, you don't need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1. You need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1= c[/itex] where c is the cardinality of the real numbers.

<br /> <br /> I differ. 2^{\aleph_0}= \mathfrak{c} because there&#039;s a bijection from an infinite bitstring to the reals, but getting 2^{\aleph_0}=\aleph_1 (or \mathfrak{c}=\aleph_1 which is the same) requires CH.

 

[Focus]

No, you don't need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1. You need the continuum hypothesis to prove that 2^{\aleph_0}= \aleph_1= c[/itex] where c is the cardinality of the real numbers.<br />

<br /> <br /> I am sorry but you do need CH to prove that identity. That is what the CH exactly states that 2^{\aleph_{\alpha}}= \aleph_{\alpha +1}.

 

[CRGreathouse]

That is what the CH exactly states that 2^{\aleph_{\alpha}}= \aleph_{\alpha +1}.

More pedantry: that's the GCH. The CH says this only for \alpha=0.

 

[Hurkyl]

I don't quite follow this. By -1/9 I take it you are denoting the number that could also be represented as the recurring decimal -0.1111 ...

No, I am not. As I said, - refers to additive inverse, and / refers to multiplication by the multiplicative inverse.

The additive inverse of 1 is ...999. (Because ...999 + 1 = 0)
The multiplicative inverse of 9 is ...8889. (Because ...8889 * 9 = 1)
Finally, ...999 * ...889 = ...111.

As I had said, the numbers in this number system are the left-infinite strings of decimal digits. -0.111... doesn't denote a number.

I'm also wondering what number the symbol oo (infinity) represents (if any). What does it mean when we say 'as x approaches oo' ?

The simplest is to define the phrase "The limit of ____ as __ approaches \infty" as a whole; the individual symbols and words aren't given any meaning. This is something you'd probably see in an elementary calculus book.

A slightly more sophisticated treatment would define a new number system (such as the extended real numbers or the extended natural numbers) that contains an element called +\infty, and then you can just use the ordinary definition of the phrase "The limit of _____ as __ approaches __".

 

[manker]

No, I am not. As I said, - refers to additive inverse, and / refers to multiplication by the multiplicative inverse...

Yes, I see this now.

The simplest is to define the phrase "The limit of ____ as __ approaches \infty" as a whole; ...

I guess I was trying to find the distinction in meaning between the symbols \infty and \aleph_0. Would I be right in thinking that while \aleph_0 is the cardinality of the set of integers (amongst others), \infty is (potentially at least) an arbitrarily large member of that set? Is there any sensible way of comparing \aleph_0 and \infty? After all the cardinality of {1, 2, 3} is also a member of the set.

 

[Hurkyl]

Yes, I see this now.

I guess I was trying to find the distinction in meaning between the symbols \infty and \aleph_0. Would I be right in thinking that while \aleph_0 is the cardinality of the set of integers (amongst others),

Essentially yes -- that is the most common intended meaning for that symbol.

\infty is (potentially at least) an arbitrarily large member of that set?

No. Scratch that -- emphatically no. Can you even define what you mean by an "arbitrarily large member of a set"?

Again, it's just another symbol. But unlike \aleph, there are actually several different mathematical structures that make use of the symbol \infty to denote an element. The most common structures you'd encounter are the projective real numbers and the extended real numbers (Look them up!) (In the latter, \infty is really just shorthand for +\infty).

Is there any sensible way of comparing \aleph_0 and \infty? After all the cardinality of {1, 2, 3} is also a member of the set.

No and yes. The most common uses of those symbols lie in different structures, without any "default" conventional way to convert between them. But that said, it is occasionally useful to observe the fact that the closed interval [0, \aleph_0] of cardinal numbers and the interval [0, +\infty] of extended natural numbers are isomorphic as ordered sets. (Equivalently, as topological spaces, when given the order topology)

 

[proton]

Hoping I may be permitted to make a belated contribution here.

Summarising the key points from some of the other posts:
The key to the problem is that the new 'number' generated by the diagonal argument will have an infinite number of non-zero digits resulting from the infinite number of leading zeros in the original numbers. Such a number is not an integer since, although the set of integers has an infinite number of members, each individual member is finite and must have a finite number of non-zero digits.

i was actually having a ton of trouble figuring out what everyone was saying until i read this post. thanks a lot!