Real numbers as infinitly wide tuples, what is aleph 2?

[TylerH]

First, can all aleph 1 sets be generalized as sets of infinitely wide tuples? As in, let a_1a_2_a3 \ldots \in \Re map to (a_1, a_2, a_3, \ldots).

Second, if countably infinite sets are n-tuples, aleph 1 sets are infinite tuples, can this pattern be generalized to even higher cardinality?

 

[pmsrw3]

First, can all aleph 1 sets be generalized as sets of infinitely wide tuples? As in, let a_1a_2_a3 \ldots \in \Re map to (a_1, a_2, a_3, \ldots).

Yup. Well, at least if the cardinality of possible a_i is small enough.

Second, if countably infinite sets are n-tuples, aleph 1 sets are infinite tuples, can this pattern be generalized to even higher cardinality?

Nope. If you have an infinite vector of infinite tuples, then that's equivalent to having an infinite vector of reals. And the cardinality of such vectors is just aleph one again.

 

[micromass]

First, can all aleph 1 sets be generalized as sets of infinitely wide tuples? As in, let a_1a_2_a3 \ldots \in \Re map to (a_1, a_2, a_3, \ldots).

Second, if countably infinite sets are n-tuples, aleph 1 sets are infinite tuples, can this pattern be generalized to even higher cardinality?

You mean tuples of natural numbers, right?? The answer is sadly that this is incorrect

Finite n-tuples of natural numbers are countables. But the set of "infinitely wide" tuples has cardinality 2^{\aleph_0}, and this is (without the continuum hypothesis) not equal to \aleph_1.

How can we see this?? Well, instead of looking at infinitely wide tuples of natural numbers, let us look at tuples such that the elements of the tuples are only 0 or 1. So, an example of such a tuple is

(0,0,1,1,1,0,1,0,1,0,...)

We will show that the set of all such tuples (a set which I will denote by A) has cardinality 2^{\aleph_0}. Indeed, there exists a bijection

A\rightarrow \mathcal{P}(\mathbb{N})

that is, every element of A determines uniquely a subset of the naturals. How do we do this?? It's an ingenious trick:

Let's say that we have (0,0,1,1,1,0,1,0,1,0,...). We will build a subset of the naturals in the following way.

The zero'th element of the tuple is 0, so 0 will not belong to our subset.
The first element of the tuple is 0, so 1 will not belong to our subset.
The second element of the tuple is 1, so 2 will belong to our subset.
The third element of the tuple is 1, so 3 will belong to our subset.
The fourth element of the tuple is 1, so 4 will belong to our subset.
The fifth element of the tuple is 0, so 5 will not belong to our subset.
The sixth element of the tuple is 1, so 6 will belong to our subset.
...

So we have formed a subset {2,3,4,6,8,...} of the naturals. In the same fashion, every infinite tuple defines a subset of the naturals, and every subset of the naturals determines a tuple. For example, {4,5} is determined by the tuple (0,0,0,0,1,1,0,0,0,0,...).

I hope that was a bit understandable. In any case, this shows that there are as much tuples of natural numbers as there are subsets of the natural numbers. And there are (by definition) 2^{\aleph_0} subsets of the naturals.

As stated by the continuumhypothesis, it cannot be proven nor disproven that \aleph_1=2^{\aleph_0}, so in the absence of the continuum hypothesis, your set will not have cardinality \aleph_1.

You may ask: which set does have cardinality \aleph_1. This question is quite difficult to answer, and I do not know any elementary sets of cardinality \aleph_1. The only sets of cardinality \aleph_1 that I know, will involve ordinals.

I hope this was a bit clear and not too technical. Feel free to ask for more explanations.

 

[micromass]

Yup. Well, at least if the cardinality of possible a_i is small enough.

Sadly, this is not possible. We can even let the possible a_i be {0,1}, and we would still have that all the infinite tuples would have cardinality 2^{\aleph_0}.

Unless I'm misunderstanding the question?

 

[pmsrw3]

Sadly, this is not possible. We can even let the possible a_i be {0,1}, and we would still have that all the infinite tuples would have cardinality 2^{\aleph_0}.

Unless I'm misunderstanding the question?

OK, then I was confused. I though aleph 1 was, by definition, the cardinality of the power set of the naturals. Apparently I was mistaken. But then, what IS aleph 1?

 

[micromass]

OK, then I was confused. I though aleph 1 was, by definition, the cardinality of the power set of the naturals. Apparently I was mistaken. But then, what IS aleph 1?

Uh, that's quite difficult to explain precisely.

The cardinality of finite sets can be 0, or 1, or 2,...
The smallest cardinality that an infinite set can have is \aleph_0, such a set is countable.
The smallest cardinality that an uncountable set can have is \aleph_1.
The smallest cardinality that an set with cardinality \geq \aleph_1 can have is \aleph_2.
And so on...

So the aleph numbers describe orders of infinite: \aleph_0 is the smallest infinity, \aleph_1 is the one after that, \aleph_2 is the one after that, and so...

It is not known (and it can never be known) where 2^{\aleph_0} falls in this hierarchy. This is the continuumhypothesis.

We can give names to the cardinality of power sets however. This yields the beth-numbers (aleph is the first letter in the hebrew alphabet, beth is the second). So

\Gamma_0=\aleph_0,~\Gamma_1=2^{\aleph_0},~\Gamma_2=2^{2^{\aleph_0}},...

(I used the symbol \Gamma here, since I don't know the LaTeX-code for Beth. See http://en.wikipedia.org/wiki/Beth_number to see what Beth looks like)

 

[pmsrw3]

Got it. OK, I guess I need to replace all the aleph's in my first post with beth's.