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

1. Jul 27, 2011

### 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?

2. Jul 27, 2011

### pmsrw3

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

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.

3. Jul 27, 2011

### micromass

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.

4. Jul 27, 2011

### micromass

Sadly, this is not possible. We can even let the possible ai 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?

5. Jul 27, 2011

### pmsrw3

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?

6. Jul 27, 2011

### micromass

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)

7. Jul 28, 2011

### pmsrw3

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