Phar2wild said:
First to clarify: I am not a mathematician, so while I find your explanation interesting I would have to struggle through some of the things that you mentioned. And I will have to look up the "Downward Skolem Lowenheim Theorem" ( First I heard of it! )
However I have been contemplating Cantor's basic work for some time and I try to be a rational ( pun intended ) being!
So I will skip, for now, some of what you have taken the time to write because I just want to make the following points in response to your post.
I did not imagine the odometer set to .00000000... to begin. To do so I would have started at zero and then your argument would make perfect sense and you would be absolutely correct.
I started with an irrational number. ( to create one would take too long! :-) ) So each turn of the odometer a new irrational is created. I suppose it will also create the rationals along the way.
I will stick to my assertion that Cantor did what I did; he created an irrational one digit at a time. And what's more he was prepared to create more than one! I request that the same opportunity be afforded to me to use this process.
If I fall, we ( Cantor and I ) fall together!
Again, many thanks for your time.
Robert H
P.S. Re: "infinite combinations of a finite number of things". Isn't this just impossible? Example?
Your odometer is an example-we have a finite number of digits (0-9) and we are arranging them in an infinite sequence.
The fallacy lies in you imagining that you will create *every* irrational number. In the language of cardinal arithmetic, if we call the cardinality of the natural numbers $N$, we have:
$10^N = 2^N > N$.
By the $>$ sign, I mean that there is no 1-1 correspondence between a set with $2^N$ elements and a set with $N$ elements.
You'll have to take my word for this (for now, you can research it if you like), but the number of elements of the power set $\mathcal{P}(A)$ (the set of all subsets of a set $A$), is $2^{|A|}$. This is easy to verify (via induction) for finite sets.
The reason $2^A$ is also used for $\mathcal{P}(A)$, is because there is a 1-1 correspondence between $\mathcal{P}(A)$ and *functions* $f:A \to \{0,1\}$.
To see this, let $S \subseteq A$, and define the function:
$\chi_S: A \to \{0,1\}$ (called the *characteristic function* of $S$) by:
$\chi_S(a) = 1$ if $a \in S$
$\chi_S(a) = 0$ if $a \not\in S$.
Cantor showed there is no 1-1 correspondence between $|A|$ and $|\mathcal{P}(A)| = 2^{|A|}$ for ANY set $A$. His most general argument runs as follows:
Suppose there were such a 1-1 correpsondence $f:A \to \mathcal{P}(A)$. Such a function would be onto $\mathcal{P}(A)$, and we will show this cannot happen.
Let $B = \{x \in A: x \not\in f(x)\}$
Now either $B$ has something in it, or it doesn't. Let's suppose there isn't any such $x$'s in $B$. This means that for EVERY $x \in A$, $x \in f(x)$. So every element of $A$ maps to some non-empty subset of $A$. So no element of $A$ maps to the empty set (which is a bona-fide element of $\mathcal{P}(A)$) ,and $f$ is thus not onto, a contradiction.
So $B$ must have something in it, then. Therefore, there is some $x \in A$ with $x \not\in f(x)$. Since $f$ is assumed onto, we must have for some $y \in A$, that $f(y) = B$.
Now if $y \in B$, then (since $f(y) = B$), then $y \in f(y)$, and thus $y \not\in B$.
However, if $y \not\in B$, then $y \in A$ (since $B \subseteq A$) and $y \not\in f(y) = B$, and thus $y \in B$.
Both of these are self-contradictions, so no such non-empty set can exist.
Since $B$ can neither be empty, nor non-empty, and is definitely *some* subset of $A$, $f$ cannot be onto, there cannot be any $y \in A$ such that $f(y) = B$.
In a bit more detail for the natural numbers case:
Let's form some subsets of the natural numbers:
$\{1\}, \{2,3\}, \{2,4,6,\dots\}, \{1,3,5,\dots\}, \{2,4,8,16,32,\dots\}, \{3,5,7\}$ and so on. Note some of these subsets are infinite (the third one is meant to be the even natural numbers, the fourth the odd natural numbers, the fifth the powers of 2 starting with 2, and the other sets are finite).
Suppose we assume we can pair these (completely) in some fashion with natural numbers:
$1 \leftrightarrow \{1\}$
$2 \leftrightarrow \{2,3\}$
$3 \leftrightarrow \{2,4,6,\dots\}$
$4 \leftrightarrow \{1,3,5,\dots\}$
$5 \leftrightarrow \{2,4,8,16,32,\dots\}$
$6 \leftrightarrow \{3,5,7\}$
Note that 1 is paired with a set that contains 1 (and so is 2), but that 3,4,5,6 are not.
Call numbers that are paired with a set that contains them "greedy", and numbers that are paired with a set that does not contain them "pure".
Consider the subset $S \subseteq \Bbb N$ consisting of all "pure" numbers. This is a subset of the natural numbers, and so is paired with some natural number $k$:
$k \leftrightarrow S$.
Is $k$ "greedy"? If so, then $k \in S$, but everything in $S$ is "pure".
Is $k$ "pure"? If so, then $k \in S$, and is thus "greedy".
Now it could be that *all* numbers are "greedy", so that $S$ is empty. In that case, no natural number maps to the empty set, and our correspondence is not complete.
So there is no natural number $k$ that can possibly be paired with $S$, violating our original assessment that such a 1-1 complete pairing could be made.
Thus the set of all subsets of natural numbers is *bigger* that the set of natural numbers, not all infinities are created equal.
