Re: Cantor's circular "proof" that evens = integers

george wrote:

On Apr 13, 12:34 am, Phil <toob-head...@xxxxxxxxxxxxx> wrote:


Well, let's make sure I can comprehend what you said ...


You did better than we had any right to expect.

Why, thank you!

But that is NOT the point. The point is that Jesse
has no business doing this. I would like to very sincerely thank
Darryl McCullough for following my unsually sadistically-
delivered advice that he NOT try THIS. ZFC has 10
axioms, some of them long and complicated, and it was
ENTIRELY PREDICTABLE that doing it THIS way would
confuse you. You said you wanted "Cantor's proof" but THAT
is even OLDER than ZFC. WHAT YOU NEED is a
SIMPLE, MODERN treatment of the issue. Unfortunately
that is not what you are going to get here, UNLESS you
talk to ME.

I'm just stepping through the posts, finally, so don't waste too much time on these, as I know you've got a more thorough treatment elsewhere. As a comment, I really think I would like to see Cantor's proof, as well as a modern one.



Let E = { n in N | (E m in N) n = 2 * m }.

Okay, this selects a set of elements


FROM N, TO E. Every element selected to BECOME
an element of E STARTED OUT as an elment of N.
Every element selected will be an element of BOTH N and E.


such that each element = 2 * a natural number,


Exactly


giving us the set of evens.


Well, of even natural numbers anyway.
E is the set of ALL and ONLY the even natural numbers.
It is the even elements of N.



The axiom of separation guarantees
that there is such a set.


Indeed, it just produces it out of N and the property.



Yes, because the evens are in N,
and have the property that they ae
divisible by 2.


No, this doesn't have ANYTHING TO DO WITH "the
evens". "The evens" is from ENGLISH.
ENGLISH is IRRELEVANT. Even if we didn't know WHAT
to call them, THIS SET WOULD *STILL* exist.
This set is GUARANTEED to exist because THIS IS AN
AXIOM. We agreed IN ADVANCE NOT to doubt any
sentence LIKE this.





Define f c N x E by f = { < m, 2 * m > | m in N }.
More precisely,
f = { x in N x E | pi_2(x) = 2 * pi_1(x) }. Again, the axiom of
separation proves f exists.

Here you're losing me.


Well, OF COURSE he is. I TOLD people NOT to TRY
this ***! ZFC has some conventions that would seem
strange to the uninitiated, which unfortunately includes
you, which, even more unfortunately, YOU DIDN'T KNOW.
The missing link here is: in ZFC,
A FUNCTION IS A SET OF ORDERED PAIRS.
The other terminological missing link is something called
"the cross-product" or "Cartesian product" of two sets.
It is called a cross-product because the symbol for it is x.
It occurs above in " N x E ". What N x E is is a set where
every element is an ordered pair, and within each ordered
pair, the first element is an element of N and the second
element is an element of E. This holds for any two sets
D and R generally: D x R is defined as the set of all (and only)
ordered pairs where the first element of the pair is in D and
the second is in R. A function whose domain is D and whose
range is R is going to be A SUBSET of this set of ordered
pairs. So in ZFC, you can always a define a function on
Dom into Rng by applying the axiom of separation to
Dom x Rng.

But this is WAY MORE THAN YOU NEEDED to know IN
ORDER to know that there is a bijection between the naturals
and their proper subset, the even naturals.
It's also less, though; I mean, first, you had to know what
a bijection was.

I really think I did know, but Jesse's symbols and terminology were totally Greek to me. Given two sets A and B, where each element in set A is mapped to only one element in B, and vice-versa, meaning there are no unmapped or linked elements, I THINK you have a bijection. And it really, really does PROVE that the two sets have equally many elements.


It looks like you'e saying that each n in N is
mapped to an even in E, specifically 2*n.


Right.


You are beginning to define a
bijection between N and E?


Not just beginning: starting, finishing and succeeding,
all in ONE axiom. For EVERY set S and EVERY(definite)proprerty P(.),
the axiom of separation guarantees that
{ x in S | P(x) } exists. Therefore, in particular,
it guarantees that
{ <n,e> in NxE | e=2*n } exists.
"The cross product of N and E" is a set, and
"the second equals 2 multiplied by the first" is a
property that any old ordered pair might or might not have.
This set that I've just told you MUST exist is a set of
ordered pairs, and it is also a function. And it is a also
a one-to-one function. And it also contains EVERY element
of E as the second member of SOME pair in it, SOMEwhere.
Because it meets ALL THREE of those criteria, this set,
in ZFC, *is* a bijection.

Here you seem to make the same mistake that Jesse made, namely ASSUMING that this set not only contains EVERY element in E (which it does), but that it also contains every element in N, which it will not IF it turns out that N and E do not have equally many elements.

You claimed to already know what
a bijection was, but whatEVER you knew it as, you did NOT
know it as THAT. Requiring you to learn all this ZFC was
a MISTAKE. The fact that infinite sets can be bijected
with proper subsets of themselves is EASIER to grasp than
this. It is UNFORTUNATE that you INSISTED on doing it
THE HARD way.



We claim that f is a function. I.e., for all m in N, if <m,n> and
<m,n'> are in f, then n = n'. Indeed, the proof follows immediately
from the fact that * is a function.

Ummm...one to one?


Jesse should've known, but didn't,
that you were too ignorant to understand this.

As I said elsewhere, I was aksing whether I had correctly interpreted his symbols and terminology, NOT asking what "one-to-one" was, but I admit my statement was open to either interpretation.

He should've begun by telling you what a bijection was.
But since you thought you knew, he thought you knew too,
and thought he could just go proving the 3 parts of the
definition. This is too much detail for you. IF you had
talked to ME then you would've gotten something you could
understand.


We claim that f is one-to-one.
Again, it follows immediately from a
proof that * is one-to-one.

Finally, f is surjective (onto E).
This is immediate by the
definition of E.

Every element in E can be found in N?


NO, dumbass, Every element in E can be found
as the SECOND element of SOME pair in * f *!

True enough, good point.

Was that IN ENGLISH ENOUGH for you????
If all of this is striking you as UNNECESSARILY
COMPLICATED then CONGRATULATIONS,
IT IS INDEED *all* that!

But that's what you GET for not talking to ME.
Of course, you will probably protest that my version
would've been unnecessarily ABUSIVE, but you would've
been wrong about that, too. Your attitude and commentary
since you got here has MERITED even WORSE abuse
than I personally am even competent to inflict.

I'll look at your proof here in a bit (have to watch "Blood Ties").

Phil
.

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