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

On May 5, 5:11 am, Phil <toob-head...@xxxxxxxxxxxxx> wrote:
Okay, first I will hopefully correctly state Jesse's proof,

No, you won't. You don't know how.
Proofs start from AXIOMS.
You would have to BEGIN by stating THE AXIOMS.
This you will never do, simply because you are too stupid
to understand them and to lazy to type out all ten of them.
Actually you can eliminate separation and pairing if you
know replacement, but that still leaves 8, which is 8 more
than you know how to parse.

and then show why it is actually just a circular argument.

NOBODY CAN EVER show a first-order proof to be "a
circular argument", DUMBASS. A PROOF is a series of lines
in which EVERY line either IS AN AXIOM, or is the result of
applying rules of inference TO EARLIER lines. NEITHER of
those things EVER causes "circularity".

Preliminary definitions:

Definitions ARE NOT preliminary, DUMBASS:
AXIOMS are preliminary. Definitions come AFTER
axioms and define NEW terms IN terms OF the OLD
(i.e. ALREADY mentioned IN THE AXIOMS) terms.

Set N is an infinite set containing all the natural numbers (for these
proofs, I still prefer the "old" naturals, which begin with 1).

Nobody GIVES A FLYING *** what YOU prefer.
It's JESSE'S proof. You have to let the PROVER state
her OWN axioms. THAT IS WHY we always TOLD YOU
to STATE YOUR AXIOMS.

In the first place: you canNOT define N as "an infinite set".
The adjective "infinite" DOES NOT OCCUR in the axioms
of ZFC (which you would've KNOWN if only you had sense
enough to write them down&out). BEFORE you define ANYthing
as "an infinite set", you FIRST have to DEFINE *finite*!
Once you have done that, there is no NEED to define anything
as an infinite set: EVERY set will ALREADY either SATISFY
the definition of "finite" OR NOT. If it satisfies it, it's finite;
if it doesn't, it's infinite.

Set E is an infinite set containing all the even natural numbers,
meaning every element e in E = 2n for some n in N.

Please be aware that if these are the only elements in E,
then E *must* be a subset of N; everything in E was in N
first. This is legitimate as an application of the axiom of
separation. However, you have already left out a step: the
axioms of ZFC do NOT include any such operation as
MULTIPLICATION. Multiplication has to be DEFINED.
If you want to define multiplication in the normal way, you
will first have to define something called the "cross-product"
or "cartestian product" of any 2 sets, and before you define
THAT, you will have to define the notion of an "ordered pair"
of two sets (unordered pairs are already with us by the axiom of
pairing).

Set N x E is the Cartesian product of N and E, the set of all the
ordered pairs that can be formed from sets N and E, where the first
element in each pair is from N.

[AND the SECOND element is from E. You can't just LEAVE
OUT STEPS like that, grasshoppa.]

Set f is a subset of N x E containing every ordered
pair from N x E with
the form <x,y> where y = 2x.

Again, this is guaranteed to be possible by the axiom of separation.

Please note that thus far, NOBODY HAS ADDED ANY AXIOMS
to the basic axioms of ZFC. Nor will they ever, which proves
your whole line is bull***, but save that for later.
The single biggest problem with your presentation so far is that
YOU HAVE NOT STATED DEFINITIONS of "finite" or "multiplication".
Thus, when the time comes to apply those definitions in the
important and critically necessary way, you are likely to confuse
an OLD definition with a NEW extra axiom.



Observations and deductions:

For any n in N, there exists an element 2n in N.

That is neither observable nor deducible from what
has gone before. The only way to make THIS happen
is to DEFINE what "2n" MEANS, given n. And given
2, for that matter; you have not defined 2 EITHER.

Therefore, set N x E contains an ordered pair of the form <n,2n>
for EVERY n in N.

For that conclusion, you would've needed "There exists an element
2n in E", not in N, as you concluded. But of course it will follow
from the fact that it is in E that it is also in N.


Set f is:

1. A function, since for any two pairs <x,y> and <x,y'>, y = y'. In
other words, for any x in f, there is one y.

Namely 2x; y and y' must both equal 2x, so they must equal
each other.


2. One-to-one, since it is also true that for any two pairs <x,y> and
<x',y>, x = x'.

It's not enough for it to just be TRUE; you have to PROVE this!

3. Onto, since for any y, there exists one x,
and (by 1. above) for any x, there exists one y.

That's not what "onto" MEANS, dumbass. Here,
it has to be onto E, specifically, so you mean "for any y IN E" --
NOT JUST "for any y" -- there exists one x, namely, y/2.
But that requires you to prove that division is a function as well,
or at least that division by 2 is. There is a LOT here that you
are skipping. If you don't include all this stuff as you go along,
then of course it is EASY to claim, when you need it later, that
it is being "added".

Therefore, there is a bijection between the infinite sets N and E,
making sets N and E equinumerous.

Where is the error? What was "added"?

In order to understand why this "proof" is
actually just a circular argument, i.e., an
argument that merely ADDS the belief that N and E are
equinumerous as an axiom to the already
existing axioms of set theory

OH, SHUT UP! EVERY line of the proof SHOWS what
axioms or prior lines it depends on! It is NOT POSSIBLE
that ANYTHING was EVER "added"!

(as opposed to proving that this
conclusion NECESSARILY FOLLOWS from the
existing axioms), we have to make
use of some basic facts concerning proofs.

Idiot, please. YOU DON'T KNOW the basic facts
about proofs. In particular, you didn't know that proving
P from A was the same as proving P from A&B.
In any case, the MOST basic fact about proofs, namely,
that every line of one is the result of applying a rule of inference
to axioms or previous lines, is one you OBVIOUSLY don't understand,
since you don't understand that it makes the kind of circularity
you are talking about IMPOSSIBLE.

.

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