Re: Question about the Diagonal Method.
- From: george <greeneg@xxxxxxxxxx>
- Date: Mon, 20 Aug 2007 07:16:21 -0700
On Aug 13, 3:05 pm, Scott <ToaTe...@xxxxxxxxx> wrote:
Let me use an example provided in a latter post. Suppose a friend
walks up and says "I have a list of everyone in China" and hands it to
me. I scan the list and compare it against everyone in China and find
their name is on the list. But just as I finish, I look over and see
John Doe standing at the border of China with one foot in China and
one foot outside of China.
This is just plain not analogous.
There is no uncertainty in this paradigm.
EVERYthing is either true or false.
Now, John Doe is creating a contradiction
because I cannot determine whether he is or is not in China.
No, that is NOT a contradiction.
THat is the OPPOSITE of a contradiction.
A contradiction would be if you could PROVE BOTH
that John Doe IS in China AND IS SIMULTANEOUSLY NOT
in China. THAT IS WHAT HAPPENS if you try to put the anti-
diagonal OF the list ON the SAME list. If you put it on at position
n,
then you can PROVE BOTH that f(n,n)=f(n,n) AND that f(n,n)=1-f(n,n).
You do NOT have a QUESTION mark: you do NOT have ZERO true
confirmed answers as opposed to the ONE you needed: You have TWO
confirmed OPPOSITE answers as opposed to the One you needed.
There are two levels here. The first level is whether John Doe is in
China or not.
Under this paradigm, he will necessarily be one or the other,
Regardless of whether you know which (or not).
The second level is whether I can put the name "John
Doe" on my list. To me, it seems like you can put the name "John Doe"
on the list, but you can't determine whether John Doe is or is not in
China.
Well, you're just being silly.
"Undefined" or "confused" is NOT a way of postponing or avoiding
the contradiction here, AS IT MIGHT BE, SAY, with A TURING MACHINE
that was trying to COMPUTE a list whose own anti-diagonal was on it.
It could just loop and never fill in the spot where the contradiction
was
going to occur. THEN, however, it would never complete a list WITH
its own anti-diagonal on it, SO YOU WOULD STILL FAIL.
However, everyone tells me that the proof of CT states just the
opposite: That I cannot put "John Doe" on the list, but it is
determined that John Doe is or is not in China (because s_d exists and
is well-defined(?)).
Right.
Every square list has an anti-diagonal.
There is nothing you can do about it.
The operation of "taking the anti-diagonal of" is well-defined.
A program or function computing THAT clearly and provably exists;
you could write the program yourself.
This is the crux of my question. Why is it the
latter rather than the former?
What am I missing that invalidates #4?
If #4 was "sets can belong to other sets regardless of their
contents" then that is what you are missing. Sets are extensional.
EVERYthing about them is true false IN and ONLY in virtue
of their contents. The axiom that enforces this is (therefore,
obviously)
called the axiom of extent or of extensionality.
In FOL with equality it says that sets with the same members are
equal.
Without, it says that sets with the same members are members of the
same sets.
.
- References:
- Question about the Diagonal Method.
- From: Scott
- Re: Question about the Diagonal Method.
- From: James Burns
- Re: Question about the Diagonal Method.
- From: Scott
- Question about the Diagonal Method.
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