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Topic: ZFC and God
Replies: 45   Last Post: Apr 18, 2013 3:47 AM

 Messages: [ Previous | Next ]
 Jesse F. Hughes Posts: 9,776 Registered: 12/6/04
Re: ZFC and God
Posted: Jan 23, 2013 11:39 AM

WM <mueckenh@rz.fh-augsburg.de> writes:

> On 23 Jan., 14:36, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> WM <mueck...@rz.fh-augsburg.de> writes:
>
>> > I know. But if you have read the discussion, you have seen that two
>> > matheologians claim just this. Why do they? Because they cannot answer
>> > the question: What paths are (as subsets of the set of nodes) in a
>> > Binary Tree that is the union of all its levels? Are there only the
>> > finite paths? Or are there also the infinite paths?
>> > Try to answer it, and you will see that you need the omegath level or
>> > must confess that it is impossible to distinguish both cases. Hence,
>> > Cantor's argument applies simultaneously to both or to none.

>>
>> I'm not interested in the web-published claims of two individuals on a
>> different topic than we're discussing.

>
> You are in error. Pause for a while and think it over.

I will not be distracted from the topic at hand until we've completed
the discussion.

>>
>> Once again, let me remind you what you claimed.  You claimed ZF was
>> inconsistent, and in particular that ZF proves that the union
>>
>>   U_n {1,...,n}
>>
>> is both finite and infinite.
>>
>> Now, we've had two competing definitions of infinite in this
>> particular discussion.
>>
>> (1) A set S is infinite if there is no natural n such that |S| = n.
>>
>> (2) A set S is infinite if it contains a number greater than every
>> natural n.
>>
>> The first definition is what mathematicians almost always mean, and
>> they *never* mean the second, but this is mere semantics.  Let's talk
>> results.

>
> You are right, mathematicians prefer (1). But matheologians use (2).
> An infinite set contains a number of elements, at least aleph_0, which
> is greater than every finite number.

You seem to have changed claims here.

Which of the following do you claim ZF proves?

(1) The set U_n {1,...,n} contains an element k greater than every
finite number.

(2) The set U_n {1,...,n} has cardinality greater than every finite
number.

The second is not controversial, and the first is unproved.

>>
>> We both agree that, using definition (1), the above union is infinite
>> and (I think) we agree that we cannot show it is finite (=not
>> infinite).  If I'm mistaken on this point, then please show me.
>>
>> On the other hand we both agree that, per definition (2), the union is
>> "finite", but I have seen no contradiction result, since you have not
>> shown that the union is "infinite" in this sense.  Nor can you find a
>> single publication in which a mathematician has claimed the union
>> above (i.e., the set N of natural numbers) contains an element larger
>> than every natural.

>
> You confuse the things. ZF claimes that the *number of elements* is
> larger than every finite number. Just this causes the contradiction. A
> union of finite initial segments cannot have a number of elements that
> is larger than every finite number.

Perhaps I misunderstood you then, but let's work with this. Here's
the new contradiction you claim ZF proves, correct?

(1) |U_n {1,...,n}| > k for every natural number k.
(2) |U_n {1,...,n}| is not greater than every natural number k.

More precisely, when we say "the number of elements of S is k", we
mean nothing more or less than "|S| = k", right? And thus we are back
to the claim that ZF proves both of the following statements:

(1) (Ak in N)(|U_n {1,...,n}| > k)
(2) NOT (Ak in N)(|U_n {1,...,n}| > k)

We may take the proof of (1) for granted. I will ask once again for a
proof of (2). I don't want to talk about paths. I don't want to
discuss anything aside from a proof of (2). Please show me that
argument or explain to me where I misinterpreted your claim and let's
move on with this discussion.

--
Jesse F. Hughes
Playin' dismal hollers for abysmal dollars,
Those were the days, best I can recall.
-- Austin Lounge Lizards, "Rocky Byways"