Date: Jan 23, 2013 8:36 AM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <> writes:

> On 23 Jan., 12:47, "Jesse F. Hughes" <> wrote:
>> WM <> writes:
>> > On 22 Jan., 21:18, "Jesse F. Hughes" <> wrote:
>> >> WM <> writes:
>> >> > That is potential infinity. That proof is not necessary, because the
>> >> > set is obviously potentially infinite. No, you shoudl give a proof,
>> >> > that there is a larger k than all finite k.

>> >> Er, no.  When I say that the union is infinite, I do not mean that it
>> >> contains an infinite number.

>> > But you mean that the tree contains infinite paths. And just that is
>> > impossible without ...

>> > In order to shorten this discussion please have a look at
>> No.  It's irrelevant.

> You are in error.

>> We're talking about whether you can prove that
>>  U_n=1^oo {1,...,n}
>> is finite.  I'm not switching topics to paths in trees (despite the
>> fact that the ignorance of your question is obvious).

> The union of FISs is finite. Yes that is my claim. But I cannot give
> an upper limit, because the finite numbers have no upper limit. This
> is called potentially infinite.

>> > There it has meanwhile turned out ... But see it with your own eyes
>> > what you would not believe if I told you.

>> > The index omega is in reach, it seems.
>> You're playing your usual little game of trying to change the topic.
>> I won't have it.
>> I take it that this new tack is so that you don't have to concede the
>> point: there is no mathematical publication which claims that the
>> above union contains elements larger than any natural, nor any
>> publication which claims that this is what it means to be infinite.

> I know. But if you hace 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.

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

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.

So, let's not get distracted by paths and whatsits. Just do what you
said you could do: show that ZF proves a contradiction. Either show
me that there is a natural k such that

| U_n {1,...,n} | = k

or show me that there is an element k in U_n {1,...,n} such that k is
larger than every natural number.

But don't distract us from the topic at hand.

Jesse F. Hughes
Mama: "I had a very good steak when I was in Bonn."
Quincy (Age 4): "A stick? I wish you brought it home. Was it very
big and did it look like a gun?"