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

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 Jesse F. Hughes Posts: 9,776 Registered: 12/6/04
Re: ZFC and God
Posted: Jan 22, 2013 11:48 AM

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

> On 22 Jan., 15:49, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> FIS: Finite Initial Segment
> FISON: Finite Initial Segment Of Naturals (or indices)
>

>> Well, it's not a "union" in the usual sense, but let's let it pass.
>
> It is a union in that sense that every FISON {1, 2, ..., n+1} added
> contains all smaler FISONs {1, 2, ..., n}.

We were speaking about the real number d, not the set N.

> This does never change. In particular the set is always finite. If
> you add with always doubling frequence, you can add all FISONs in
> finite time - given that an "all" is meaningfull here. But at the
> end, you think, we cannot follow so quickly, and abracadabra we get
> something larger than every FISON? Not in mathematics!

Look, you need to offer an actual proof that

U_n=1^oo {1,...,n} is finite.

I'll give you a hint, by showing you the proof that

U_n=1^oo {1,...,n} is infinite.

Just so you know what a proof looks like.

Here's the proof. First, let me be clear what I mean by infinite. I
mean that there is no natural k such that |U_n=1^oo {1,...,n}| = k.

Let k be a natural number. Then |{1,...,k+1}| = k+1 > k. Also,

{1,...,k+1} c U_n=1^oo {1,...,n},

and hence

k < k + 1 = |{1,...,k+1}| <= |U_n=1^oo {1,...,n}|.

Thus, since k was arbitrary, we see that

(A k in N) |U_n=1^oo {1,...,n}| != k.

Hence, by definition, U_n=1^oo {1,...,n} is infinite.

Now, I'm sure you're familiar with that argument. I'd like to see
your claim to the contrary proved with just as much detail and I'd
like for you to be willing to answer any questions I raise about any
steps I don't understand.

Until then, I simply cannot see your reasoning that the union is
finite.

Just to be clear, as well, let's define what we mean by

U_n=1^oo {1,...,n}.

It is a set, so it's sufficient for us to specify precisely what
elements are in that set.

x in U_n=1^oo {1,...,n} iff there is an n in N such that
x in {1,...,n}.

frequency or anything like that. I simply defined the union in the
usual manner. The union of a set S of sets is defined by

x in US iff (E s in S)(x in s).

Thus,

x in U{ {1,...,n} | n in N } iff there is an n in N such that
x in {1,...,n}.

>> Here's the problem. Let d(n) be the n'th digit of d, and similarly
>> b_k(n) the n'th digit of b_k. Let d_k be the finite initial segment
>> of d consisting of the first k digits.
>>
>> We can prove, as you say, that
>>
>> (An)(d(n) != b_n(n)).

>
> From this "for every n" you claim "for all". And that is wrong for
> infinite sets.

I have no idea what difference between "for every" and "for all" you
have in mind, but feel free to read the above as "for every n".

>>
>> From this, you claim that it follows that
>>
>> (Ek)(An)(d_k != b_n) (*)

>
> No I do not confuse quantifiers. I simply know that for FIS of the
> diagonal there exist a FIS in the list such that:
>
> (An)(Ek) (d_1,d_2, ..., d_n) = (b_k1, b_k2, ..., b_kn)

Yes, obviously (though you've change notation to a more confusing
form). On this we agree. Every finite initial segment of d is in the
list, i.e., is equal to some b_k.

We disagree on your claim that, if d differs from each b_k in the kth
position, then some finite initial segment of d is not in the list.

>>
>> I simply do not see how that follows.

>
> It follows from the purported completeness of the list (or decimal
> tree) that contains all possible FIS of decimal representations of
> reals. Up to any finite n, such a FIS (b_k1, b_k2, ..., b_kn) is
> present in a complete list. And we know that there is no infinite n.

Again, you're focusing on our point of agreement, not our point of
disagreement.

>> > Do you agree that the actually infinite path {1, 2, ...} does not
>> > differ from B, such that "behind" every natural nothing can happen? In
>> > particular the inclusion of {1, 2, ...} does not change anything in B?
>> > Therefore {1, 2, ...} cannot differ from the union B?

>>
>> I'm not at all sure what your question means, but I agree (of course)
>> that the identity stated in (B) is correct. The set N is the union of
>> all the sets {1,2,...,n} (using the convention that N's least element
>> is 1, of course).

>
> And all these sets are finite. In particular no FIS contains more
> natural indices than every FIS. So the union cannot contain more
> either.

>
>> > By the way, every FISON, which stands for finite initial segment of
>> > the naturals, is finite. This does not change by unioning as many as
>> > are available. The union never gets larger than every FISON. Can you
>> > understand that?

>>
>> Of course that claim is simply nonsense, but let's focus on the matter
>> at hand. Are you claiming that the "reason" that some d_k is missing
>> from the list of b_n's is that

>
> There is no FIS (d_1,d_2, ..., d_n) of the diagonal missing, since
> the list is complete!

I repeat, the claim that I disagree with is the following:

it is impossible that the diagonal differs from all entries b_n at a
finite place. If this was possible, the list would have a gap,
namely a finite initial segment of d.

We both agree that every finite initial segment d_k of d is in the
list. You claimed

(An)(d(n) != b_n(n)) -> (Ek)(An)(d_k != b_n).

This implication does not seem to follow.

--
Jesse F. Hughes
"The future is a fascinating thing, and so is history. And you people
are a fascinating part of history, for those in the future."
-- James S. Harris is fascinating, too