
Re: ZFC and God
Posted:
Jan 24, 2013 8:16 AM


WM <mueckenh@rz.fhaugsburg.de> writes:
> On 24 Jan., 13:36, "Jesse F. Hughes" <je...@phiwumbda.org> wrote: >> WM <mueck...@rz.fhaugsburg.de> writes: > >> Well, what you present below is *not* a proof of (*). > > That is wrong. You have no reason to believe that your definition of > proof is correct or the only one.
This argument doesn't involve U_n {1,...,n} *at all*!
But let's let it pass. > >> >> Clearly, for all j, d(j) != t_j(j) and hence d != t_j for any j in >> N. >> >> Is this what you mean up 'til now? > > Yes. > >> >> > 4) Certainly you agree that, since all t_i = (t_i1, t_i2, ..., t_in) >> > have only a finite, though not limnited, number n of digits, the >> > diagonalization for every t_i yields a finite d_i =/= t_ii. >> > (The i on the left hand side cannot be larger than the i on the right >> > hand side. In other words, "the list" is a square. Up to every i it >> > has same number of lines and columns. ) >> >> No idea what you mean by the parenthetical remark. > > You will have have recognized that here the diagonal argument is > applied. It is obvious that up to every line = column the list is a > square.
It is clear that, for all j, d(j) != t_j(j) and hence d != t_j. If that's what you mean by the diagonal argument, great!
Once again, however, you say something that has no clear meaning to me. Can you clarify "It is obvious that up to every line = column the list is a square?" I've no clue what it means.
>> >> I do agree that d_i is defined for every i in N. In particular, (d_i) >> is an infinite sequence of digits. Is this what you're claiming, too? >> You've lost me. I don't know what you mean when you say, "everything >> here happens among FISs." And I'm also puzzled by the meaning of the >> next sentence. > > Every t_i is finite. Hence, in a square, if the width is finite, also > the length must be finite.
Again, no idea what you're talking about here.
We have a function
t: N > R
or, if you prefer
t: N x N > {0,...,9}
taking (i,j) to the j'th digit of the i'th terminating decimal, in the usual order. From this, we construct
d: N > {0,...,9}
which is clearly the decimal representation of a real which has no terminating representation.
All this nonsense of a square seems beside the point. It is clear that
(An)(Em)( len(t_m) > n ),
so I've no idea why you think that the "width" of the "square" is finite. The list looks like this:
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.11 0.12 ... 0.99 0.101 0.102 0.103 ... 0.999 0.1001 0.1002 ...
>> >> Here are some obvious things. >> >> d(j) is defined for every j in N. >> d(j) != 0 and d(j) != 9 for any j in N. >> >> Hence the number d does not have a terminating decimal >> representation. > > Neither the set of t_i does have a largest element. Nevertheless there > is no t_i of actually infinite length.
Correct, to both claims. So what? The digit d(j) is defined for every j, and is neither 0 nor 9, so d is a real number which has no terminating decimal representation.
Do you agree with that (obvious) claim or not?
>> >> This looks like I do *not* agree with your claim that "d cannot be >> longer than every t_i". > > A sequence of squares will never result in a square such that all > sides are finite but the diagonal d is infinite. The overlap of d and > t_i cannot be larger than t_i.
Sorry, I could've sworn you were talking about what ZF proves. Can you state this in the language of ZF and prove it? Much thanks!
> In particular, what would be changed in the length of d if we admitted > also nonterminating t_i (of infinite length)?
Nothing would change. So what?
Anyway, let's get back to ZF. You said that you could prove ZF was inconsistent, so I'm eager to see how you state these claims in the language of ZF and prove them.
 Jesse F. Hughes
Baba: Spell checkers are bad. Quincy (age 7): CHEKERS ARE BAD.

