|
|
Re: ZFC and God
Posted:
Jan 23, 2013 1:04 PM
|
|
WM <mueckenh@rz.fh-augsburg.de> writes:
> On 23 Jan., 18:25, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>> "Jesse F. Hughes" <je...@phiwumbda.org> writes:
>>
>> > We're talking about whether ZF proves a contradiction, by proving that
>>
>> > NOT (A k in N)(U_n {1,...,n} > k).
>>
>> Should be NOT (A k in N)(|U_n {1,...,n}| > k). (Missed the
>> cardinality symbols).
>
> You have not yet understood our topic at all. *Every* k in N can be
> surpassed by the union of FISs. But not all:
> For every k in N we have infinitely m > k in N.
> For all k in N we have no m > k in N.
Well, you seem to be speaking in an ambiguous manner, but I think you
mean this:
(Ak in N) { m in N | m > k } is infinite
NOT (E m in N)(A k in N) ( m > k ).
But, of course, both of those claims are obvious and neither
contradict one another nor the claim that
U_n {1,...,n} is infinite.
(NOTE: Perhaps the second statement should be
NOT (E m)(A k in N) ( m > k ) (dropped "in N" from first quant),
in which case that is *not* obviously true, but it seems that it is
irrelevant to the issue at hand and I will postpone discussing that
until it is obviously necessary.)
> Understand the Binary Tree. After you will have understood it, you
> will understand, why it is important.
No, let's first settle the point at hand. Can you show that ZF proves
U_n {1,...,n}
is finite?
If not, then simply tell me I misunderstood your claim, you do not
claim that you can show this, and we can move on. But not until I'm
certain of your claim.
You did say that you could show ZF proves a contradiction involving
the claim that U_n {1,...,n} is infinite, right?
--
"So yeah, do the wrong math, and use the ring of algebraic integers
wrong, without understanding its quirks and real mathematical
properties, and you can think you proved Fermat's Last Theorem when
you didn't." -- James S. Harris on hobbies
|
|
