```Date: Jan 23, 2013 1:04 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

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 youmean 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 neithercontradict 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 isirrelevant to the issue at hand and I will postpone discussing thatuntil 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 notclaim that you can show this, and we can move on.  But not until I'mcertain of your claim.You did say that you could show ZF proves a contradiction involvingthe claim that U_n {1,...,n} is infinite, right?-- "So yeah, do the wrong math, and use the ring of algebraic integerswrong, without understanding its quirks and real mathematicalproperties, and you can think you proved Fermat's Last Theorem whenyou didn't." -- James S. Harris on hobbies
```