"Sylvia Else" <sylvia@not.here.invalid> wrote ... > On 18/06/2010 3:03 PM, |-|ercules wrote: >> "Sylvia Else" <sylvia@not.here.invalid> wrote >>> On 18/06/2010 10:40 AM, Transfer Principle wrote: >>>> On Jun 17, 6:56 am, Sylvia Else<syl...@not.here.invalid> wrote: >>>>> On 15/06/2010 2:13 PM, |-|ercules wrote: >>>>>> the list of computable reals contain every digit of ALL possible >>>>>> infinite sequences (3) >>>>> Obviously not - the diagonal argument shows that it doesn't. >>>> >>>> But Herc doesn't accept the diagonal argument. Just because >>>> Else accepts the diagonal argument, it doesn't mean that >>>> Herc is required to accept it. >>>> >>>> Sure, Cantor's Theorem is a theorem of ZFC. But Herc said >>>> nothing about working in ZFC. To Herc, ZFC is a "religion" >>>> in which he doesn't believe. >>> >>> Well, if he's not working in ZFC, then he cannot make statements about >>> ZFC, and he should state the axioms of his system. >> >> Can you prove from axioms that is what I should do? >> >> If you want to lodge a complaint with The Eiffel Tower that the lift is >> broken >> do you build your own skyscraper next to the Eiffel Tower to demonstrate >> that fact? >> > > That's hardly a valid analogy. > > If you're attempting to show that ZFC is inconsistent, then say that you > are working within ZFC. > > If you're not working withint ZFC, then you're attempting to show that > some other set of axioms is inconsistent, which they may be, but the > result is uninteresting, and says nothing about ZFC. > > Sylvia.
That would be like finding a fault with the plans of The Leaning Tower Of Piza.
I might look at ZFC at some point, but while you're presenting Cantor's proof in elementary logic I'll attack that logic.
Instead of 'constructing' a particular anti-diagonal, your proof should work equally well by giving the *form* of the anti-diagonal.
This is what a general diagonal argument looks like.
For any list of reals L.
CONSTRUCT a real such that An AD(n) =/= L(n,n)
Now to demonstrate this real is not on L, it is obvious that An AD(n) =/= L(n,n)
Therefore [ An AD(n) =/= L(n,n) -> An AD(n) =/= L(n,n) ] proves superinfinity!
And THAT is Cantor's proof!
Want to see his other proof? That no box contains the box numbers (of boxes) that don't contain their own box number?
