Re: Calculus XOR Probability

David R Tribble said:
And it's trivially easy to convert "n is finite" into an inequality
that is true for all n in N:
1. 1 < oo.
2. If n < oo, then succ(n)=n+1 < oo.
3. Therefore, n < oo for all n in N.


Tony Orlow wrote:
Yes, and that inequality holds in all finite cases, and fails in the infinite
case, which was my point. Inequalities may be based on differences with a limit
of 0 as n->oo, in which case there is no longer an inequality, and the proof
fails for infinite n.


David R Tribble said:
My proof above shows that n<oo for every n in N.

But if what you claim is true, then there must exist some n where the
inequality
succ(n)=n+1 < oo
is false.

So which n is that "infinite case"?


Tony Orlow wrote:
For any infinite n.

Which is defined how? With Peano you've got 0 and succ(n). Where do
you get infinite(n)?


There is no finite n with infinite successor. So what?
Nobody ever claimed that gap was bridged by any single application of successor().
That gap is bridged by the infinite number of succession in the uncountable set.

That means nothing until you can formalize it. What is "the infinite
number of successions" and "the uncountable set"? You're not talking
about N, that's for sure.

.

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