keiths' infinity "proof"

-
More like poof-

1.Definition: Set A is infinite if every finite subset of A is also a proper subset of A.

That just moves the problem back to defining "finite". But anyway...

2. Proof that infinite sets exist:

a. Assume that infinite sets do not exist. By the definition above, that means that every set has at least one finite subset that is not a proper subset.

OK.

b. Consider all finite subsets S1, S2, etc. of the set N of natural numbers. Every such subset contains a greatest element Ln (if there were any elements greater than Ln, then one of them would be the greatest element instead of Ln).

OK

c. Ln is a finite natural number. Therefore Ln+1 exists and is also a finite natural number.

Ln + 1 would be the new Ln. You just covered that in "b".

d. Ln + 1 is not a member of Sn, but it is a member of N.

No, it would be a member of Sn. Ln+1=Ln, just as YOU said in "b"

Therefore every Sn is a proper subset of N, thus contradicting a).

Not if one follows YOUR directions.


OK, do it for SLn.

SLn= {0,1,2,3,4,...,Ln}

SLn + 1 = {0,1,2,3,4,...,Ln}

N= {0,1,2,3,4,...,Ln}


SLn finite subset of N that is also an improper subset.

keiths flails away:

The only improper subset of N is N itself, and N is infinite, not finite.

Hey asswipe, you cannot use that which you are trying to prove as proof. No one can prove N is infinite. If N is the set of natural NUMBERS and infinity is NOT a number, then it doesn't belong in the set. However the largest number does belong in the set- largest number Ln, ie greatest element.

And keiths, if Ln = largest number, then Ln + 1 is still the largest number.

SLn is a finite subset of N. Ln= largest number {0,1,2,3,4,...,Ln} the Ln is INSIDE of the brackets keiths. It is the largest NUMBER. So SLn would be the name of the finite set that starts at 0 and goes to the largest number. And the set of natural numbers is also the set that starts at 0 and goes to the largest number.