Least Upper Bound Property

[Deinosuchus383]

Hi,

I was reading the Appendix to Serge Lang's "Short Calculus", and I was having trouble understanding the proof:

Let "a" be a number such that
0≤ a < 1/n
for every positive integer n.
Then a=0. There is no number b such that b≥n for every positive integer n

Proof: Suppose there is a number a≠0 such that a<1/n for every positive integer n. Then n<1/a for every positive integer n. Thus to prove both our assertions it will suffice to prove the second.

Suppose there is a number b such that b≥n. for every positive integer n. Let S be the set of positive integers. Then S is bounded, and hence has a least upper bound. Let C be this least upper bound. No number strictly less than C can be an upper bound. Since 0<1, we have C<C+1, whence C-1<C. Hence there is a positive integer n such that C-1<n This implies that C<n+1 and n+1 is a positive integer. We have contradicted our assumption that C is an upper bound for the rest of the set of positive integers so no such upper bound can exist.

Perhaps I just have trouble understanding proofs, but why does the proof utilize 0<1? And how does it know to use n<C-1, or that n+1 is a positive integer?

 

[Ishuda]

"why does the proof utilize 0<1"...

If you have
a < b
then you can add something (in this case C) to both sides and still have the inequality stand.

...And how does it know to use n<C-1, or that n+1 is a positive integer?..."

You have that backwards. If C-1 is not an upper bound for the set of positive integers, then there must be a positive integer n such that
C-1 < n
[If there weren't such an n, then C-1 would be larger than all positive integers and thus be an upper bound]. Now use the same thing we did before and add something to each side (in this case 1) and we have
C < n + 1.
Since n is a positive integer, n+1 is a positive integer and C is not the least upper bound, etc., etc., etc.