Question

Let S be in R be nonempty. Prove that if a number u in R has the following properties 1. for every n in N u - 1/n is not an upper bound of S and the number u + 1/n is an upper bound then u = supS.

Answer 1

Ok I can see it now.
Yes it is right

Answer 2

positive. here is how i did it. first if u = sup S then it must be an upper bound and the smallest.

Answer 3

by contradiction if u is not an upper bound then u + x must be in s and u + x < u + 1/n. this leads to a contradiction with the achemedian prop.

Answer 4

so you have u- 1/n this tends to u as n tends to infinity. (from below)
If this is not an upper bound that means that there is a number in S that is higher.
u + 1/n tends to u as n tends to infinity (from above).
As this is an upper bound there cannot be any number in S that is higher.
By the sandwich theorem u must be supS

Answer 5

Yours looks cool too. I dont really like this part of maths. Bit boring