
Upper Bound
S = {x  x in R , x>=0, x^2 < c}
a. Show that c+1 is an upper bound for S and therefore, by completeness axiom, S has a least upper bound that we denote by b.
b. Show that if b^2>c, we can show choose a suitably positive number r such that br is also an upper bound for S, thus contradicting the choice of b as an upper bound.
c. If b^2< r, then we can choose a suitable positive number r such that b+r belongs to S, thus contradicting the choice of b as an upper bound of S.
All I could do here was to show that
if c<=1, then x<1<1+c. So 1+c is an upper bound for S
and
if c>1, x<Sqrt(c)<c<1+c. So 1+c is an upper bound for S.
How is the rest of part(s) proved?!

b) If $\displaystyle b^2 > c, b^2 = c + d$ for some positive number d.
Now $\displaystyle (br)^2 = b^2  2rb + r^2 = c + d  2rb + r^2$
Note that if $\displaystyle d  2rb + r^2 > 0$, then we have $\displaystyle (br)^2 > c$.
Since $\displaystyle \lim_{r \to 0} r^2  2rb = 0$, there exists a positive r such that $\displaystyle r^2  2rb < d$.

Suppose that $\displaystyle \left( {\exists y \in S} \right)\left[ {c + 1 < y} \right]$ then $\displaystyle (c+1)^2<y^2<c^2$. WHY?
What is wrong with that picture?
How does that contradiction show that $\displaystyle c+1$ is an upper bound on $\displaystyle S$?