# Upper Bound

• Feb 20th 2010, 12:50 PM
harish21
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 b-r 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?!
• Feb 20th 2010, 01:26 PM
icemanfan
b) If $b^2 > c, b^2 = c + d$ for some positive number d.

Now $(b-r)^2 = b^2 - 2rb + r^2 = c + d - 2rb + r^2$

Note that if $d - 2rb + r^2 > 0$, then we have $(b-r)^2 > c$.

Since $\lim_{r \to 0} r^2 - 2rb = 0$, there exists a positive r such that $r^2 - 2rb < d$.
• Feb 20th 2010, 01:28 PM
Plato
Suppose that $\left( {\exists y \in S} \right)\left[ {c + 1 < y} \right]$ then $(c+1)^2. WHY?
What is wrong with that picture?
How does that contradiction show that $c+1$ is an upper bound on $S$?