Thread: Upper Bound

1. 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.

2. Originally Posted by harish21
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.
$\displaystyle (1+c)^2=1+2c+c^2\geqslant c^2$