# Proving that a set is unbounded above

• May 20th 2013, 09:07 AM
AW63
Proving that a set is unbounded above
Prove that the set (-2, ) is unbounded above and has no minimum element.

• May 20th 2013, 09:24 AM
Plato
Re: Proving that a set is unbounded above
Quote:

Originally Posted by AW63
Prove that the set (-2, ) is unbounded above and has no minimum element.

Suppose that $\displaystyle \mathcal{S}=(-2,\infty)$.

unbounded above. If $\displaystyle t\in\mathcal{S}$ we know that $\displaystyle 0<1$ so $\displaystyle t<t+1$.

has no minimum element.. If $\displaystyle t\in\mathcal{S}$ then $\displaystyle -2<t$.

So $\displaystyle -2<\frac{t+2}{2}<t$. Can $\displaystyle t$ be minimal in $\displaystyle \mathcal{S}~?$
• May 20th 2013, 09:58 AM
HallsofIvy
Re: Proving that a set is unbounded above
S has no upper bound
Proof by contradiction: Suppose S has an upperbound, a. Certainly, 0 is in S so we must have 0< a. Then a< a+ 1 so a+ 1 is not in S. But S contains all numbers larger than -2 so that is a contradiction

S has no smallest member.
Proof by contradicton: Suppose S has smallest member b. Since b is in S, -2< b. Now, -2< (b- 2)/2< b is larger than -2 so in S but less than b which contradicts the assumption that b is the smallest member of S.