# Proving that a set is unbounded above

• May 20th 2013, 10:07 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, 10:24 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 $\mathcal{S}=(-2,\infty)$.

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

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

So $-2<\frac{t+2}{2}. Can $t$ be minimal in $\mathcal{S}~?$
• May 20th 2013, 10:58 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.