Upper Bounds and Supremums

• September 23rd 2012, 07:51 AM
lovesmath
Upper Bounds and Supremums
Suppose that A is contained in the set of all real numbers and is bounded above. Prove that if A contains one of its upper bounds, then this upper bound is sup A.

It seems like it should be a simple problem, but I'm lost.
• September 23rd 2012, 08:34 AM
Plato
Re: Upper Bounds and Supremums
Quote:

Originally Posted by lovesmath
Suppose that A is contained in the set of all real numbers and is bounded above. Prove that if A contains one of its upper bounds, then this upper bound is sup A.

Suppose that $a \in A\; \wedge \;\left( {\forall x} \right)\left[ {x \in A \Rightarrow x \leqslant a} \right]$.

Now prove that $a=\sup(A)$.

Hint: If you assume that $a<\sup(A)$ there is an intermediate contradiction.