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.
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 $\displaystyle a \in A\; \wedge \;\left( {\forall x} \right)\left[ {x \in A \Rightarrow x \leqslant a} \right]$.
Now prove that $\displaystyle a=\sup(A)$.
Hint: If you assume that $\displaystyle a<\sup(A)$ there is an intermediate contradiction.