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 .
Now prove that .
Hint: If you assume that there is an intermediate contradiction.