Supremum of Real Numbers [closed]

Let $A$ be any non-empty subset of $\mathbb{R}$. Then $s = \text{sup}A$ iff $s$ has the following properties:

(1) $s ≥ a$ for every $a \in A$

(2) If $t < s$, then there exists an $a \in A$ such that $a > t$.

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What is the question? – Michael Greinecker Sep 30 '12 at 18:48
up vote 0 down vote favorite Let A be any non-empty subset of R. Then s=supA iff s has the following properties: (1) s≥a for every a∈A (2) If t<s, then there exists an a∈A such that a>t.Prove it? – user43165 Oct 1 '12 at 9:45

closed as not a real question by Ayman Hourieh, Thomas, Noah Snyder, Norbert, Jason DeVitoOct 8 '12 at 15:56

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

I take it that the problem is showing the given definition to be equivalent to the one of the supremum as the smallest upper bound. Well, (1) says it is an upper bound, so one has to show $s\leq b$ for every upper bound. Suppose not. Then there is an upper bound $b$ of $A$ such that $b< s$. Then $(2)$ contradicts $b$ being an upper bound.
 Well, to me this looks like only one of the directions..even though the other direction is trivial… i.e. showing that if a number fulfills (1) and (2) then it is an upper bound of $A$ and smaller then every upper bound $b$ of $A$. – user22705 Sep 30 '12 at 19:07