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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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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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.
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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. |
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