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Homework Statement
Let [itex] S \subset \mathbb{R} [/itex] be bounded above. Prove that [itex] s \in \mathbb{R} [/itex] is the supremum of [itex] S [/itex] iff. [itex] s [/itex] is an upper bound of [itex] S [/itex] and for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] |s - x| < \epsilon [/itex].
Homework Equations
**Assume I have only the basic proof methods, some properties of inequalities, and the Completeness Axiom at my disposal**
Assume I have already proved that if [itex] s \in \mathbb{R} [/itex] is the supremum of [itex] S [/itex], then for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] s - x < \epsilon [/itex].
The Attempt at a Solution
The [itex] \Rightarrow [/itex] direction:
By the definition of the supremum, [itex] s [/itex] is an upper bound.
Thus [itex] s \geq x [/itex] for all [itex] x \in S [/itex], so [itex] s - x \geq 0 [/itex]. Then [itex] |s - x| = s - x < \epsilon [/itex], by the result mentioned above.
The [itex] \Leftarrow [/itex] direction:
Suppose that [itex] s \in \mathbb{R} [/itex] is an upper bound of [itex] S [/itex] and that for all [itex] \epsilon > 0 [/itex], there exists [itex] x \in S [/itex] such that [itex] |s - x| < \epsilon [/itex]. Now suppose for a contradiction that [itex] s [/itex] is not the supremum of [itex] S [/itex]. Then there exists an upper bound [itex] u [/itex] of [itex] S [/itex] with [itex] u < s [/itex].
But since [itex] S [/itex] is bounded above, it does have a supremum. Call the supremum [itex] t [/itex]. We have that [itex] t \leq s [/itex], and for every [itex] \epsilon > 0 [/itex], there exists [itex] y \in S [/itex] such that [itex] |t - y| < \epsilon [/itex].I'm wondering whether there are any logical errors so far, and whether someone could give me a tiny hint as to whether I'm on the right track, and what I might think about next. I'm posting here rather than Math.SE because I would prefer "Socratic" help rather than an answer. Thanks!