Proof of points arbitrarily close to supremum

[Vale132]

Homework Statement

Let S \subset \mathbb{R} be bounded above. Prove that s \in \mathbb{R} is the supremum of S iff. s is an upper bound of S and for all \epsilon > 0, there exists x \in S such that |s - x| < \epsilon.

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 s \in \mathbb{R} is the supremum of S, then for all \epsilon > 0, there exists x \in S such that s - x < \epsilon.

The Attempt at a Solution

The \Rightarrow direction:
By the definition of the supremum, s is an upper bound.

Thus s \geq x for all x \in S, so s - x \geq 0. Then |s - x| = s - x < \epsilon, by the result mentioned above.

The \Leftarrow direction:
Suppose that s \in \mathbb{R} is an upper bound of S and that for all \epsilon > 0, there exists x \in S such that |s - x| < \epsilon. Now suppose for a contradiction that s is not the supremum of S. Then there exists an upper bound u of S with u < s.

But since S is bounded above, it does have a supremum. Call the supremum t. We have that t \leq s, and for every \epsilon > 0, there exists y \in S such that |t - y| < \epsilon.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!

 

[Samy_A]

Homework Statement

Let S \subset \mathbb{R} be bounded above. Prove that s \in \mathbb{R} is the supremum of S iff. s is an upper bound of S and for all \epsilon > 0, there exists x \in S such that |s - x| < \epsilon.

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 s \in \mathbb{R} is the supremum of S, then for all \epsilon > 0, there exists x \in S such that s - x < \epsilon.

The Attempt at a Solution

The \Rightarrow direction:
By the definition of the supremum, s is an upper bound.

Thus s \geq x for all x \in S, so s - x \geq 0. Then |s - x| = s - x < \epsilon, by the result mentioned above.

The \Leftarrow direction:
Suppose that s \in \mathbb{R} is an upper bound of S and that for all \epsilon > 0, there exists x \in S such that |s - x| < \epsilon. Now suppose for a contradiction that s is not the supremum of S. Then there exists an upper bound u of S with u < s.

But since S is bounded above, it does have a supremum. Call the supremum t. We have that t \leq s, and for every \epsilon > 0, there exists y \in S such that |t - y| < \epsilon.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!

The ⇒ direction seems correct.

For the ⇐ direction:
You have $s$ and the supremum $t$.
If $s=t$, you are done.
$s<t$ is impossible, as $t$ is the supremum, the least upper bound.
That leaves the case $s>t$.
Now try to prove that then there must exist an $x \in S$ satisfying $x>t$, which contradicts $t$ being a supremum.
Hint: set $\epsilon=\frac{s-t}{3}$.