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 $\epsilon=\frac{s-t}{3}$.