Proof of points arbitrarily close to supremum

In summary, to 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, we use the basic proof methods, properties of inequalities, and the Completeness Axiom. In the \Rightarrow direction, we use the definition of supremum to show that s is an upper bound. In the \Leftarrow direction, we suppose for a contradiction that s is not the supremum of S and show that this leads to a contradiction. We set \epsilon = \frac{s-t}{3} and use the fact that for every \epsilon
  • #1
Vale132
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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!
 
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  • #2
Vale132 said:

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!
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}##.
 
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FAQ: Proof of points arbitrarily close to supremum

1. What is the definition of "Proof of points arbitrarily close to supremum"?

The proof of points arbitrarily close to supremum refers to a mathematical concept in which a set of points is shown to approach a specific value known as the supremum, which is the least upper bound of the set. In other words, the points in the set get closer and closer to the supremum but never actually reach it.

2. Why is it important to prove points arbitrarily close to supremum?

Proving points arbitrarily close to supremum is important because it allows us to establish the existence of the supremum for a given set of points. This is crucial in many mathematical proofs and allows us to make precise statements about the behavior of a set of numbers.

3. How is the proof of points arbitrarily close to supremum typically done?

The proof of points arbitrarily close to supremum is typically done using the definition of the supremum. This involves showing that for any given distance between a point and the supremum, there exists a point in the set that is within that distance from the supremum. This can be proven using the Archimedean property of real numbers.

4. Can you give an example of a proof of points arbitrarily close to supremum?

Sure, let's consider the set of real numbers between 0 and 1. The supremum of this set is 1. To prove that there are points arbitrarily close to 1, we can take any distance, say 0.5, and show that there exists a point in the set that is within 0.5 of 1. In this case, we can choose 0.999 as our point, which is clearly within 0.5 of 1. This can be done for any distance, thus proving the existence of points arbitrarily close to 1.

5. Are there any real-world applications of the proof of points arbitrarily close to supremum?

Yes, the concept of the supremum and the proof of points arbitrarily close to supremum are widely used in fields such as economics, physics, and computer science. They are particularly useful in optimization problems, where finding the supremum of a set of values can help determine the most optimal solution.

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