How can we prove that A is less than or equal to B using only one element of S?

[Mr Davis 97]

Homework Statement

Suppose that $S$ is a non-empty bounded subset of the real numbers. Show that $\forall s \in S (\inf S \le s \le \sup S)$ implies that $\inf S \le \sup S$.

Homework Equations

The Attempt at a Solution

How do deduce this logically? All I can say is that it is obvious that, since it is always the case that $\inf S \le s \le \sup S$, and since there exists at least one $s$ for which this is true, it must always be true, so that $\inf S \le \sup S$. Is there any way to make this deduction more logical and rigorous?

 

[fresh_42]

Homework Statement

Suppose that $S$ is a non-empty bounded subset of the real numbers. Show that $\forall s \in S (\inf S \le s \le \sup S)$ implies that $\inf S \le \sup S$.

Homework Equations

The Attempt at a Solution

How do deduce this logically? All I can say is that it is obvious that, since it is always the case that $\inf S \le s \le \sup S$, and since there exists at least one $s$ for which this is true, it must always be true, so that $\inf S \le \sup S$. Is there any way to make this deduction more logical and rigorous?

This depends on what you've not written under section $2$. How is $\leq$ defined in your specific case? All what you can say is, that $\inf S \leq s \leq \sup S$ means $\inf S \leq s \,\wedge \, s \leq \sup S$ where you only need one element $s \in S$ here, which is given by $S \neq \emptyset$. Now $\inf S \leq \sup S$ is a direct consequence from the transitivity property of $\leq$ which should be part of the definition. However, I cannot know, whether it is part of your definition.

 

[member 587159]

Yes, this follows directly from transitivity, if S contains at least one element. His definition doesn't matter, the transitivity property should have been proven before working with sup and inf.

 

[Mr Davis 97]

Yes, this follows directly from transitivity, if S contains at least one element. His definition doesn't matter, the transitivity property should have been proven before working with sup and inf.

Why is it sufficient that we see it's true for only one element? Why don't all elements of $S$ matter in this case?

 

[Stephen Tashi]

Why is it sufficient that we see it's true for only one element? Why don't all elements of $S$ matter in this case?

Suppose we are given that $S$ is a non-empty bounded set of real numbers and that , for some numbers $A$ and $B$, $\exists x \in S$ such that $A \le s \le B$. We can prove $A \le B$ We don't need to consider all the elements in $S$ to accomplish the proof.

Also, as you have written the question, the details of the definitions of "inf" and "sup" are not needed. We only need to know that they are real numbers.