2
$\begingroup$

I have the following problem:

Let $A, B \subseteq\mathbb{R}, A \neq \emptyset$ and $B \neq\emptyset$ such that $$(\forall a \in A \wedge b \in B): a \leq b$$ Prove that $\sup(A) \leq \inf(B)$.

My attempt:

We know by definition that $A$ is bounded from below and $B$ is bounded from above, therefore by Completeness Axiom, we know there exist $\sup(A)$, and by theorem there exists $\inf(B)$.

Let $a \in A$ and $b \in B$, then $a \leq \sup(A)$ and $\inf(B) \leq b$. Hence, $a + \inf(B) \leq \sup(A) + b$

However, I don't have idea how to get $\sup(A) \leq \inf(B)$.

Hope you can help me :)

$\endgroup$
3
$\begingroup$

$\sup A $ is the least upper bound of $A $, that is, it is the least number which is greater than or equal to all elements of $A $ [1]. So by definition, $\sup A\leq b $ for all $b\in B $. Now, $\inf B $ is the greatest lower bound of $B $. And since $\sup A\leq b,\, \forall b\in B $, we have $\sup A\leq \inf B $.

$\endgroup$
0
$\begingroup$

Assume $y= \inf (B) \lt \sup(A) = x$. By the definition of $\inf, \forall \epsilon \gt 0 \exists b \in B~(y \leq b \lt y+ \epsilon)$. By the definition of $\sup, \forall \epsilon \gt 0 \exists a \in A~(x - \epsilon \lt a \leq x).$ Choose $\epsilon = \frac{x-y}{3}.$ Then $b \lt y+ \epsilon \lt x - \epsilon \lt a$, so $\exists a \in A, b \in B \text{ with } b \lt a$ contradicting our hypothesis.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.