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 🙂
Best Answer
$\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 $.