How to prove $\sup(A) \leq \inf(B)$?

Question

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 :)

Answer 1

$\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 $.

Answer 2

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.

Answer 3

Let $\sup A = s$ and $\inf B = d$, and suppose, for the sake of contradiction, that $s > d$. Then $s-d > 0$. Given $\varepsilon > 0$, there exist $x \in A$ and $y \in B$ such that $y-x < 2\varepsilon + d-s$. Choosing $\varepsilon =-\dfrac{d-s}{2}$, we have $y-x < -2 \cdot \dfrac{d-s}{2} + d-s = 0 \Rightarrow y < x$, which is a contradiction since $x \geq y$. Thus, $d \geq s$.