On the proof that if $a \leq b$ for all $a \in A$ and $b \in B$ then $\sup(A)\leq\inf(B)$ 
Suppose $A$ and $B$ are subsets of $\mathbb{R}$, both nonempty, with the speical property that $a \leq b$ for all $a \in A$ and for all $b \in B$. Prove: sup$(A)$ $\leq$ inf$(B)$. 

I have an objection to this problem. What if $A$ is a monotone increasing sequence that is not bounded above and $B$ is also monotone increasing sequence that is not bounded above but bounded below? Then sup$(A)$ $= \infty$, inf$(B)=M$ for some $M \in \mathbb{R}$. Hence, the inequality won't make sense. Do I have to assume that these subsets are bounded? 
 A: if $A$ is not bounded above then the sets $A$ and $B$ cannot fulfill the required conditions, because if we pick $b\in b$ then this $b$ is not an upper bound for $A$, and hence there is $a\in A$ with $b<a$.
Another way to verify that your counterexample does not work is to prove the theorem.
A: If $A$ is not bounded above, you cannot have $a\le b$ for all $a\in  A$ ($b$ fixed).
A: 
Suppose $A$ and $B$ are subsets of $\mathbb{R}$, both nonempty, with the speical property that $a \leq b$ for all $a \in A$ and for all $b \in B$. Prove: sup$(A)$ $\leq$ inf$(B)$. 

Steps for the proof


*

*Since $B\neq\emptyset$, let $b$ be some element of $B$, show that
$$
\sup(A)\leq b\tag{1}.
$$ 

*Show that (1) is true for any $b\in B$. It follows that $\sup(A)$ is a lower bound for $B$.

*Finish the proof. 




I have an objection to this problem. What if $A$ is a monotone increasing sequence that is not bounded above and $B$ is also monotone increasing sequence that is not bounded above but bounded below? Then sup$(A)$ $= \infty$, inf$(B)=M$ for some $M \in \mathbb{R}$. Hence, the inequality won't make sense. Do I have to assume that these subsets are bounded? 

Exercise:
Show that by the assumptions of the problem, $A$ must be bounded above and $B$ must be bounded from below. 
HINT:
Since for any $a\in A$ and any $b\in B$, $a\leq b$, it is in particular true that for any $a\in A$ and some $b\in B$, $a\leq b$.
A: If sup A > inf B then a>inf B for some a, and thus a>b for some b (for if it wasn't then inf B would not be the infimum of B)
