How to prove $a \lt b$ for all $a \in A, b \in B$ implies $\sup A \leq \inf B$? If $A, B$ are two non empty sets of real numbers and for every $a$ from $A$ and $b$ from $B$

$a < b$,

how can I prove that 

$\sup A \leq \inf B$?

 A: I don't think this is true. Let $A=(0,1)$ and $B=(1,2)$ then $\operatorname{sup}A=1$ and $\operatorname{inf}B=1$ but $A$ and $B$ satisfy your conditions.
A: With the edited problem: if $a\lt b$ for all $a\in A$ and $b\in B$, why is it that $\sup(A)\leq\inf(B)$? Is every element of $A$ a lower bound for $B$? If so, what does that tell you about the elements of $A$ relative to $\inf(B)$? And given that, what does that tell you about $\sup(A)$?
A: Here's the solution:
Let us suppose that $\sup(A) > \inf(B)$. So, there exists some $x∈A$, such that $\inf(B) < x$. Again, as $\inf(B)< x$, there exists $y∈B$, such that $y<x$. which is a contradiction as we have been given that $A<B$ (i.e, $x<y$ for all $x∈A$, $y∈B$). So, our assuption that $\sup(A) > \inf(B)$ is WRONG. So, $\sup(A) \leq \inf(B)$. 
Hence Proved...... :)
A: Consider $b\in B$, by hypothesis is a upper bound of $A$, so by definition of sup (the least upper bound), $\sup A\leq b,$ $ \forall b\in B$. Now the $\sup A$ is a lower bound of $B$, so by defnition of inf(the grater lower bound), $\sup A\leq \inf B$ .
