Prove: $A,B\in\mathscr P(\Bbb R)\setminus\emptyset ,a \lt b\ \forall a \in A,\forall b \in B\implies \sup A \lt \inf B$ This is a true or false question that I need to prove. I deduced that if $a \leq u$ being $ u = \sup A $ and $ v \leq b $  being $v = \inf B $ so $ a \leq u \leq v \leq b$. But I do not know if it is the right demostration or it needs more explanation.
 A: A=[0,1)  and B=[1,2]. Now you can deduce. @Anurag solved it.
A: 
But I do not know if it is the right demostration or it needs more explanation.

It needs a little more explanation.
For every $a \in A$ then $a < b\in B$ for all $b$ so every $a$ is lower bound of $B$ and $a \le \inf B \le b$ for all $a\in A, b\in B$.
And for every $b \in B$ then $b > a\in A$ for all $b$ so every $b$ is an upper bound of $A$ and $a \le \sup A \le b$ for all $a\in A, b\in B$.
So we have the three following possibilities:

*

*$a\le \inf B < \sup A \le b$ for all $a\in A; b\in B$.

*$a \le \sup A = \inf B \le b$ for all $a\in A; b\in B$

*$a \le \sup A < \inf B\le b$ for all $a \in A; b\in B$.

1 is impossible.  $\inf B< \sup A$ so $\inf B$ cant be an upper bound of $A$ so $a\le \inf B$ for all $B$ is not possible.  Likewise $\inf B < \sup A$ so $\sup A$ can't be a lower bound of $B$ so $\sup A \le b$ for all $b\in B$ is not possible.
3 is certainly possible.  We could just have $A= \{1\}$ and $B = \{2\}$ and $a \le \sup A = 1 < 2= \inf B = b$ for all $a\in A, b\in B$.
But is 2) possible or impossible?
Well if $u= \inf B= \sup A$.  We can show that $u$ can't be in both $A$ and $B$ as then you have $b=u \in B$ and $a=u \in A$ but $u=a \not < b=u$.  But there's no reason that $\inf B$ must be in $B$ or that $\sup A$ must be in $A$.
If we can think a bounded above set that might or might not contain its $\sup$, for example $A= (m,n)$ where $\sup A = n\not \in A$.  And a set that is bounded below where which might not contain it's $\inf$, for example $B = (j,k)$ where $\inf B=j\not \in B$.  But have the $\sup$ of one be the $\inf$ of the other, for instance $A = (m,n)$ and $B=(j,k)$ and $n=j$; that'd be just fine.
Example:  If $m < n < k$ and $A= (m,n)$ and $B=(n,k)$ then for all $a\in A; b\in B$ we have $m < a < n < b < k$ and that $\sup A = \inf B = n$.
So the statement $\sup A < \inf B$ is not true.
The true statement is $\sup A \le \sup B$.  The inequality is not strict.
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And interesting problem could be if $A$ and $B$ finite then $\sup A < \inf B$. Or if $A$ and $B$ are both closed then $\sup A < \inf B$.
