$\sup A < \inf B$ implies existence of $c \in \mathbb{R}$ such that $a < c < b$ Related: What are necessary and sufficient conditions such that sup A < inf B implies there is a c such that a<c<b for all a,b in their respecitve sets
Working through some revision Analysis questions and got stumped on this. Asked to determine whether the following statement is true or not (and if not, a counterexample);
If $\sup A < \inf B$ for sets $A, B$ in $\mathbb{R}$, then there exists $c \in \mathbb{R}$ such that $ a < c < b$ for all $a \in A$, and $b \in B$.
An answer to the above question linked indicates this is true, but I haven't gotten any further to proving it other than noting that $\forall \epsilon > 0$, there are $a$ and $b$ such that $\sup A < a + \epsilon$ and $b-\epsilon < \inf B$. Hints?
 A: This is clearly wrong. Take $A=[0,1]$ and $B=[0,2]$. As $1 \in A > 0 \in B$, there can not exist such a $c$.
A: After the typo correction by OP, the question is trivial.
If $A, B$ are not empty sets of real numbers such that $\sup A < \inf B$ then there exists a real number $c$ with the property that $a < c < b$ for all $a\in A, b\in B$.
Why is this trivial? Simply because we can take any number $c$ such that $\sup A < c <\inf B$ (such a $c$ exists because given any two distinct real numbers there lie an infinity of real numbers between them). And then for all $a \in A, b\in B$ we have $$a\leq \sup A < c < \inf B \leq b$$ by definition of $\inf, \sup$.
Note that if $\sup A \nless \inf B$ then such a $c$ can't be found and it is easy to construct example sets $A, B$ showing the non-existence of such a $c$.

Also understand that most results and theorems in (introductory) analysis use the density of real numbers and the results of these kind remain valid in rational number system also and such results can be considered trivial (i.e based on direct application of definitions). Most of the problems involving $\epsilon-\delta$ gymnastics also fall under this category and are made difficult only by presence of greek symbols. The really significant results of analysis deal with the completeness property of real numbers and they can not be solved using common sense or algebraic techniques. For such problems you need to have an understanding of order relations and completeness of real numbers.
A: it is not true:
A={1, 5, 10}
B={1, 3, 15}. 
supA=10
supB=15. So far all conditions are met.
Now choose a=1 ;b=1. And the staetement is not possible.
