Let A and B be two non-empty bounded subsets of $\mathbb{R}$.
$\forall{a\in A,\, b\in B} \mid a\le b$
Prove that: $\sup A \le \inf B$
My solution goes as follows:
Suppose $\sup A \gt \inf B$:
$\forall {a\in A,b\in B}\,\,\exists{\varepsilon>0}\mid (a + \varepsilon \gt \sup A)\land (b - \varepsilon \lt \inf B) $.
Therefore, $\space a + \varepsilon \gt b - \varepsilon \space\space \rightarrow \space\space a + 2\varepsilon \gt b$
Why did I not get a contradiction?