A is a subset of B, both sets are open.
Is the set B/A open?
My answer would be no, because the complement of an open set is always closed. Is this the correct definition in this case?
You must understand that $\displaystyle B\setminus A=B\cap A^c$ and is known as a relative complement.
That is the complement of A relative to B.
You see $\displaystyle A$ is an open set in the underlying space which may not be $\displaystyle B$.
In the real numbers $\displaystyle ((0,2)\setminus (0,1)=[1,2)$ which is not open.
Yes, that's true. But "B\A" is NOT the "complement" of A, which is, by definition, the set all points not in A, whether they are in B or not. If $\displaystyle B= (0, 1)\cup (2, 3)$ and $\displaystyle A= (0, 1)$, then $\displaystyle B\A= (2, 3)$ an open set. The [b]complement of A is $\displaystyle (-\infty, 0]\cup [1, \infty$, a closed set.