Let $A = \{1,2,3,4,5,6\}$ and let $B = \{1,2,3\}$
Let $R$ be a relation such that $R=\{(x,y) \in P(A) \times P(A): x \cap B = y\cap B\}$
How many equivalence classes are possible?
I'm kinda stuck at this problem... can't seem to visualize this or find the correct path to calculate this.
Some help? Thanks.
$x\cap B$ is a subset of $B$ and $B$ has $8$ subsets. For each of these subsets there is one equivalence class. So the required number is $8$.