Let $X = \{ 1, 2, 3, \ldots , 2015\}$ and $Y = \{ 1, 2, 3,\ldots, 271 \}$. Let S be the relation on power set (X) defined by
For all A, B in $\mathcal{P} (X)$, $(A, B) \in S$ if and only if $|A \cap Y | = |B \cap Y |$.
(a) Prove that S is an equivalence relation on $\mathcal{P} (X)$.
$a \sim a$ so that $|A \cap Y| = |A \cap Y|$, thus it's reflexive $a \sim b$ and $b \sim a$ , because $A,B$ are subsets of $X$, $[a] \cap [b]$ = empty set, or $[a]=[b]$, $|A \cap Y| = |B \cap Y|$ and vise versa, thus, it's symmetric. I am not sure I have done it right but will try to do transitive on my own if I am at least going into the right direction. Let me know, because I am not sure at all, if this is right..
You're correct on reflexivity, like you say $A \in \mathcal{P}(X)$. Then $A \sim A$ because $|A \cap Y| = |A \cap Y|$.
You're also correct on symmetry. If we have $A,B \in \mathcal{P}(X)$ and $(A,B) \in S$ then $|A \cap Y| = |B \cap Y| \implies |B \cap Y| = |A \cap Y| \implies (B,A) \in S $
The last property for a equivalence relation is transivity. That is if $(A,B) \in S$ and $(B,C) \in S$ then $(A,C) \in S$. Assume $(A,B) \in S$ and $(B,C) \in S$ then we have $(1)$ $|A \cap Y| = |B \cap Y|$ and $(2)$ $|B \cap Y| = |C \cap Y|$. Substituting the left hand side of $(2)$ into the right hand side of $(1)$ we get $|A \cap Y| = |C \cap Y| \implies (A,C) \in S$
To conclude, $S$ defines an equivalence relation on $\mathcal{P}(X)$