Problem: Let $A = \lbrace1,2,3,4,5,6\rbrace$ and $B = \lbrace2,4,6\rbrace$. The relation $R$ on $\mathcal{P}(A)$ defined by $\forall x,y \in \mathcal{P}(A), xRy \Longleftrightarrow x-B = y-B$
Prove that $R$ is an equivalence relation
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Does this imply that $X$ and $Y$ will essentially have to be the same set?
I'm mostly confused by how to visualize $xRy \Longleftrightarrow X-B = Y-B$. I read it as $x$ is only related to $y$ iff the set $X - B$ equals to $Y - B$
Your interpretation is correct. An example of to unequal, but $R$-equivalent sets $x, y$ in ${\cal P}(A)$ is: $$ x = \{1, 2, 3, 4\}, \quad y = \{1, 3, 4, 6\}. $$ For visualization, it is helpful to give the following definition of $R$, equivalent to the one above: $$ x \cap (B^c) = y \cap (B^c). $$ Now, $$ B^c = A - B = \{1, 3, 5\}. $$ So, two sets in ${\cal P}(A)$ are $R$-equivalent if and only if they contain the same odd numbers.