Suppose $B \subseteq A$ and define a relation $R$ on $\mathscr P(A)$ as follows: $$ R = \{(X,Y) \in \mathscr P(A) \times \mathscr P(A) \mid X\Delta Y \subseteq B \}$$ Where $\Delta$ stands for symmetric difference.
Prove that $R$ is an equivalence relation on $\mathscr P(A)$
My attempt:
We need to show that $R$ is:
- Reflexive
- Symmetric
- Transitive
1.
Take arbitrary $X \in \mathscr P(A)$, because $X \Delta X = \emptyset$, we have $X \Delta X \subseteq B $, and thus $(X,X) \in R$, which means that set is reflexive.
2.
Take $(X,Y) \in R$. Since $X \Delta Y = Y \Delta X$, we have $(Y,X) \in R$. Hence set is symmetric.
3.
Take $(X,Y), (Y,Z) \in R$ Need to show that $X \Delta Z \subseteq B$. Take arbitrary $a \in X \Delta Z$. It implies that either
- $a \in X$ and $a \notin Z$
or
- $a \notin X$ and $a \in Z$
Consider first case.
Suppose $a \in Y$. Then $a \in Y \Delta Z$ and $a \in B$.
Suppose $a \notin Y$. Then $a \in X \Delta Y$ and hence $a \in B$.
Consider second case.
Suppose $a \in Y$. Then $a \in X \Delta Y$ and hence $a \in B$.
Suppose $a \notin Y$. Then $a \in Y \Delta Z$ and then $a \in B$.
Therefore, provided that $(X,Y), (Y,Z) \in R$, we have $X \Delta Z \subseteq B$ and $(X,Z) \in \mathscr P(A)$, Hence $R$ is transitive.
We've shown that $R$ is reflexive, symmetric and transitive, and thus we can conclude that $R$ is equivalence relation on $A$. $\Box$
Is it correct?
Your proof looks fine to me. I didn't catch any mistakes.
Let $R$ be a set. We say $a\sim b$ is an equivalence relation on $I$ iff the following properties are satisfied:
1) Reflexivity: $\forall a\in I, (a,a)\in R$
2) Symmetry: if $(a,b)\in R,$ then $(b,a)\in R$
3) Transitivity: if $(a,b),(b,c)\in R,$ then $(a,c)\in R.$
Your proof follows the definitions very closely, which is standard for these kinds of proofs.
However, I believe you could shorten your proof for $3.$ We know $X\Delta Y$ is the set of elements in exactly one of $X$ and $Z.$ The same goes for $X\Delta Y$ and $Y\Delta Z.$ If both $X\Delta Y$ and $Y\Delta Z$ are in $R,$ then exactly the elements that are contained in one of $X,Y,Z$ are contained in $R,$ so $X\Delta Z\in R\Rightarrow (X,Z)\in R.$