Problem: For two subsets $A$ and $B$ of some set $X$, we define \begin{align*} A \triangle B = (A \cup B) \setminus (A \cap B). \end{align*} We now define a relation $R$ on $P(X)$ (power set of $X$) by \begin{align*} R = \left\{(A,B) \in P(X) \times P(X) \mid A \triangle B \ \text{is countable}\right\}. \end{align*}
a) Prove $R$ is an equivalence relation.
b) Suppose $X$ is uncountable, and that $x_0 \in X$. Is the equivalence class $[\left\{x_0\right\}]_{R}$ of $\left\{x_0\right\} \in P(X)$ then a finite, countably infinite or uncountable set?
Solution:
For a), I know we have to prove reflexivity, symmetry and transivity. But I'm confused with the notation here. For reflexivity, do I take just one set $A \in P(X)$ or $A \in R$, or two sets? And I don't know what to do about the condition: $A \triangle B$ is countable.
(reflective) For each $A\in\wp\left(X\right)$ we have: $A\Delta A=\emptyset$ and $\emptyset$ is by definition a countable set. In the notation: $(A,A)\in R$ for each $A\in\wp(X)$.
(symmetric) Follows directly from $A\Delta B=B\Delta A$. If $A\Delta B$ is countable then so is $B\Delta A$. In the notation: $(A,B)\in R\Rightarrow (B,A)\in R$
(transitive) $A\Delta C\subseteq\left(A\Delta B\right)\cup\left(B\Delta C\right)$ so countablity of $A\Delta B$ and $B\Delta C$ implies countability of $A\Delta C$. In the notation: $(A,B)\in R\wedge(B,C)\in R\Rightarrow (A,C)\in R$.
$\left[\left\{ x_{0}\right\} \right]=\left\{ A\in\wp(X)\mid A\Delta\left\{ x_{0}\right\} \text{ is countable}\right\} $. If $x_{0}\in A$ then $A\Delta\left\{ x_{0}\right\} =A-\left\{ x_{0}\right\} $ and if $x_{0}\notin A$ then $A\Delta\left\{ x_{0}\right\} =A\cup\left\{ x_{0}\right\} $ so we find that $A\Delta\left\{ x_{0}\right\} $ is countable iff $A$ is countable. So $\left[\left\{ x_{0}\right\} \right]=\left\{ A\in\wp(X)\mid A\text{ is countable}\right\} $.
If $X$ is uncountable then it has uncountable subsets that are countable.