If $A\dot{-} B$ is countable and $B \dot{-} C$ is countable then $A\dot{-} C$ is countable?

Question

Prove that: If $A\dot{-} B$ is countable and $B\dot{-} C$ is countable then $A \dot{-} C$ is countable?

If not give a counter-argument

Answer 1

Symmetric difference is associative. Therefore, $(A\mathop{\triangle}B)\mathop{\triangle} C=A\mathop{\triangle}(B\mathop{\triangle}C)$.

Moreover, $B\mathop{\triangle}B=\varnothing$, and $B\mathop{\triangle}\varnothing=B$, for any set $B$. Combine these and we have that:

$$A\mathop{\triangle}C=(A\mathop{\triangle}\varnothing)\mathop{\triangle}C=A\mathop{\triangle}(B\mathop{\triangle}B)\mathop{\triangle}C=(A\mathop{\triangle} B)\mathop{\triangle}(B\mathop{\triangle}C)\subseteq(A\mathop{\triangle}B)\cup(B\mathop{\triangle}C)$$

Therefore $A\mathop{\triangle}C$ is a subset of the union of two countable sets, and thus countable.

Answer 2

It is countable.

$A \bigtriangleup C = ( A \setminus C ) \cup ( C \setminus A )$.

Then $A \setminus C = (A \setminus C \cap B) \cup ( A \setminus C \cap B^c)$. We see that $A \setminus C \cap B \subset B \setminus C \subset B \bigtriangleup C $ and $A \setminus C \cap B^c \subset A \cap B^c \subset A \bigtriangleup B $. Hence $A \setminus C \cap B$ is countable.

The same analysis, mutatis mutandis, applies to $C \setminus A $.