$ P( A \mathrel{\triangle} B ) \le P(A \mathrel{\triangle} C) + P(B\mathrel{\triangle} C)$

Question

Show that $$ P( A \mathrel{\triangle} B ) \le P(A \mathrel{\triangle} C) + P(B\mathrel{\triangle}C)$$ where $\mathrel{\triangle}$ indicates the symmetric difference

I cannot write my idea, because it's trivial. Please how to show this inequality. Thank you for helping.

Answer 1

Check that $A \mathrel{\triangle}B \subset (A \mathrel{\triangle}C) \cup (B \mathrel{\triangle}C)$: if $x \in A \mathrel{\triangle} B$, then $x \in A \setminus B$ or $x \in B \setminus A$. Suppose WLOG that $x \in A \setminus B$. If $x \in C$, then $x \in C \setminus B$ and hence $x \in B \mathrel{\triangle}C$ and we're done. If $x \not\in C$, then $x \in A \setminus C$ and hence $x \in A \mathrel{\triangle}C$ and we're done in the same way. Concluding:

$$A \mathrel{\triangle}B \subset (A \mathrel{\triangle}C) \cup (B \mathrel{\triangle}C) \implies P(A \mathrel{\triangle}B) \leq P(A \mathrel{\triangle}C)+P(B \mathrel{\triangle}C).$$

Answer 2

Hint

$$A \mathrel{\triangle}B= (A \mathrel{\triangle} C) \mathrel{\triangle} (C \mathrel{\triangle}B) \subset (A \mathrel{\triangle}C) \cup (C \mathrel{\triangle}B)$$

Answer 3

Let $x\in A$ and $x \not\in B$. If $x\in C$, $x\in B\mathrel{\triangle} C$. If $x\not\in C$, then, since $x\in A$, $x\in A\mathrel{\triangle} C$.

Can you do the rest?