I would like to know if my procedure was correct in proving the next property ($\oplus \equiv$ symmetric difference):

$$(A_1\cup A_2)\oplus (B_1\cup B_2)\subset (A_1\oplus B_1)\cup (A_2\oplus B_2)$$

$$Let\ x\in(A_1\cup A_2)\oplus (B_1\cup B_2)$$

$$i.e.\ x\in(A_1\cup A_2)\setminus (B_1\cup B_2)\ \lor\ x\in(B_1\cup B_2)\setminus(A_1\cup A_2)$$

$$i.e.\ [x\in(A_1\cup A_2)\ \land\ x\notin(B_1\cup B_2)]\ \lor\ [x\in(B_1\cup B_2)\ \land\ x\notin(A_1\cup A_2)]$$

$$i.e.\ [(x\in A_1\lor x\in A_2)\land(x\notin B_1\land x\notin B_2)]\ \lor\ [(x\in B_1\lor x\in B_2)\land(x\notin A_1\land x\notin A_2)]$$

$$i.e.\ (x\in A_1\land x\notin B_1\land x\notin B_2)\lor(x\in A_2\land x\notin B_1\land x\notin B_2)\ \lor\ (x\in B_1\land x\notin A_1\land x\notin A_2)\lor (x\in B_2\land x\notin A_1\land x\notin A_2)$$

so, particularly, we have

$$\Rightarrow (x\in A_1\land x\notin B_1)\lor(x\in A_2\land x\notin B_2)\lor(x\in B_1\land x\notin A_1)\lor(x\in B_2\land x\notin A_2)$$

and, since $\lor$ is asociative, we have

$$\Rightarrow ((x\in A_1\land x\notin B_1)\lor(x\in B_1\land x\notin A_1))\lor((x\in A_2\land x\notin B_2)\lor(x\in B_2\land x\notin A_2))$$

$$i.e.\ (x\in A_1\setminus B_1\ \lor\ x\in B_1\setminus A_1)\lor(x\in A_2\setminus B_2\ \lor\ x\in B_2\setminus A_2)$$

$$i.e.\ (x\in A_1\oplus B_1)\lor(x\in A_2\oplus B_2)$$ $$i.e.\ x\in (A_1\oplus B_1)\cup (A_2\oplus B_2)$$

therefore $(A_1\cup A_2)\oplus (B_1\cup B_2)\subset (A_1\oplus B_1)\cup (A_2\oplus B_2)$.

There's also another way using this result: $\ x\in A\oplus B\ iff\ 1_A (x)\neq 1_B (x)\ $but don'y know how to aprroach with this one. Any ideas?


1 Answer 1


Yes, it seems correct.

  1. It can also be done without referring to elements, using only the set operations and their properties. $\def\1{{\bf1}}$
  2. $x\in (A_1\cup A_2)\oplus(B_1\cup B_2)\ \iff\ \1_{A_1\cup A_2}(x)\ne \1_{B_1\cup B_2}(x)$.
    Now use that $\1_{A_1\cup A_2}(x)=\max(\1_{A_1}(x),\,\1_{A_2}(x))$ to conclude that at least one of the inequalities $\1_{A_i}(x)\ne \1_{B_i}(x)$ hold ($i=1,2$), hence $x\in (A_1\oplus B_1)\cup (A_2\oplus B_2)$.

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