symmetric difference proof

Question

I have just started to learn set theory and logic, and I find it difficult how to make proofs , could you show me the method to prove :

1. Given the sets A,B,C show that:

$A\triangle B = A\triangle C$

if and only if

$B=C$

2. Given the equation $A \cap X=B$ and $A,B \in P(\mathbb{N})$ A,B are sets

for what Values the euation will have only one solution

thanks

Answer 1

For Q2, for subsets $A,B$ of $\Bbb N$, the existence of a unique $X\subset \Bbb N$ such that $A\cap X=B$ implies (i) at least one such $X$ exists, and (ii) there cannot be more than one such $X$.

Now (i) implies $B\subset A,$ so (i) and (ii) together imply $A=\Bbb N$. Because if $B\subset A\subset \Bbb N$ and $n \in \Bbb N\setminus A$ then $X_1=B$ and $X_2=B\cup \{n\}$ are two different solutions for $X$.

So for $X$ to be exist and be unique it is necessary that $A=\Bbb N.$

And $A=\Bbb N$ is also sufficient. Because if $A=\Bbb N$ then $X=B$ is a solution, and for any $X\subset \Bbb N$ such that $X\ne B$ we have $A\cap X=\Bbb N\cap X=X\ne B.$

Answer 2

Hint (for 1.):  the symmetric difference is associative (e.g. here) and $\,A \Delta A = \emptyset, A \Delta \emptyset = A\,$, so:

$$ A \Delta B = A \Delta C \implies A \Delta\big(A \Delta B\big) = A \Delta\big(A \Delta C\big) \implies \big(A \Delta A\big) \Delta B=\big(A \Delta A\big) \Delta C \implies B=C $$

Answer 3

Here is a solution using indicator functions. The proof uses the fact that the map $ A\to I(A)$ is a bjection from $P(\Omega)$ to $2^{\Omega}$. First observe that $I(A\Delta C)=|I(A)-I(C)|=(I(A)+I(C)) \mod2$. Now $$ \begin{align} A\Delta B=A\Delta C \iff& (I(A)+I(B))\mod2= (I(A)+I(C))\mod {2}\\ \iff& I(A)+I(B)\equiv I(A)+I(C) \pmod{2}\\ \iff& I(B)\equiv I(C) \pmod{2}\\ \iff& B=C. \end{align} $$