elementary set theory - Let $ADelta Csubseteq ADelta B$. Prove $Acap B subseteq C$. (Proof.v)

Let $A\Delta C\subseteq A\Delta B$. ($\Delta$ denotes symmetric difference.)

Prove $A\cap B \subseteq C$.

I am getting ready for a test and I could really use proof verification and any help with this.

Proof: Let us look at the indicators, $x_{A\Delta C}=x_A+x_C-2x_Ax_C$, $x_{A\Delta B}=x_A+x_B-2x_Ax_B$, $x_{A\cap B}=x_Ax_B$.

Let $x_{A\cap B}(a)=1$. Then $x_{A\Delta B}(a)=0$ which means $x_{A\Delta C}(a)=0$. $x_A(a)=x_B(a)=1$ and therefore $x_C(a)$ must be 1. Therefore $x_{A\cap B}(a)=1\Rightarrow x_C(a)=1$ $\Rightarrow A\cap B \subseteq C$.

Answer

Your argument is flawless.

Depending on how much you did with indicators during your classes, you might want to elaborate on some of the steps, like:

• $x_{A \Delta B}(a)=0 \to x_{A \Delta C}(a) = 0$


• $x_{A \Delta C}(a) =0, x_A(a) = 1 \to x_C(a) = 1$