I wanted to ask about this problem.
My book states:
Let the symmetric difference be defined as $A + B=(A\setminus B) \cup (B \setminus A)$.
It proceeds to define a power set as
$P_D= \left\{ A:A \subseteq D\right\}$
It then asks me to show:
- Prove that there is an identity element with respect to the operation $+$, which is ______.
- Prove every subset $A$ of $D$ has an inverse with respect to $+$, which is ______. Thus, $\left\langle P_D,+\right\rangle$ is a group!
Here's what I got:
-
Suppose $E$ is the identity set. We want to find $E$ such that $A + E=A$ (we don't need to check $E+A=A$ because $\cup$ is commutative–it will work). Hence $E=\varnothing$ because $(A\setminus \varnothing) \cup (\varnothing \setminus A) = A \cup \varnothing = A$.
-
Let $I$ denote the inverse set that satisfies $A + I = E$. The solution to $A+I=\varnothing$ is $I=A$. We can verify this: $A+A = ( A\setminus A)\cup(A\setminus A)=\varnothing$.
Obviously, in my answer I did not specifically prove that every subset $A$ of $D$ has an inverse. Since I do not know how to prove this, I cannot prove that "$\left\langle P_D,+\right\rangle$ is a group", as they asked.
How should I fix my proof?
Thank you.
Best Answer
Since you already solve the problem, this answer should play as a comment but I put it here for proper format.
$P_D$ can be identified with the set of functions $$f:D\to \Bbb Z/2,$$ from $D$ to the integers mod $2$, where each subset $A$ is identified to its characteristic function $1_A$ such that $$1_A(a)=1\iff a\in A.$$ Then the operation $A+B$ translates to $1_A+1_B$ in normal sense.
It follows easily that