I tried Exercise K from chapter 2 of General Topology by Kelley. Is this enough to show isomorphism between Boolean rings? This is part (d).
Let $X$ be a set and let $\Bbb Z_2^x$ be the family of all functions on $X$ to $\Bbb Z_2$ , Define addition and multiplication of functions pointwise:
- $(f + g)(x) = f(x) + g(x)$, and
- $(f \cdot g) (x) = f(x) \cdot g(x)$.
Then $(\Bbb Z_2^x ,+,\cdot)$ is a Boolean ring with unit and is isomorphic to $(\mathcal a =\mathcal P(X),\Delta, \bigcap)$.
Assume $(\mathcal a =\mathcal P(X),\Delta, \bigcap)$ is Boolean ring (followed from previous exercise).
Proof. Observe that any function of $\Bbb Z_2^x$ has value of either $0$ or $1$.
So let $f,g\in \Bbb Z_2^x$, then for any $x\in X$ sum of the functions and multiplication is in $\{0,1\}$ for any $x\in X$:
- If $f(x)=0$ then $f(x)+f(x)=0+0=0,$
- if $f(x)=1$ then $f(x)+f(x)=1+1=2 \underbrace{=}_{\text{(mod 2)}}0$.
So in all cases, the addition satisfies definition of Boolean ring, which is $r+r=0 \quad \forall r\in R$.
Likewise with multiplication in Bool. ring: $r\cdot r=r$. If $f(x)=0$, then $0\cdot 0=0=f(x)$, and $1\cdot 1=1$ for $f(x)=1$.
Now to show isomorphism between $(\Bbb Z_2^x ,+,\cdot)$ and $(\mathcal a,\Delta, \bigcap)$ let
$$\varphi: \Bbb Z_2^x \to \mathcal a.$$
Then $\forall f\in \Bbb Z_2^x$:
$\begin{align} \bullet \varphi(f(x)+f(x))=\varphi(0)=\emptyset&=A\Delta A \\ &=\varphi(f(x)) \Delta \varphi(f(x)),\\ \bullet \varphi(f(x)\cdot f(x))= \varphi(f(x))=A&=A\bigcap A\\ &=\varphi(f(x))\bigcap \varphi(f(x)) \\ &= \varphi(f(x)\cdot f(x)). \square \end{align}$
That $\Bbb Z_2^X$ is actually a Boolean ring looks OK.
The second part of the proof consists of actually specifying the isomorphism and checking it. I don't see you define $\phi$ at all, just state the properties it should have.
So define $\phi: \Bbb Z_2^X \to \mathscr{P}(X)$ by $$\phi(f) = f^{-1}[\{1\}] =\{x \in X: f(x)=1\} \in \mathscr{P}(X) \text{ for } f \in \Bbb Z_2^X$$
One directly checks that $\phi(\overline{0}) = \emptyset$ and $\phi(\overline{1})=X$.
Then note that $x \in \phi(fg)$ iff $f\cdot g(x)=1$ iff $f(x)=1 \land g(x)=1$ iff $x \in \phi(f) \cap \phi(g)$
so that $\phi(fg)=\phi(f) \cap \phi(g)$ as sets.
Also $x \in \phi(f+g)$ iff $(f+g)(x)=1$ iff $f(x)+g(x)=1$ iff exactly one of $\{f(x), g(x)\}$ is $1$ iff $x$ is in exactly one of $\phi(f)$ or $\phi(g)$ iff $x \in \phi(f) \Delta \phi(g)$
so that $\phi(f+g)=\phi(f) \Delta \phi(g)$ as sets as well.
This shows that $\phi$ preserves ring operations, it finally remains to show that $\phi$ is a bijection between $\Bbb Z_2^X$ and $\mathscr{P}(X)$, do you see why this holds?