Algebra - Prove this is a Ring

Question

Let $S$ be a nonempty set. Prove that the power set $P(S)$ whose elements are all subsets of $S$, forms ring under the following operations:

\begin{align*} a + b &= (a \cup b) \setminus (a \cap b)\\ a \cdot b &= a \cap b \end{align*}

For proving it's an abelian group first, I've gotten that the identity is $\emptyset$.

\begin{align*} a + \emptyset &= (a \cup \emptyset) \setminus (a \cap \emptyset)\\ &= a \setminus \emptyset\\ &= a \end{align*}

Each element is it's own inverse.

\begin{align*} a + a &= (a \cup a) \setminus (a \cap a)\\ &= a \setminus a\\ &= \emptyset \end{align*}

It's certainly commutative. \begin{align*} a + b &= (a \cup b) \setminus (a \cap b)\\ &= (b \cup a) \setminus (b \cap a)\\ &= b + a \end{align*}

But I can't get associative. That is, $(a + b) + c = a + (b + c)$. Anyone want to aid?

The properties for multiplication aren't too bad either. The only one that is throwingme a curve ball is the left/right distribution. That is,

$a(b + c) = ab + ac$ and $(a + b)c = ac + bc$.

While this questions is asked elsewhere on this site and has answers to most parts in proving that it is a ring, it does not explain or show the associativity portion for addition in detail. That is $(a + b) + c = a + (b + c)$ is left very vague.

Answer 1

It's much easier if you compute the truth table for $+$: $$ \begin{array}{cc|c} a & b & a+b \\ \hline T & T & F \\ T & F & T \\ F & T & T \\ F & F & F \end{array} $$ Then the truth table for $(a+b)+c$ is $$ \begin{array}{ccc|c} a & b & c & a+b & (a+b)+c \\ \hline T & T & T & F & T \\ T & T & F & F & F \\ T & F & T & T & F \\ T & F & F & T & T \\ F & T & T & T & F \\ F & T & F & T & T \\ F & F & T & F & T \\ F & F & F & F & F \end{array} $$ You can clearly see that a return of $T$ corresponds to an odd number of $T$’s in the data and $F$ to an even number of $T$. This is independent on the order $a$, $b$ and $c$ are considered in. Hence the truth table for $(b+c)+a$ is the same, which means $$ (a+b)+c=(b+c)+a=a+(b+c) $$ using the obvious commutativity of $+$.

You can do similarly for the distributive property: $$ \begin{array}{ccc|c} a & b & c & a+b & (a+b)c \\ \hline T & T & T & F & F \\ T & T & F & F & F \\ T & F & T & T & T \\ T & F & F & T & F \\ F & T & T & T & T \\ F & T & F & T & F \\ F & F & T & F & F \\ F & F & F & F & F \end{array} $$ and computing the one for $ac+bc$.

A different strategy is to consider indicator functions. The indicator function for $a$ is the map $\chi_a\colon S\to\{0,1\}$ defined by $\chi_a(x)=1$ if $x\in a$, $\chi_a(x)=0$ if $x\notin a$. Two subsets of $S$ are equal if and only if they have the same indicator function.

If we use modulo $2$ arithmetic, the indicator function of $a+b$ is easily seen to be $\chi_{a}+\chi_b$ (pointwise sum). Thus the indicator function for $(a+b)+c$ is $\chi_a+\chi_b+\chi_c$.

For the distributive property, observe that the indicator function of $ab$ is $\chi_a\chi_b$ (pointwise product).

Answer 2

Hint: $a+b=\ldots =a\backslash b\cup b\backslash a$

Second hint: $x\backslash y=x\cap y^c$

Now write out both $(a+b)+c$ and $a+(b+c)$ independently in terms of $\cap$, $\cup$ and compare.

Answer 3

Here is an answer using Damian's hints, if you want to go that route for some reason. We have: \begin{align}a + b &= (a \cup b)\setminus(a\cap b) \\ &= (a\cup b) \cap (a\cap b)^c \\ &= (a\cup b) \cap (a^c \cup b^c) \\ &= (a\cap b^c) \cup (b\cap a^c) \\ &= (a\setminus b) \cup (b\setminus a)\end{align} So, in excruciating detail, we see that: \begin{align} (a+b)+c &= ((a\setminus b) \cup (b\setminus a)) + c \\ &= [((a\setminus b)\cup (b\setminus a))\setminus c] \cup [c \setminus ((a\setminus b)\cup (b\setminus a))] \\ &= [((a\cap b^c)\cup (b\cap a^c))\cap c^c] \cup [c\cap ((a\cap b^c) \cup (b\cap a^c))^c] \\ &= [(a\cap b^c \cap c^c) \cup (b\cap a^c \cap c^c)]\cup [c\cap (a^c \cup b)\cap (b^c \cup a)] \\ &=(a\cap b^c \cap c^c) \cup (b\cap a^c\cap c^c) \cup (c\cap a^c \cap b^c) \cup (c\cap b \cap a) \\ &= (c\cap a^c \cap b^c) \cup (b\cap a^c \cap c^c) \cup (a \cap b^c \cap c^c) \cup (a\cap b \cap c)\\ &= [(c\cap a^c \cap b^c) \cup (b\cap a^c \cap c^c)]\cup [a \cap (c^c \cup b)\cap (b^c \cup c)] \\ &= [((c\cap b^c)\cup (b\cap c^c))\cap a^c] \cup [a\cap ((c^c\cap b)\cup (b^c\cap c))^c] \\ &= [((c\setminus b) \cup (b\setminus c))\setminus a]\cup [a\setminus ((c\setminus b) \cup (b\setminus c))] \\ &= a + ((c\setminus b) \cup (b\setminus c)) \\ &= a + (b + c) \end{align}

For distribution property, we have, in less detail \begin{align} a(b + c) &= a\cap ((b\setminus c) \cup (c\setminus b))\\ &= ((a\cap b)\setminus c) \cup ((a\cap c)\setminus b) \\ &= ((a\cap b)\setminus (a\cap c))\cup ((a\cap c)\setminus (a\cap b)) \\ &= (a\cap b) + (a\cap c) \\ &= ab + ac \end{align} and similarly for the other way around.