Why is this a boolean algebra

Question

Let $A = \{a,b\}$. The $\mathcal P(A) = \{\emptyset,\{a\},\{b\},A\}$.
Let $+$ be $\cup$, $\cdot$ be $\cap$, complement be set complement, $1$ be $A$, and $0$ be $\emptyset$.

I need to explain why this description is a Boolean algebra.

I am thinking that I can use the identity axiom and say there there exists two elements, $0$ and $1$, in $A$ such that for every $a\in A$

1. $a + 0 = a$
   $a \cup \emptyset = a$ - using substitution

2. $a \cdot 1 = a $
   $a \cap 1 = a $ - using substitution
   $a \cap A = a $ - since $1 = A$
   

The same argument could be used $\{b\}$.

I am not sure if I am doing this correctly or not. Any help is appreciated.

Thanks

Answer 1

You're on the right track! In order to show that a set and some operations are a boolean algebra you have to show that they satisfy all of the laws for a Boolean Algebra(or at least the ones in your text book).

You have shown the identity law applies for $\cup$ and $\cap$. Now you just have to show the rest.

Also, checkout this for more about the boolean algebra of power sets with $\cup$, $\cap$, and compliment.