Get Peanos definition of a boolean algebra from the definition of Huntington.

Question

I have the following definition:

A boolean algebra is a set $B$ with two operations on $B$, so that for all elements $a \in B$, $b \in B$ and $c \in B$ holds:

Commutativity: \begin{equation}\label{1.1.1} a \land b = b \land a \end{equation} \begin{equation}\label{1.1.2} a \lor b = b \lor a \end{equation} Distributivity: \begin{equation}\label{1.1.3} a \land \left(b \lor c \right) = \left(a \land b\right) \lor \left(a \land c \right) \end{equation} \begin{equation}\label{1.1.4} a \lor \left(b \land c \right) = \left(a \lor b\right) \land \left(a \lor c \right) \end{equation} Existence of neutral elements: There are elements $0 \in B$ and $1 \in B$, so that: \begin{equation}\label{1.1.5} a \land 1 = a \end{equation} \begin{equation}\label{1.1.6} a \lor 0 = a \end{equation} Existence of complements: For every $a\in B$ there is $\neg a\in B$, so that: \begin{equation}\label{1.1.7} a \land \neg a = 0 \end{equation} \begin{equation}\label{1.1.8} a \lor \neg a = 1 \end{equation}

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I want to prove, that set $B$ only can have two elements - namely $0$ and $1$.

I was able to prove that there is one and only one neutral $0$ and only one neutral $1$. I was also able to prove that there is only one and only one complement for every $a \in B$.

I thought that I could try a proof by contradiction from here with the assumption that there is a $x \in B$ with $x \neq 0$ and $x \neq 1$ but I don't know how to get the contradiction.

Could one give me a hint?

Answer 1

You'd better not prove that a Boolean algebra can only have two elements. Consider any nonempty element $X$ and let $\mathcal P(X)$ be its powerset. Now let, for $a,b \in \mathcal P(X)$

• $0 := \emptyset$,
• $1 := X$,
• $-a := X \setminus a$
• $a \wedge b := a \cap b$ and
• $a \vee b := a \cup b$.

Then $(\mathcal P(X); \vee, \wedge, -, 0,1)$ satisfies the above requirements on a Boolean algebra - as you may easily verify - and has $2^{\operatorname{card}(X)}$ elements.