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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}

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?

Andrés E. Caicedo
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Harald Meisner
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    What makes you think that $B$ can have only two elements? – bof Nov 02 '16 at 18:22
  • If i didn't misunterstood the german wikipedia article of a boolean algebra, it say's that you can get the definition of peano from the huntington definition and the peano definition is about a set with 2 elements. – Harald Meisner Nov 02 '16 at 18:24
  • Your axioms are all in the form of identities. Any set of identities which has a nontrivial model (more than oine element) has infinite models. – bof Nov 02 '16 at 18:25
  • The Wikipedia article (in German) is a bit misleading. It does say that the set has two elements. However, this is meant to say that it has at least two distinct elements. – Stefan Mesken Nov 02 '16 at 18:27
  • So where exactly does the thing with "exactly two elements" come from? – Harald Meisner Nov 02 '16 at 18:29
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    I'm not sure what you mean. There is a (unique) Boolean algebra with exactly $2$ elements - which is labeled as the 'most important' Boolean algebra within the Wikipedia article (a statement that I find hard to justify - labeling it as the 'trivial' Boolean algebra would be more appropriate). There are however many other Boolean algebras and they play an important role in both model theory and set theory. – Stefan Mesken Nov 02 '16 at 18:46
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    So the boolean algebra with 2 elements is nothing more than a special of the far more general concept of boolean algebras? – Harald Meisner Nov 02 '16 at 18:51
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    Yes, that's right. – Stefan Mesken Nov 02 '16 at 19:54

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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.

Stefan Mesken
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  • @HaraldMeisner You're most welcome (: – Stefan Mesken Nov 02 '16 at 18:41
  • Good, but there is no need to exclude the empty set case. The powerset of the empty set is still a Boolean algebra; in this case, it has only one element—$\varnothing$. Note also that it is still a power of $2$, as it is the case for the cardinality of any finite Boolean algebra. – amrsa Nov 02 '16 at 20:24
  • @amrsa Sure, note however that (just like with rings) authors sometimes demand explicitly that $1 \neq 0$. By excluding the case $X = \emptyset$, I wanted to avoid confusions/concerns about this convention. – Stefan Mesken Nov 02 '16 at 20:29
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    That's right. I suppose it's mostly with fields that they demand $0 \neq 1$, but maybe some also with rings. However, if we make such demand, we no longer get an equational class (the same with rings). And in the mentioned wikipedia article (at least in english) they don't make that demand and explicitly include the powerset of the empty set as a Boolean algebra. – amrsa Nov 02 '16 at 20:49