Boolean algebra and closure axiom

Question

A Boolean algebra is an algebraic system (B,$∨$,$∧$,$¬$), where $∨$ and $∧$ are binary, and $¬$ is a unary operation.

One of the Boolean algebra axiom is: If $a$ and $b$ are elements of $B$, then $(a ∨ b)$ and $(a ∧ b)$ are in $B$.

i.e. the set $B = (1111,0011,0110,1010,0000,1100,1001,0101)$ I can't use as carrier for Boolean algebra, because the result of operation $0011 ∨ 0110 = 0111$ is't in set $B$.

Is it correct? Do I correctly think about closure for $∨$ and $∧$ operators?

@RobertIsrael may be you can send me link to book which explains how can I prove, that some set with operations, defined on it, is Boolean algebra? — V. Gai, Dec 20 '18 at 12:58

score 1 · Accepted Answer · answered May 05 '20 at 16:37

The problem is how to define the $B$. The $B$ is not a concrete collection that has some concrete elements. For help get around this, you should think the $B$ is a collection that be formalizing defined, such as $\mathbb{N}$, $\mathbb{R}$, etc. You can say $x \in \mathbb{B}$ but you cannot list all the elements of $\mathbb{B}$.

So the closure axiom can be denote: For all $a$, $b$ if $a, b \in$ $\mathbb{B}$, then $(a \wedge b )\in \mathbb{B}$

Boolean algebra and closure axiom

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