Cardinality of a subalgebra of a Boolean algebra. I have been trying to prove the following.
Suppose $B$ is a finite non-trivial Boolean algebra. Consider the product $B^{B^B}$. Define for $b\in B$, $\pi_b\colon B^B\rightarrow B$ given by $\pi_b(f)=f(b)$ for all $f\in B^B$. Then note that $\pi_b\in B^{B^B}$; so consider the generated Boolean subalgebra $\langle\pi_b\mid b\in B\rangle$, and show that this has cardinality $2^{2^{|B|}}$.
I suppose one could try to establish a bijection between the subalgebra and the set $2^{2^{B}}$, but I cannot see a natural way to define it. Any help would be appreciated!
 A: It is enough to show that the algebra has $2^{|B|}$ atoms.
Now you have $|B|$ projections and each atom has the shape
$$\bigwedge_{b\in B}\pi_b^{\sigma(b)},$$
where $\sigma\in\{-1,1\}^B$, and $\pi_b^1=\pi_b$, and $\pi_b^{-1}=\pi_b'$.
Since each $\sigma\in\{-1,1\}^B$ gives a different atom, the number of atoms is $|\{-1,1\}^B|=2^{|B|}$.

Edit.
Here I justify that, according to the notation above, each $\sigma$ gives rise to a different atom of the Boolean algebra.
Let us fix $\sigma_1 \in \{-1,1\}^B$ and define $f:B\to B$ by
$$
f(x) =
\begin{cases}
1, &\text{if} &\sigma_1(x)=1,\\
0, &\text{if} &\sigma_1(x)=-1.
\end{cases}
$$
It follows that, for each $b \in B$,
$$\pi_b^{\sigma_1(b)}(f)=1,$$
because if $\sigma_1(b)=1$, then $f(b)=1$, whence $\pi_b^{\sigma_1(b)}(f)=\pi_b(f)=f(b)=1$;
if $\sigma_1(b)=-1$, then $f(b)=0$, and $\pi_b^{\sigma_1(b)}(f)=\pi_b'(f)=(f(b))'=0'=1$.
Thus, we conclude that
$$\left( \bigwedge_{b\in B} \pi_b^{\sigma_1(b)} \right)(f) = 1.$$
Now, if $\sigma_2\neq\sigma_1$, then there exists $b_0\in B$ such that $\sigma_1(b_0) \neq \sigma_2(b_0)$.
Using the same $f$ as above, we see that
$$\pi_{b_0}^{\sigma_2(b_0)}(f)=0$$
because if $\sigma_2(b_0)=1$, then $\sigma_1(b_0)=-1$ and $f(b_0)=0$, whence
$$\pi_{b_0}^{\sigma_2(b_0)}(f)=\pi_{b_0}(f)=f(b_0)=0;$$
if $\sigma_2(b_0)=-1$, then $\sigma_1(b_0)=1$ and $f(b_0)=1$, whence
$$\pi_{b_0}^{\sigma_2(b_0)}(f)=\pi_{b_0}'(f)=(f(b_0))'=1'=0.$$
Hence
$$\left( \bigwedge_{b\in B} \pi_b^{\sigma_2(b)} \right)(f) = 0.$$
So if $\sigma_1 \neq \sigma_2$ then we constructed $f:B\to B$ such that
$$\bigwedge_{b\in B}\pi_b^{\sigma_1(b)}(f) \neq \bigwedge_{b\in B}\pi_b^{\sigma_2(b)}(f),$$
and therefore we can conclude that
$$\bigwedge_{b\in B}\pi_b^{\sigma_1(b)} \neq \bigwedge_{b\in B}\pi_b^{\sigma_2(b)}.$$
