Number of subalgebras of the power set algebra

Question

Let $X=\{a,b,c\}$ and $\mathcal{P}X=\{\emptyset,X,\{a\},\{b\},\{c\},\{a,b\}.\{a,c\},\{b,c\}\}$.

I can only see 4 subalgebras of $\mathcal{P}X$, namely:

$\mathcal{F}_0=\{\emptyset,X\}$

$\mathcal{F}_1=\{\emptyset,X,\{a\},\{b,c\}\}$

$\mathcal{F}_2=\{\emptyset,X,\{b\},\{a,c\}\}$

$\mathcal{F}_1=\{\emptyset,X,\{c\},\{a,b\}\}$

Is there any rule to determine the number of subalgebras of a power set algebra one can form?

How do you define a subalgebra? There are other sublattices isomorphic to a Boolean algebra. — Matt Samuel, Jan 15 '15 at 20:09

score 2 · Accepted Answer · answered Jan 15 '15 at 20:40

For a finite set $X$ of cardinality $n$, the boolean subalgebras of $\mathcal P X$ correspond to the equivalence relations on $X$, i.e. partitions of $X$ into "atoms". The number of these (including $\mathcal P X$ itself, which you didn't include in your four for $n=3$) is the Bell number $B_n$.

Number of subalgebras of the power set algebra

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