Cardinality of Sub(A) is smaller than ${\aleph_0}$ or equal $2^{\aleph_0}$? Let $\mathbb{A}=(A,\mathcal{F}^\mathbb{A})$ be a countable algebra. I need to prove that $|Sub(\mathbb{A})|\leqslant\aleph_0$ or $|Sub(\mathbb{A})|=2^{\aleph_0},$ where $Sub(\mathbb{A})$ is the lattice of subuniversums of $\mathbb{A}.$
I suppose I have to use topology and Cantor set.
 A: Give the set $2$ the discrete topology, and then give $2^{\omega}$
the product topology. The resulting space is (or is homeomorphic to,
depending to your definition) the Cantor space.
If $A$ is an algebra with universe $\omega$, then any
subalgebra $S\leq A$ is a subset of $A$, hence has a
characteristic function $\chi_S\in 2^{\omega}$.
The collection $X$ of these characteristic functions
is in bijective correspondence with Sub($A$),
so it suffices to prove that the set $X$
of characteristic functions is countable or
of cardinality $2^{\aleph_0}$.
$X$ is a closed subset of $2^{\omega}$, since subalgebra
generation is an algebraic closure operator.
By the Cantor–Bendixson theorem, a closed subset
of the Cantor space is the union of a countable set
and a perfect set. So, if $X$ is uncountable,
then it contains a perfect subset, from which one gets
$|X|\geq 2^{\aleph_0}$. (Perfect subsets of the Cantor set have size continuum.) 
The reverse inequality follows
from $X\subseteq 2^{\omega}$.
