Let $A$ be a Boolean algebra that has infinitely many atoms.
Let $X$ be the set of atoms.
Define the map from $A$ to $\wp(X)$ as follows, $a \mapsto \{x \in X : x \leq a\}$.
(a) $f$ is 1-1. (b) $f$ is an isomorphism (c) $f$ is surjective (d) None of the last
here is my thought process: I think the answer is (a) because the set $A$ is atomic but is not complete so is it not an isomorphism
since for a Boolean algebra $A$ and a power set algebra $\wp(At (A))$, $f : A \to \wp(At (A))$, where $f(b) = \{a \in At (A) | a \leq b\}$, is a homomorphism.
If $A$ is atomic, $f$ is an embedding and if $A$ is complete, $f$ is an epimorphism.
Let $\mathbf A = \wp(Y) \times \mathbf B$, where $Y$ is any infinite set and $\mathbf B$ an atomless Boolean algebra.
Then $\mathbf A$ has infinitely many atoms: the elements of the form $(a,0)$ with $a$ an atom of $\wp(Y)$ (that is, a singleton of $Y$).
Now let $b,b' \in B \setminus \{0\}$. Then there is no atom of $\mathbf A$ below either $(0,b)$ or $(0,b')$.
So $f(0,b) = f(0,b')=\varnothing$, and so the map is not injective.
Now, let $\mathbf A = FC(\mathbb N)$, the set of finite and co-finite subsets of $\mathbb N$.
Then $\mathbf A$ has infinitely many atoms: the singletons of $\mathbb N$ are the atoms of $\mathbf A$.
Let $B = \{\{2n\} : n \in \mathbb N\}$, a subset of the atoms of $\mathbf A$.
Suppose there exists $a \in A$ such that $f(a) \supseteq B$; it follows that $\{2n\} \subseteq a$, for all $n \in \mathbb N$, and so $E \subseteq a$, where $E$ is the set of even numbers.
Now, it's clear that $E$ is not finite, and so $a$ can't be finite, and therefore, if $a$ is an element of $\mathbf A$, then $a$ must be co-finite, that is, $a = \mathbb N \setminus C$, where $C$ is a finite set of odd numbers.
But this means that $a$ contains odd numbers (indeed, infinitely many), and so $f(a) \neq B$.
Hence, the map is not surjective.