Show that atom of Boolean algebra B inside B' is also the atom of B' where B' is subalgebra of B.

Question

Let B := (B, ≤, ∨, ∧, c , 0, 1) be a Boolean algebra and B' := (B' , ≤, ∨, ∧, c , 0, 1) be a Boolean subalgebra of B. Show that an element of B0 that is an atom of B must also be an atom of B' . However, there can be atoms of B' that are not atoms of B.

I know some basic concept of atoms and its existence that for an atom a of BA B ... please guide me in steps...

Answer 1

If $a$ is an atom in $B$, then there's no smaller nonzero element in $B$, so neither can such exist in $B'$.

For the other direction, consider e.g. a subalgebra of the Boolean algebra $P(\{1,2,3,4\})$.