A math logic book gives a puzzling solution to this problem.
Let set $A = \{a, b\}$ . Determine the power set P(A). Determine $|A|$ and $|P(A)|$ .
Solution: $P(A) = \{\emptyset, {a}, {b}, \{a, b\}\}$ . So $|A| = 2, |P(A)| = 4$.
The empty set is a subset of every set. But here it's true only for $P(A)$ , but not for $A$ . Why?
If I explicitly include the empty set as one of the element of $A$ will it increase the cardinality of $A$ ?
Best Answer
Yes, the empty set is a subset of every set. However, it is not an element of each set. That's a different thing.
$\varnothing\notin A$, but $\varnothing\subseteq A$, which makes $\varnothing\in P(A)$.