Why is $\emptyset$ an element of the power set of a set? Take a set $S$. Why is the empty set part of the power set of $S$?
Intuitively speaking, the power set is the collection of all possible subsets of $S$. How is $\emptyset$ such a subset then? Why is not selecting any element and still forming a subset allowed?
 A: Suppose $B$ is some set.
We could change the rules and say that $\emptyset$ is not a subset of $B$.  But it would be inconvenient.  Here  are just two of many examples.
Suppose that $P$ and $Q$ are subsets of $B$.   We would like to be able to conclude that $P\cap Q$ is a subset of $B$. This is intuitively appealing, because if every element of $P$ is in $B$, and every element of $Q$ is also in $B$, then surely every element that is in both $P$ and $Q$ is in $B$.
But with your proposal, we cannot say that  $P\cap Q$ is a subset of $B$.  Because $P\cap Q$ might be empty.
Here is another example. It is tempting to say that $X$ is always a subset of $X\cup Y$.  Because surely anything in $X$ is also in $X\cup Y$, which is defined to be the set of things that are in $X$ or that are in $Y$.
But with your proposal, we cannot say this, because $X$ might be empty.
Instead of making a lot of annoying and useless exceptions like those, we find that it's simpler all around to agree that the empty set is a subset of every set.
A: Simply put, because the empty set is a subset of every set.
The definition of $A \subseteq B$ is as follows:
$\forall x (x \in A \rightarrow x \in B)$
Where $A = \emptyset$ this holds trivially; there are no $x \in \emptyset$ and so it satisfies the conditional vacuously.
Hence $\emptyset$ is a subset of every set, and hence it's in the powerset of every set.
A: This is a very natural usage.
Consider the set {a, b}. You form its subsets from all combinations of the rules "take a/leave a" and "take b/leave b".
There are four combinations: {a, b} (Ta/Tb), {b} (La/Tb), {a} (Ta/Lb) and... {} (La/Lb).
This is why a powerset has $2^N$ elements ($2^N$ combinations).
You can look at it differently, by using the binary notation, one bit per element: $11$, $01$, $10$, $00$. These are all binary numbers of two bits, there is no reason to exclude $00$.
