The power set $P(S)$ of a set $S$ is the set of all subsets of $S$, including the empty set $\emptyset$ and $S$ itself.

Whether $P(S)$ contains an element $s$ depends on whether $s$ is a subset of $S$:

• If $s$ is an element of $S$ (i.e., $s \in S$), then $s$ itself is not in $P(S)$. Instead, the singleton set $\{s\}$ (which is a subset of $S$) is in $P(S)$.
• $S$ itself is always in $P(S)$ since it is trivially a subset of itself.
• If $S$ is a set containing just one element, say $S = \{s\}$, then $P(S) = \{\emptyset, \{s\}\}$, meaning $s$ is not in $P(S)$, but $\{s\}$ is.

Example:

Let $S = \{a, b\}$. Then,

$P(S) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$

Here, neither $a$ nor $b$ appear alone in P(S) ), but \( \{a\} and $\{b\}$ do.

Conclusion:

$P(S)$ contains subsets of $S$, not the individual elements of $S$. It does contain $\{S\}$ (if $S$ is considered as an element of a larger set), but not $s$ directly unless $s$ is itself a subset of $S$.

Would you like a deeper explanation or more examples?

Related Questions:

1. How do you determine the number of elements in $P(S)$?
2. Can a power set ever contain elements that are not subsets of $S$?
3. What is the power set of an empty set?
4. How does the power set relate to Boolean algebra?
5. How can the power set be used in probability theory?

Tip:

The number of elements in $P(S)$ is always $2^{|S|}$, where $|S|$ is the number of elements in $S$.