Is both $\{\}$ and $\{\{\}\}$ same element $?$ Let $P(S)$ denotes the power set of set $S$. Which of the following is always true$?$


*

*$P(P(S)) = P(S)$

*$P(S) ∩ P(P(S)) = \{ Ø \}$

*$P(S) ∩ S = P(S)$

*$S ∉ P(S)$



I try to explain $:$
If $S$ is the set $\{x, y, z \},$ then the subsets of S are:
${}$ (also denoted $\phi,$ the empty set)
$\{x\},
\{y\},
\{z\},
\{x, y \},
\{x, z \},
\{y, z \},
\{x, y, z \}$
and hence the power set of S is i.e.,
$P(S) =$ $\{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\}$.
Similarly , 
$P(P(S))=\{\{\{\}\}, \{\{x\}\}, \{\{y\}\}, \{\{z\}\}, \{\{x, y\}\}, \{\{x, z\}\}, \{\{y, z\}\}, \{\{x, y, z\}\}\}.....\}$
therefore , 
$P(S) ∩ P(P(S)) = \{ Ø \}$
Note that $\{ Ø \}$ is always element of powerset of a set , and also $\{ Ø \}$ is the subset of a set , in other words all subset of a set is a powerset .


My question is $:$  both $\{\}$ and $\{\{\}\}$ same element ?

 A: No, $\{\}$ and $\{\{\}\}$ are two different sets. You can see that by noting that there's something that is an element of one but not of the other:
$$ \{\}\notin \{\} \qquad\text{but}\qquad \{\}\in\{\{\}\} $$
$\{\}$ is a set that has no elements. It is the same set that is also notated $\varnothing$.
$\{\{\}\}$ is a set which has exactly one element, namely $\{\}$. Since $\{\}$ is the same as $\varnothing$, a different name for $\{\{\}\}$ would be $\{\varnothing\}$.
A: No, $\{\}$ is the empty set, and $\{ \{ \} \}$ is the set containing the empty set.  They aren't the same -- $\{\}$ has no elements, and $\{ \{ \} \}$ has one element.
By the way, in $P(P(S))$, you forgot to add the element $\{ \}$, since the empty set is a subset of every set.
A: No, $\{\}$ and $\{\{\}\}$ are not the same. Think of sets as bags. $\{\}$ is an empty bag. $\{\{\}\}$ is a bag with an empty bag within it. Therefore the outermost bag is not empty: It has a bag inside it.

For the original question, you are wrong, lets consider all four possibilities.

*

*No. By Cantor's theorem, we always have $|P(P(S))| > |P(S)|$. Hence they cannot be equal. Alternatively, let $S=\{\}$ and write out.


*Consider $S=\{\{\}\}$. Then $P( S)=\{\{\},\{\{\}\}\}$ and $P(P(S))=\{\{\},\{\{\}\},\{\{\{\}\}\},\{\{\},\{\{\}\}\}  \}$, hence $P(S) \cap P(P(S)) = \{\{\},\{\{\}\}\}$.


*No. Note that $A \cap B \subseteq B$ for all sets $A,B$. Hence  $S \cap P(S) \subseteq S$, hence $$|S \cap P(S)| \leq |S| < |P(S)|$$
hence they are not equal.


*By definition $S \subseteq S$ and thus $S \in P(S)$.
Therefore, none of the statements is true.
