# Is every element in a set also an element of the powerset?

From my understanding of powersets, a powerset contains all the possible subsets of a set. So if we want the power set of {1} then the powerset of {1} is {{}, {1}}.

I was speaking to my tutor who was sure that if 1 is an element of the original set {1} (which is true) then that 1 is also an element of the power set {{}, {1}}. I was sure that it isn't an element, but the set containing that element is.

Can someone confirm or deny my suspicions?

• I think you are right and your tutor is wrong. – Tunococ Jun 21 '13 at 11:32
• The elements of $\{\{\},\{1\}\}$ are $\{\}$ and $\{1\}$. Apparently, $1$ is not among them. – azimut Jun 21 '13 at 12:08
• You need a new tutor. – PyRulez Jun 21 '13 at 15:04

However, there are sets $S$ and elements $e ∈ S$ for which $e$ is also in the powerset. Take $S = \{1,\{1\}\}$ and $e = \{1\}$. On the other side, in this example $e' = 1$ is not in the powerset of $S$, since $\mathcal{P}(S) = \{∅,\{1\},\{\{1\}\},S\}$, and none of the listed elements is $1$.
Another thing: There are sets $S$ with the property $S ⊂ \mathcal{P}(S)$, i.e. that every element of $S$ is indeed an element of the power set of $S$. Take for example $S = ∅$, $S = \{∅\}$ or $S = \{∅,\{∅\}\}$. I’m sure these sets have a name, but I don’t know it. Edit: As Andres mentioned in a comment, these sets are called transitive sets.
It is also common to define $0 := ∅$ and inductively any natural number $n+1$ as $n+1 := n ∪ \{n\}$, e.g. $1 = \{0\} = \{∅\}$ and $2 = \{0,1\} = \{∅,\{∅\}\}$ and so forth. Using this definition, it is certainly true that: $$0 ∈ 1 ∈ 2 ∈ 3 ∈ 4 ∈ …\quad \text{and} \quad 0 ⊂ 1 ⊂ 2 ⊂ 3 ⊂ 4 ⊂ …$$ such that “$∈$” essentially becomes “$<$”, whereas “$⊂$” becomes “$≤$”. So since any element of a so-defined natural number $n$ is also a subset of $n$, these natural numbers are indeed very prominent examples of sets whose elements are also members in their power sets, i.e. of transitive sets.
• The name you are looking for in the second paragraph is transitive. A set $A$ is called transitive if every element of $A$ is also a subset of $A$, that is, if whenever $y\in A$, then $y\subset A$, so $y\in\mathcal P(A)$. (The reason for the name "transitive" is perhaps explained by noticing that the condition is saying that whenever $x\in y\in A$, then $x\in A$.) – Andrés E. Caicedo Jun 21 '13 at 13:33