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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?
Your tutor is wrong and you are right.
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.
