TITLE: Is it possible, in "intuitive set theory" to have a set whose elements are subsets of the set?
QUESTION [11 upvotes]: Note, this is not homework.  I have not seen the inside of a classroom in decades.
I decided to brush up on set theory and logic and stumbled on  this:
Exercise 1-3.7. 
Give several examples of a set $\mathcal{X}$ such that each element
of $\mathcal{X}$ is a subset of $\mathcal{X}$.
Stoll, Robert R.. Set Theory and Logic (Dover Books on Mathematics)
(Kindle Locations 482-483). Dover Publications. Kindle Edition. 
This is from the introductory chapter on intuitive set theory.
Terms such as predicate and symbols such as $\implies$ and
$\iff$ have not yet been introduced. Nor has any formal logic.
So this is my assessment of the situation:
A set is a collection of all entities which satisfy some predicate
$P$. That is $\mathcal{X}=\left\{ x\vert P\left[x\right]\right\} $
is a set if $P\left[x\right]$ is always either true or else false. 
The statement $\mathcal{Y}$ is a subset of $\mathcal{X}$ may be written
symbolically as$\mathcal{Y}\subseteq\mathcal{X}\iff y\in\mathcal{Y}\implies y\in\mathcal{X}$.
Here is an example of a set whose elements are not subsets of the
set. 
$\mathcal{A}=\left\{ a_{1},a_{2}\right\} $. The subsets of $\mathcal{A}$
are $\emptyset$,$\mathcal{A}$, $\left\{ a_{1}\right\} $ and $\left\{ a_{2}\right\} $,
so $a_{1}$ and $a_{2}$ are not subsets of $\mathcal{A}$.
I can write the expression $\mathcal{S}=\left\{ x\vert x\subseteq\mathcal{S}\right\} $,
and claim it meets the criteria. But, if I enumerate the elements
$\mathcal{S}=\left\{ x_{1},x_{2},\dots\right\} $, the subsets might
be understood to be $\mathscr{P}\left[\mathcal{S}\right]=\left\{ \emptyset,\mathcal{S},\left\{ x_{1}\right\} ,\left\{ x_{2}\right\} ,\left\{ x_{1},x_{2}\right\} ,\dots\right\} $.
But, since it was predicated that $x_{i}\subseteq\mathcal{S}$, perhaps
$\mathscr{P}\left[\mathcal{S}\right]=\left\{ \emptyset,\mathcal{S},x_{1},x_{2},\dots,\left\{ x_{1}\right\} ,\left\{ x_{2}\right\} ,\left\{ x_{1},x_{2}\right\} ,\dots\right\} $. 
Assume $\mathcal{S}$ has a finite number of elements, say two, for
simplicity. Then $\mathscr{P}\left[\mathcal{S}\right]=\left\{ \emptyset,\mathcal{S},x_{1},x_{2},\left\{ x_{1}\right\} ,\left\{ x_{2}\right\} \right\} $,
so the power set of $\mathcal{S}$ has $6$ elements. But the power set of a set of two elements should on have $4$ elements.
That indicates to me that any set satisfying the criteria of the exercise
would have to be infinite.
Even if I consider infinite sets, I'm still not comfortable with including
the elements as subsets.
Is this a trick question? Or is there some way around this seeming
contradiction?

REPLY [17 votes]: This isn't a trick question; there's several examples of such sets, both finite and infinite, and they are called transitive sets. First, the simplest examples is $\varnothing$, every element of $\varnothing$ is a subset of $\varnothing$, vacuously. So, next, let us consider the set $1$, which I am going to define as $1 = \{\varnothing\}$. This set has one element, $\varnothing$, and $\varnothing$ is trivially a subset of $1$, so $1$ is also transitive. Next, I define $2 = \{\varnothing, 1\} = 2 = \{\varnothing, \{\varnothing\}\}$.  It's another quick calculation that this set is transitive. Similarly, $3 = \{\varnothing, 1, 2\}$ is transitive, as is $4 = \{\varnothing, 1, 2, 3\}$, $n = \{\varnothing, 1, 2, \dots, n - 1\}$. The first infinite example I'll give is $\omega = \{\varnothing, 1, 2, \dots, n, n + 1, \dots \}$. What I've essentially done here is construct the first few ordinal numbers.