Is it possible, in "intuitive set theory" to have a set whose elements are subsets of the set?

Solution 1:

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.