2.6: Sets of sets
Sets are heterogeneous — a single set can contain four universities, seven integers, and an ahi tuna — and so it might occur to you that they can contain other sets as well. This is indeed true, but let me issue a stern warning: you can get in deep water very quickly when you start thinking about “sets of sets." In 1901, in fact, the philosopher Bertrand Russell pointed out that this idea can lead to unresolvable contradictions unless you put some constraints on it. What became known as “Russell’s Paradox" famously goes as follows: consider the set \(R\) of all sets that do not have themselves as members 1 . Now is \(R\) a member of itself, or isn’t it? Either way you answer turns out to be wrong (try it!) which means that this whole setup must be flawed at some level.
The good news is that as long as you don’t deal with this kind of self-referential loop (“containing yourself as a member") then it’s pretty safe to try at home. Consider this set: \[V = \{~3, 5, \{~5, 4~\}, 2~\}.\] This set has four (not five) members. Three of \(V\) ’s members are integers: 2, 3, and 5. The other one is a set (with no name given). That other set, by the way, has two members of its own: 4 and 5. If you were asked, “is \(4 \in V\) "? the answer would be no .
As a corollary to this, there’s a difference between \[\varnothing\] and \[\{~\varnothing~\}.\]
The former is a set with no elements. The latter is a set with one element: and that element just happens to be a set with nothing in it.
- For instance, the set Z of all zebras is a member of R, since Z itself is a set (not a zebra) and so Z /∈ Z. The set S, on the other hand, defined as “the set of all sets mentioned in this book,” is not a member of R, since S contains itself as a member.