First things first. I need to explain what a set is. Basically, a set is just a collection of items. However, there are two caveats:

1. The order of a set’s items doesn’t matter. So, for example, {1, 2, 3} = {1, 3, 2}.

2. A set can’t contain the same item more than once. So, for example, {1, 1, 1} is just {1}.

Other than that, a set is just a collection.

By the way, the collection can be empty. The *null set* is the set that has nothing in it.

If an item is in a set, then we say that it’s a *member* of the set. So, for example, 1 is a member of {1, 2, 3}.

A set can have members that are themselves sets. For example, {{1, 2}, 3} is a set that has a set as a member. {{1, 2}, {3}} is a set that has two sets as a member.

Now that you know that, let’s consider the set of all sets that are not members of themselves. So that I don’t have to continually type “the set of all sets that are not members of themselves”, let’s call that set “S”.

So, the question is: Is S a member of itself?

Let’s suppose that S is a member of itself. The only sets in S are sets that are not members of themselves. So, if S is in S, then S is not a member of itself. So, if S is a member of itself, then S is not a member of itself.

Now let’s suppose that S is not a member of itself. S is the set of all sets that are not members of themselves. So, if S is not a member of itself, then S is a member of itself.

So, S is a member of itself if and only if it is not a member of itself.

Any satisfactory theory of sets has to resolve this paradox.

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