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Hello to all math lovers. I was studying Russell's paradox in set theory and came across something ambiguous that I couldn't justify no matter how hard I tried. The strange thing was: sets that are members of themselves and sets that are not members of themselves. This statement threw me thousands of meters away from the paradox itself. What does Russell mean by this? We know that when something is a part of a set, we say it is a member of that set. But how can a set be a part of itself or not? What unique characteristic makes a set a member of itself or not? I don't understand the concept of being a member of a set in itself. I would appreciate your guidance. Thank you, a math enthusiast :)

Connor W
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Mostafa Zeinodini
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    Imagine, if you can, the set of all sets. It's a member of itself, right?! – José Carlos Santos Jul 22 '23 at 08:44
  • Or if that is too big, then try to imagine "the set of sets which can be defined in English with fewer than a hundred letters" which took fewer than 100 letters to state and would have a finite number of elements – Henry Jul 22 '23 at 09:23
  • Related: https://math.stackexchange.com/questions/1046863/how-can-a-set-contain-itself – Henry Jul 22 '23 at 09:26
  • @JoséCarlosSantos , Well, I imagined it, but what is the characteristic of that set that makes it a member of its own? I do not understand this . Thanks :) – Mostafa Zeinodini Jul 22 '23 at 11:53
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    @Henry , I read it but can't understand :)))) can you tell me an example for a set it contain itself ?? . Thanks – Mostafa Zeinodini Jul 22 '23 at 11:56
  • There's a useful application of sets that are members of themselves. See Quine Atoms. – Chad K Jul 22 '23 at 15:52
  • You can consider Russell's own words in the post linked above: "it seemed to me that a class sometimes is, and sometimes is not, a member of itself. The class of teaspoons, for example, is not another teaspoon, but the class of things that are not teaspoons, is one of the things that are not teaspoons. There seemed to be instances that are not negative: for example, the class of all classes is a class." – Mauro ALLEGRANZA Jul 25 '23 at 09:59
  • "usual" sets are not members of themselves: consider the set $\mathbb N$ of all naturals: it is not a natural number, and thus $\mathbb N \notin \mathbb N$. But consider instead, as Russell did, the collection $V$ of all sets: IF it is a set, clearly we must have $V \in V$. – Mauro ALLEGRANZA Jul 25 '23 at 10:01

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The first step that might help is to start thinking about sets that contain other sets. For example, the set $\{\{1, 2\}, \{3, 4\}\}$ has two elements - the set $\{1, 2\}$ and the set $\{3, 4\}$.

After that, you can start thinking about more complicated nesting of sets. For example, we can start with the empty set $\varnothing = \{\}$ and then look at the set that contains just the empty set $\{\varnothing\}$. Then let's make a set that contains those two sets, $\{\varnothing, \{\varnothing\}\}$. And we can keep going, putting those three sets in a set together to get $\{\varnothing, \{\varnothing\}, \{\varnothing, \{\varnothing\}\}\}$. You can even take that to infinitely many sets, giving you something that looks like: $\{\varnothing, \{\varnothing\}, \{\varnothing, \{\varnothing\}\}, \ldots\}$.

So let's consider a set $U$. And this is going to be a special set, because $U$ is a set that contains every set. Name a set, it's an element of $U$. The empty set? That's in $U$. The set of natural numbers? It's in there. The set containing the set containing the empty set? Sure. So is $U$ an element of $U$? Well, $U$ is a set, and every set is an element of $U$, so therefore $U$ must be an element, and hence $U$ is a set that contains itself.

ConMan
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