2

So, I though of the following set: E := {{...{...{...}...}...}}. In other words, a set that contains one element which is a set that contains one element which is a set that ..., and so on. Given |E| = 1 it is not empty , but given its only rather odd element, it seems/feels "almost" empty (also in comparison with attributes of the empty set).

My question is now, has anybody come across this (I found something called "quine atoms" but not much more)? I just want to check it out because I think it's a funny object. Or perhaps, is my construction flawed, and this object isn't interesting or even existing at all? :D

frudnick
  • 37
  • 1
  • 5
    If it is meaningful, you have $E={E}$. – Henry Jan 06 '23 at 14:22
  • 7
    And it should be noted that in ZFC it is not allowed for a set to be an element of itself as per the axiom of regularity and axiom of pairing. As such, your object either is not a set, does not exist, or you are working in a set theory different than what most others do. – JMoravitz Jan 06 '23 at 14:27
  • In the standard set theory (ZFC axioms) such set doesn't exists because the axiom of regularity. In other alternative set theories such set can be considered. – jjagmath Jan 06 '23 at 14:28
  • Related: https://math.stackexchange.com/questions/1445251 – Anne Bauval Jan 06 '23 at 14:32
  • I doubt that this is a well-defined set because of the pointed out equality that would have to hold for the "limit-set". – Peter Jan 06 '23 at 14:35
  • Do you consider a set with one element to be an "almost-empty" set ? – Peter Jan 06 '23 at 14:40
  • 1
    Historically, types of sets such as the one you have pointed out, which contain themselves, were directly responsible for motivating modern set theory to develop a framework which deals with such 'paradoxes'. – Alborz Jan 06 '23 at 14:46
  • 1
    I think JMoravitz's comment should be an answer, as it fully answers the question, no? – Adam Rubinson Jan 06 '23 at 14:48
  • Yes, or close as a duplicate of the related question above. – Anne Bauval Jan 06 '23 at 14:59
  • 2
    Just because you can write an approximation of something in symbols doesn't mean it means anything. Most set theories try to avoid such sets, or at least leave agnostic whether such sets exist (which is to say, the theory doesn't let you define such a set, although models might exist where there is an $E={E}.$) – Thomas Andrews Jan 06 '23 at 15:10
  • For the sake of notation, let's call your set $E_0$ and the unique element of $E_i$ is $E_{i+1}$. I'm not sure why people are assuming that $E_0 = {E_0}$. By extensionality, this is only true if the unique element of $E_0$ is the unique element of ${E_0}$, in other words $E_1 = E_0$. By extensionality, this only happens if $E_2 = E_1$ and so on and so forth. The $E_i$ could be equal, but they don't have to be. – TomKern Jan 06 '23 at 15:35
  • Your infinite chain of inclusion $E_0 \ni E_1 \ni E_2 \ni E_3 \cdots$ also doesn't violate the axiom of foundation, since at no point did you require that ${E_0, E_1, E_2, \ldots}$ be a set. To its dismay, the axiom of foundation cannot prevent infinite descending chains that aren't sets. It's only if you assume all of the $E_i$ are equal that this set must exist. – TomKern Jan 06 '23 at 15:36
  • 4
    Check this out, I found this totally amazing real number $x=x-1$. But now, you might argue, I have found nothing. I wrote an equation, but did not prove that it has a solution, let alone a unique solution. In fact, you can prove it has none. Well, you already wrote an equation, but you did not prove that it has a solution, and I can prove that it has none. – Asaf Karagila Jan 06 '23 at 18:25
  • @TomKern ZFC proves that any set has a transitive closure. If the OP's "set" were a set, its transitive closure would be ${E_0,E_1,E_2,\dots}$ (in your notation), which violates the axiom of foundation. – Alex Kruckman Jan 07 '23 at 19:49
  • @TomKern On the other hand, your comment is correct that the assumption $E_0 = {E_0}$ is unjustified. Under AFA, we can indeed prove that if $E_i = {E_{i+1}}$ for all $i$, then $E_i = E_{i+1}$ for all $i$. But other non-well-founded set theories fail to prove this. – Alex Kruckman Jan 07 '23 at 19:59
  • Ah. Here I was thinking about a descending chain under $\in$ whose length is a nonstandard natural number, but that would be notated more like ${{\ldots\ldots{{}}\ldots\ldots}}$ – TomKern Jan 07 '23 at 22:14

0 Answers0