2

Why is the set $\{ x \mid x = x\} $ not defined?

Since, $x=x$ is always true, the set is actually "the set of everything".
But why is it illegal to be defined as a set?

Cortizol
  • 3,669
AnnieOK
  • 2,920
  • 1
    There probably are several discussions of this (or very similar questions). See, for example, http://math.stackexchange.com/questions/162/why-is-the-set-of-all-sets-a-paradox – Martin Sleziak Apr 09 '14 at 10:07
  • 1
    It is a well-defined class; it's just not a set. –  Apr 09 '14 at 10:08
  • It is a "long story", that goes back to Cantor himself and the "founding fathers" of set theory; see user2357112's answer. The perfectly natural idea that for all "property" that we may imagine we can "collect together" all objects that satisfy that property (all human being into the collection called "humanity") gives us (gave them : the founding fathers) troubles. Thus, we have to repress this "natural impulse" and refrain from asserting the existence of every "imaginable" collection. There are only some "allowed" way of "collecting together": those permitted by your preferred set theory. – Mauro ALLEGRANZA Apr 09 '14 at 10:29

4 Answers4

7

ZFC has no method of set construction that allows us to construct the set of all sets with a given property. If it did, we could define $\{ x | x \notin x\}$, the set of all sets that don't contain themselves, and we would have Russell's paradox; that set would have to both contain and not contain itself.

Instead, ZFC has the axiom schema of restricted comprehension. For any set $S$ and predicate $p$, we can define

$$\{ x \in S \mid p(x)\}$$

the set of all elements of $S$ such that $p$ holds for those elements. We have to restrict the scope to elements of another set; we can't perform a set comprehension over all sets.

user2357112
  • 2,401
5

While it is quite possibly to prove that this collection is not a set via Russell's paradox, here's an easier conclusion from the axiom of regularity (and pairing).

Suppose $A=\{x\mid x=x\}$ is a set, then in particular $A=A$ so $A\in A$. However this is a contradiction to the axiom of regularity, since now $\{A\}$ is a set and $A\cap\{A\}\neq\varnothing$ contrary to the requirement that $\in$ is well-founded.

Asaf Karagila
  • 393,674
  • Your answers are always useful, also on topics like this, were "all has been told". But ... how we explain that the "natural" idea that for every imaginable property we cannot "collect together" all objects satisying it is "unsound" because it does not fit with regularity? Where is the source of our "feeling" about its "evidence", apart from the "regressive" (see M.Potter, Set Theory and Its Philosophy: A Critical Introduction) support ? I'll find it in the "foundational" chapter of your future textbook on s-t? – Mauro ALLEGRANZA Apr 09 '14 at 10:38
  • Agreed, but I think the argument using the Russell paradox is actually a bit easier (and the comprehension axiom seems to me more indispensable than regularity). – Marc van Leeuwen Apr 09 '14 at 10:39
  • @Mauro: My future text, eh? I'll think about it. I'm not disputing the fact that the various paradoxes of set theory do deserve merit and mention. But it's just easier to appeal to regularity, I think. Since this axiom gives you as a direct consequence that $x\notin x$. – Asaf Karagila Apr 09 '14 at 10:53
  • 1
    @Marc: I'm not saying that the Russell paradox argument is difficult. But I do feel that regularity, when understood, is actually easier to use. You don't have to argue that if one case fails the other must hold and that is a contradiction. You can just prove directly that $\sf ZFC$ proves $\forall x(x\notin x)$. If ${x\mid x=x}$ is a set then it is a member of itself, so it's not a set. I find it easier (and in fact always found it easier, even when I started with set theory). – Asaf Karagila Apr 09 '14 at 10:54
  • @AsafKaragila: But if you show first that $(P\leftrightarrow\lnot P)\rightarrow\bot$ for any proposition $P$, then you don't need any case distinction any more, and the Russell paradox becomes immediate. But I guess it remains a matter of taste. – Marc van Leeuwen Apr 09 '14 at 11:05
1

Using the axioms of ZFC, you can't prove the existence of it because is a proper class (using the NBG terminology).

Martín-Blas Pérez Pinilla
  • 41,546
  • 4
  • 46
  • 89
  • 2
    A bit of nitpicking. Shouldn't you write: You can prove non-existence instead of You can't prove existence? (Hypothetically, ZFC might be inconsistent.) – Martin Sleziak Apr 09 '14 at 10:11
1

Because there is no pre-existing set that it is a subset of. To construct a set from a predicate we require the axiom of comprehension/specfication.

Dan Brumleve
  • 17,796