-2

My question is: which axiom of ZFC the collection defined by $$ R=\{x: x\notin x\} $$ violate in order to does not to be a set?

Edit 1: Let me made a clarification. I know the separation axiom there exists to avoid some paradoxes. Here is what I would like to understand: In some expositions I have found the following argument: By its definition $R\in R$ and $R\notin R$, absurd! Therefore $R$ is not a set.

I expected a little more explanation of why $R$ is not a set. See for instance this video of Professor Philip Welch,

Set theory lecture 2

Edit 2:

I think I'm not being able to explain my issue in a way that makes me understand right now.

In the above mentioned video, Professor Welch states:$R=\{x: x\notin x\}$ and give a proof of that fact, by exhibiting the classical argument of Russel paradox. To me this makes no sense, once he is taking in consideration an obeject which the construction already violates the separation axiom, and tha is what I am trying to understant

Asaf Karagila
  • 393,674
Eduardo
  • 957
  • 1
    There is no axiom which let’s you construct sets this way. You can, for some set $A$ construct a set $${x\in A\mid x\notin x}$$ but you can’t define a set without basing it on an existing set, unless another axioms assert it exists. – Thomas Andrews Aug 05 '21 at 14:20
  • Where do the elements $x$ live? If you want $R$ to be a set it should have the form $R = {x \in A : x \notin x}$ for some set $A$. – Umberto P. Aug 05 '21 at 14:20
  • @ThomasAndrews beat me by 5 seconds. – Umberto P. Aug 05 '21 at 14:20
  • Basically, the notation: $${x\in A\mid P(x)}$$ is for defining subsets of existing sets. There are other ways to define sets, but none of this general sort. That doesn’t mean we can’t have this set. If we did, of course, set theory would be inconsistent, but we can’t prove set theory is consistent. But the bullet-headed effort to define the set with this notation won’t work in ZFC. – Thomas Andrews Aug 05 '21 at 14:26
  • I have made a further edit in the question – Eduardo Aug 05 '21 at 14:36
  • @HansLundmark In the above mentioned video, Professor Welch states:$R={x: x\notin x}$ and give a proof of that fact, by exhibiting the classical argument of Russel paradox. To me this makes no sense, once he is taking in consideration an obeject which the construction already violates the separation axiom, and tha is what I am trying to understant – Eduardo Aug 05 '21 at 14:42

1 Answers1

1

The ZFC axioms consist only of a small handful of valid ways to construct new sets, usually out of other sets you already have. The construction $\{x\mid x\notin x\}$ simply isn't one of those ways.

The closest you can get within ZFC is the axioms of comprehension. One of those axioms say that if you have a set $A$, you can make the set $\{x\in A\mid x\notin x\}$. This will turn out to be equal to $A$, as the axiom of regularity implies all sets $x$ satisfy $x\notin x$.

Arthur
  • 199,419
  • 2
    It is worth noting that the axiom of comprehension in the axiomatizations of set theory prior to ZFC actually did allow for this construction, but that was changed due to Russell's paradox and similar phenomena. – Rushabh Mehta Aug 05 '21 at 14:36
  • 1
    @DonThousand Indeed, it was paradoxes such as Russel's that necessitated the more limited comprehension that only lets you build subsets, not new sets from scratch. – Arthur Aug 05 '21 at 14:37
  • In the above mentioned video, Professor Welch states:$R={x: x\notin x}$ and give a proof of that fact, by exhibiting the classical argument of Russel paradox. To me this makes no sense, once he is taking in consideration an obeject which the construction already violates the separation axiom, and tha is what I am trying to understant – Eduardo Aug 05 '21 at 14:42
  • 2
    @Eduardo My guess is that, as my previous comment alludes to, Professor Welch is using an older system of set theory that permits this sort of construction and showing its limitations. – Rushabh Mehta Aug 05 '21 at 14:43