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If we define the set of all sets to be $A$,

Then why can't I say that the set $R=\{G \in A | G \notin G\}$ is not well defined?

If it was well defined then $R\in A$ and the condition $G\notin G$ must be well defined if we substitute $G=R$

but it isn't well defined because "Russell's paradox"

so you can change the condition to be $ \begin{equation} \begin{cases} true\ \ \ \ \ \ \ \ G=R\\ G\notin G\ \ \ \ G\neq R \end{cases} \end{equation} $ and then it will be well defined

Where is my misunderstanding?

btw, the same idea can be included if you try to disproof the existing of the sets of all sets with Cantor's theorem. If you use the Wikipedia's proof then we can choose $f=Id$ then B is exactly $\{G \in A | G \notin G\}$ and its not well defined as i said before.

Edit:

Lot of peoples referencing this question but this question is about "why we care about Russell's paradox", but my question is about "Russell's paradox is not a paradox"

I got the answer to the question: Russell's paradox apply on the standard way we define sets-a condition that all element must met to be part of the set. but I thought about sets in more specific way-an infinite/finite nesting of sets.

for example, the set $\{X|X\in X\land\emptyset\in X\land|X|=2\}$ is like {{},{{},{...}}}

This answer shows a good explanation.

Ofek
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    If you downvoted, please tell me what is the issue with the question, and i will fix it. or remove it if its not related to the site. – Ofek Dec 23 '21 at 19:58
  • I think these questions and answers will explain it. – Nikolaj Pedersen Dec 23 '21 at 20:09
  • In naive set theory there is nothing preventing you from forming the set $R$. Russel's paradox shows precisely that you cannot make naive set theory work without paradoxes, so that you do need to make some restrictions on how you form sets so that such paradoxes and similar ones cannot occur. In naive set theory, there is no restriction on how you form sets, and the set you wrote is defined, it just leads to a paradox, which once again, is what made mathematicians develop more rigorous versions of set theory. I'm not sure what you mean about changing the condition. – Snaw Dec 23 '21 at 20:26
  • @NikolajPedersen i didn't asked what is the implications of Russel's paradox, i asked why can't we just say that Russel's set is an illegal set and the set of all sets is a legal set. – Ofek Dec 23 '21 at 20:31
  • @Snaw thanks for the answer, i guess that my approach was different, i can define the set of all sets to be any unlimited nesting of infinite sets. and every set is a nesting of limited/unlimited sets. so the standard approach is that a set is a condition that must be met for the element to be in a group. – Ofek Dec 23 '21 at 20:36
  • @Snaw by changing the condition i mean the right part of the definition of R $R={...|change\ this}$ – Ofek Dec 23 '21 at 20:37
  • The problem is that you really want to be able to have an axiom schema such as Comprehension. It is, to say the least, not obvious how you would alter that schema to avoid encountering Russell's Paradox while still retaining the structures you do want. – Robert Shore Dec 23 '21 at 20:40
  • @RobertShore as i said here in the comments, probably my approach was different, i thought about sets as a limited/unlimited nesting of sets. something like A={{},A} is the same as {{},{{},{...}}} and set of all sets is the set of all possible limited/unlimited nesting of sets. and Russel's set can't be shown as a nesting of sets. – Ofek Dec 23 '21 at 20:49
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    See my answer. – Asaf Karagila Dec 23 '21 at 21:46
  • @AsafKaragila thanks, that's really help me understand the point. – Ofek Dec 23 '21 at 21:56
  • @HennoBrandsma I already saw this question before I asked this question. and also Nikolaj Pedersen already sent a link to this question. (second comment) – Ofek Dec 23 '21 at 21:58
  • I understand set theory at an intuitive level only, and in my mind, the resolution to Russell’s paradox is “there’s no such set”. I think the paradox reveals something we must be careful about when developing set theory formally, or axiomatically. – littleO Dec 23 '21 at 22:39
  • Te technical term is "set", not group (which has a different technical algebraic meaning). – Arturo Magidin Dec 23 '21 at 22:40

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