See this.
RonL
Can someone answer this:
What the Russell Paradox is (a complete proof that at some point you will want to prove that certain assumption leads to a contradictiion). And then Explain why it is so bad that a contradiction can be proved somewhere.( why ncant we just ignore the fact that in some remote corner of set theory this contradiction arises?), And Finally my 3rd question abotu russel paradox iii) Assuming a Statement B, which happens tobe a contradiction, prove that 6=21.
The article referred to by CaptainBlack is nice. I will however try to discuss it in easier, if perhaps more imprecise terminology.
Frege's original conception of a set was essentially as follows: anything you can describe is a set. The set of all natural numbers. The set of all elephants. The set of all tea cups. The set of all things that are green. etc, etc.
So, let R denote "the set of all sets which are not members of themselves." There is nothing a priori in Frege's original conception which would disallow such a thing. Now ask the following question: is R a member of itself?
However you try to answer this question leads to an inevitable contradiction. Suppose R is a member of itself. Then, by definition, R is not a member of itself (since it contains precisely those sets which are not members of themselves). Conversely, if R is not a member of itself, then by definition it is a member of itself. There is the paradox.
Now you ask, what is so bad about a paradox in a formal system? This is called inconsistency. And the problem is, you can prove (and disprove) absolutely anything once you've encountered one.
Suppose you have a statement A, and that you've established the contradiciton that both A and ~A (read NOT A) are true. Let P be any other statement. The statement "A or ~A" is a tautology. Thus "~P implies (A or ~A)" is also a tautology. This is equivalent to, by contraposition, "~(A or ~A) implies P". But "~(A or ~A)" is equivalent to "A and ~A". Thus we have "(A and ~A) implies P". But "A and ~A" is a true statement! And since "(A and ~A) implies P", we must conclude that P is true.
So we see that a paradox is never self-contained. Once you have both a statement and its negative being true, you can prove absolutely any other statement (and its negative).
ok well nop1 seemed to answer this so maybe u cansee if my proof is correct ore there is some mistake somewhere
PROOF.
Step 1. let p(x) be the statement "2=7"
step 2. then let S be the set such that S = {x: p(x)}
s3. Assume p(x)
s4. Then p(x) is such an x such that the set S is true, but we know 2 doesnt equal 7. Then S is not true which results in a contradiction. This is an example of the Russel Paradox.
Brain_103 gave you a very clear discussion of this problem.
I will just add a few comments.
If you have access to a mathematics library find NAÏVE SET THEORY by Paul Halmos. That book has one of the clearest discussions of the problem. As Halmos shows that with proper axioms one can prove that “nothing contains everything”. In other words, the whole problem is one of extensionality just as the Brain said.
To answer TPH’s question: paradoxes are never settled. We just improve the axioms to avoid the problem. In the newer axiom systems, most use ZF or Quine, both avoid the Russell paradox.
Having read you question a number of times, I still am unsure about #3.
The sentence, If then is a true statement.
You see a of property of implication is: A false statement implies any statement.
Thus, if we allow any paradox to remain in the system then any is true.
I think I understand what you said. I just want to be sure on it. You are saying that the paradoxes are not settled but they do not arise because the ZFC axioms forbid such a set to exist. Thus, we did not settle the paradox rather we just ignore it and it never bothers us because of the consistency of the axioms?
There is a problem with the "woman" answer.
The original statement of the paradox says:
The problem statement clearly specifies that the barber is a man.Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men who do not shave themselves.
Under this scenario, we can ask the following question: Does the barber shave himself?
-Dan