Russell's paradox and logical errors in the proof

  • #26
Dear Stephen Tashi,

Your arguments show that R cannot exist because assuming it exists, contradicts the axiom that "If S is a set and x is an element then either x is a member of S or x is not a member os S".

My arguments are to demonstrate that there is a logical error in each of the two logical arguments in Russell's paradox. :) Please see my message#20 in this thread :)

Yours,

Dan
 
  • #27
rubi
Science Advisor
847
348
Dear Dan,

I don't understand what you are trying to say. I have shown (using conventional predicate logic) that the statement [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex] implies a contradiction ([itex](\exists R\forall x (x\in R \leftrightarrow x\notin x))\rightarrow\bot[/itex]).

Now i need you to answer the following question: Do you agree that this derivation is correct?

If the answer is no: Which step in the derivation is wrong? (You should be able to answer this question in one sentence, because you only need to point at one single step!) To make it even more clear, here are is the derivation again:
1. [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex]
2. [itex]\forall x (x\in S \leftrightarrow x\notin x)[/itex] (from 1; if there exists such an R, i can assign it to an unused variable)
3. [itex]S\in S \leftrightarrow S\notin S[/itex] (from 2; if it holds for all x, it also holds for R)
4. [itex]\bot[/itex] (from 3; it's a contradiction)
5. [itex](\exists R\forall x (x\in R \leftrightarrow x\notin x))\rightarrow\bot[/itex] (by the decuction theorem)

If the answer is yes, you have the following options: Either you accept that [itex]\exists R\forall x (x\in R \leftrightarrow x\notin x)[/itex] is false or you are willing to accept contradictory statements in your axiomatic system. Which one of these options do you choose?
 
Last edited:
  • #28
Dear Rubi,

You correctly applied the axiom of predicate calculus.

Have you read my message #20 above? Have you understood everyithing there?

Yours,

Dan
 
  • #29
rubi
Science Advisor
847
348
Have you read my message #20 above?
Yes, i have.

Have you understood everyithing there?
No i haven't. That's why i want you to answer my question. I'm not willing to spend more time on this if you refuse to answer my questions.
 
  • #30
Dear Rubi,

I answered your first question as well as I could.
"∃R∀x(x∈R↔x∉x) is false" - this is correct, of course :) There is no such R :). I am not making any alternative "my" axiomatic system )))).

What exactly (which line) have you failed to understand in my message#20?

Yours,

Dan
 
  • #31
rubi
Science Advisor
847
348
"∃R∀x(x∈R↔x∉x) is false" - this is correct, of course :) There is no such R :).
Then we are done, because we came to the same conclusion as Russel did. There is no way to define the set [itex]R = \{x:x\notin x\}[/itex] in usual predicate logic without implying a contradiction. This is exactly Russels paradox.

What exactly (which line) have you failed to understand in my message#20?
"Both assumptions (R∈R is true and R∈R is false) contradict the definition of R" doesn't make sense, because you can't contradict a definition. Your use of language is wrong.
 
  • #32
Dear Rubi,

Then we are done, because we came to the same conclusion as Russel did. There is no way to define the set [itex]R = \{x:x\notin x\}[/itex] in usual predicate logic without implying a contradiction. This is exactly Russels paradox.

Russell's paradox is like:
Let R be the set of all sets that are not members of themselves. Then it is a member of itself if and only if it is not a member of itself. - paradoxical incoherence.

The usual conclusion that we make from the paradox is that there is no such R.

"Both assumptions (R∈R is true and R∈R is false) contradict the definition of R" doesn't make sense, because you can't contradict a definition. Your use of language is wrong.

"Let ABC be a triangle. Suppose it has four angles." :) - well, this is a sort of joke ))
If seriously, enter "contradicts the definition" into google serach and find some math texts ;)

Yours,

Dan
 
  • #33
rubi
Science Advisor
847
348
Russell's paradox is like:
Let R be the set of all sets that are not members of themselves. Then it is a member of itself if and only if it is not a member of itself. - paradoxical incoherence.
We proved this and you already said that you accept the derivation.

The usual conclusion that we make from the paradox is that there is no such R.
That is true. If a statement leads to a contradiction, then it must be false. You also agreed on this.

If you really agree that one can derive a contradiction from the assumption of the existence of the Russel set, then I don't see what problems remain.

"Let ABC be a triangle. Suppose it has four angles." :) - well, this is a sort of joke ))
If seriously, enter "contradicts the definition" into google serach and find some math texts ;)
What I was trying to say was that it is irrelevant for the proof whether something contradicts the definition of the Russel set. [itex]R\in R[/itex] is a well-formed formula and thus you are allowed to use it according to the rules of predicate logic.
 
  • #34
Dear Rubi,

I said you correctly used the axiom of predicate logic ))

Actually, I suspect you did not even try to read my message#20 after the sentence "Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves)"
Which makes further discussion with you senseless ))

Yours,

Dan
 
  • #35
rubi
Science Advisor
847
348
I said you correctly used the axiom of predicate logic
So either you agree that the existence of R implies [itex]R\in R\leftrightarrow R\notin R[/itex] or you claim that using predicate logic is not a valid way of reasoning. I have still not found out, which of these two options you advocate.

Actually, I suspect you did not even try to read my message#20 after the sentence "Both assumptions (R∈R is true and R∈R is false) contradict the definition of R (Let R be the set of all sets that are not members of themselves)"
Which makes further discussion with you senseless
I have read it, but i don't see what the problem is. Let's look at this part for example:
Argument 1
Premise 1: Let R be the set of all sets that are not members of themselves R = {x: x∉x}
Premise 2 (assumption): Suppose R ∈ R.
Conclusion: Then, according to its definition, R ∉ R.
We can easily phrase this in predicate logic:
1. [itex]\forall x (x \in R \leftrightarrow x\notin x)[/itex] (this is your first premise)
2. [itex]R \in R[/itex] (this is your second premise)
3. [itex]R \in R \leftrightarrow R\notin R[/itex] (from 1 by setting [itex]x=R[/itex])
4. [itex]R \notin R[/itex] (from 2 and 3 by modus ponens)

This is completely valid reasoning. [itex](\forall x (x \in R \leftrightarrow x\notin x))\rightarrow R\notin R[/itex] is a theorem.
 
  • #36
Dear Rubi,

Yes, in predicate logic Argument 1 is a theorem. I know it ))

My point is that assumption R ∈ R contradicts the definition of R because if R ∈ R, R includes a member that is included in itself (R itself is such a member).
Or, symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R

That is the second premise contradicts the first one. "Contradictory premises" logical error.
Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).
The reasoning on `legless Dan' contains the same logical error as the Russell's paradox does: the second premise contradicts the definition which is the first premise though to make the conclusion both premises are used.

Yours,

Dan
 
  • #37
rubi
Science Advisor
847
348
My point is that assumption R ∈ R contradicts the definition of R
And why is that a bad thing? You can assume whatever you want. The entire point of Russels paradox is to show that you end up with contradictions. If you don't want contradictions, then just don't assume that the Russel set exists.

symbolically: R ∈ R → ∃ y: y ∈ R ∧ y ∈ y → R ≠ {x: x∉x}: indeed, such y exists as we can take y = R
This is just yet another contradiction that you can derive if you assume the existence of the Russel set. Remember that if you have one contradiction in your axioms, you can derive every statement that you can think of.
 
  • #38
22,129
3,297
Well-know American logician H. Curry once expressed the opinion that in spite of the fact that it seemed to be absolutely impossible to explain Russell's paradox in terms of conventional 19th century logic, it may happen in modern days that some error would be identified.

Can you give a reference where Curry said that?
 
  • #39
Dear Micromass,

Hi! Remember me? ))

The reference is in the reference section of the paper)) It is a book by a Russian logician Ivin. He mentions this. By the way I am not quoting.

Yours,

Dan
 
Last edited:
  • #40
Dear Rubi,

It is a logical error to make conclusion such a way )) Which is not a good thing )))

Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).

Well, if you consider the example to be OK, Russell's paradox is OK as well :smile:

Yours,

Dan
 
Last edited:
  • #41
rubi
Science Advisor
847
348
It is a logical error )) Which is not a good thing )))
It just shows that you can derive logical if you assume the existence of the Russel set, you can derive contradictions. This is nothing new. It has been known for more than 100 years now. The conclusion is that you should not assume the existence of the Russel set or any axiom that implies the existence of the Russel set.

Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).
You just showed that if you assume contradictory statements, then you end up with contradictions.

Well, if you consider the example is OK, Russell's paradox is OK as well :smile:
Depends on what you mean by "Russel's paradox is OK". If you mean that one can safely assume the existence of the Russel set, then it's not OK. But if you mean that the existence of the Russel set implies contradictions, then it's perfectly OK, because nobody forces us to assume it's existence. We can just deny it and everyone is happy until someone finds a new paradox. You can easily disprove the existence of the Russel set in modern set theory.
 
  • #42
Dear Rubi,

Reasoning on legless Dan is an example of "contradictory premises" logical error.

The same one as in Russell's paradox.

Yours,

Dan
 
  • #43
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,872
417
Let's see if I understand what this thread is about...

We all agree that the following statement is a theorem: There's no set ##R## such that ##R=\{x\,|\,x\notin x\}##.

(I proved it in #13, and rubi did it in several of his posts).

Dan is arguing that some attempted proofs are flawed, because they're making two contradictory assumptions. The two assumptions are ##R\in R## and ##R\notin R##.

In that case, I think Dan's concern is very easy to answer: No one is assuming that both of those statements are true.

Dan, I don't know how can you continue to claim that there's an error in "Russell's paradox" after agreeing that there are valid ways to prove the theorem. Do you mean something different from that theorem when you say "Russell's paradox"?
 
  • #44
rubi
Science Advisor
847
348
Reasoning on legless Dan is an example of "contradictory premises" logical error.

The same one as in Russell's paradox.
You can apply the deductive rules of predicate logic to any set of axioms you like. Some choices for your axioms might give you contradictions, though. The conclusion is that you should choose other axioms. This is exactly what happened: We found that naive set theory implies a contradiction (Russels paradox), so we rejected it and invented ZFC instead. Now we hope that nobody finds a contradiction in ZFC anymore.
 
  • #45
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,872
417
Dan, I'm giving you one last chance to answer what I said in post #43 before I close the thread. Don't copy and paste from earlier. Is your entire paper, and this entire thread, based on the idea that some people are assuming both ##R\in R## and ##R\notin R##?
 
  • #46
Dear Fredrik,

No!!! You missed the point, sorry. Please try to read my message#20 ))
I thought you missed it - that is why I wanted to put it here again...

Yours,

Dan
 
  • #47
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,872
417
I read post #20 again. OK, I think I see what you're saying: The two premises that you say are contradictory are not ##R\in R## and ##R\notin R##. It's (in argument 1) ##R\in R## and ##R=\{x\,|\,x\notin x\}##.

If that's what you meant, then my answer is that this is irrelevant, since no one considers argument 1 (or argument 2) to be a complete proof. ##R\in R## is not one of the assumptions that go into the proof. There's only one assumption (which is made only to obtain a contradiction), and that is that ##R=\{x\,|\,x\notin x\}##.

Also, no one is saying that arguments 1 and 2 together prove the theorem, because then we'd have three assumptions that don't agree with each other.

Do you agree that there is a valid proof of the theorem I stated in #43? (I'm thinking of the proof I posted in #13 and the similar proofs posted by rubi).
 
  • #48
22,129
3,297
Like in the below:
Premise 1: Let Dan be a completely legless man
Premise 2: Suppose, Dan’s right ankle is severely bleeding
Conclusion: Then, according to his definition, Dan should be taken to an emergency for legless people (for his ankle bleeding).

These are not well-formed formula's, so I don't see the point of considering these statements. They imply nothing about math. Math only deals with well-formed formulas.
 
  • #49
Dear Fredric,

If that's what you meant, then my answer is that this is irrelevant, since no one considers argument 1 (or argument 2) to be a complete proof.

Nor do I! My point is that each of them separately is a fallacious, that is, containing a logical error, argument.

Do you agree that there is a valid proof of the theorem I stated in #43? (I'm thinking of the proof I posted in #13 and the similar proofs posted by rubi).

Well it depends on what you mean by valid...
As to me it contains logical errors but at the same time it is a theorem in predicate logic (that is quite deducable, valid and correct thing in this formal system).

Yours,

Dan
 
  • #50
Dear Micromass,

This is an example of the logical error.

I could not invent the example of this logical error in math different from Russell's paradox-like things...

Yours,

Dan
 

Related Threads on Russell's paradox and logical errors in the proof

  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
13
Views
7K
  • Last Post
Replies
10
Views
1K
Replies
1
Views
3K
Replies
5
Views
3K
Replies
9
Views
4K
Replies
9
Views
7K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
Top