Russell's paradox and logical errors in the proof

[rubi]

"∃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 R = \{x:x\notin x\} 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.

 

[DanTeplitskiy]

Dear Rubi,

Then we are done, because we came to the same conclusion as Russel did. There is no way to define the set R = \{x:x\notin x\} 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

 

[rubi]

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. R\in R is a well-formed formula and thus you are allowed to use it according to the rules of predicate logic.

 

[DanTeplitskiy]

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

 

[rubi]

I said you correctly used the axiom of predicate logic

So either you agree that the existence of R implies R\in R\leftrightarrow R\notin R 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. \forall x (x \in R \leftrightarrow x\notin x) (this is your first premise)
2. R \in R (this is your second premise)
3. R \in R \leftrightarrow R\notin R (from 1 by setting x=R)
4. R \notin R (from 2 and 3 by modus ponens)

This is completely valid reasoning. (\forall x (x \in R \leftrightarrow x\notin x))\rightarrow R\notin R is a theorem.

 

[DanTeplitskiy]

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

 

[rubi]

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.

 

[micromass]

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?

 

[DanTeplitskiy]

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

 

[DanTeplitskiy]

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

Yours,

Dan

 

[rubi]

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

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.

 

[DanTeplitskiy]

Dear Rubi,

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

The same one as in Russell's paradox.

Yours,

Dan

 

[Fredrik]

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"?

 

[rubi]

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.

 

[Fredrik]

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 $R\notin R$?

 

[DanTeplitskiy]

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

 

[Fredrik]

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).

 

[micromass]

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.

 

[DanTeplitskiy]

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

 

[DanTeplitskiy]

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

 

[micromass]

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

A logical error must be written in the language of logics. That is: it must be written as a well-formed formula. If you can't do this, then you're not doing math.

The fact that you can't write your arguments in well-formed formulas should already tell you that your entire point is invalid.

 

[DanTeplitskiy]

Dear Micromass,

Are you sure that when we taik about logical errors we have to talk about well-formed formula (otherwise we can not do it) !? That is a thing I can not agree with))

Yours,

Dan

 

[micromass]

Dear Micromass,

Are you sure that when we taik about logical errors we have to talk about well-formed formula (otherwise we can not do it) !? That is a thing I can not agree with))

Yours,

Dan

If you don't agree with it, then you're not doing math. It's simple as that.
So this means that your paper is not math, but rather something else. Nothing wrong with that, but it's good to realize.

 

[DanTeplitskiy]

Dear Micromass,

If you don't agree with it, then you're not doing math. It's simple as that.

It is your way of thinking that one can not give an example of a logical error in math paper with things that are not well-formed formula. Sorry ))

Yours,

Dan

 

[micromass]

Dear Micromass,



It is your way of thinking that one can not give an example of a logical error in math paper with things that are not well-formed formula. Sorry ))

Yours,

Dan

Sure, it could be due to my way of thinking. Or it could be due to the fact that the entire paper is complete crackpottery. You had this paper for close to 2 years and you still didn't manage to publish in a reputable math journal. This makes me think that the fault is due to your paper and not due to me. Sorry to say it.

 

[DanTeplitskiy]

Dear Micromass,

Are you sure that I have been trying to place it all the time since then?

Yours,

Dan

 

[micromass]

Dear Micromass,

Are you sure that I have been trying to place it all the time since then?

Yours,

Dan

I could indeed see you capable of arguing over this thing on internet forums for 2 years. Not sure I see the point though. Well, if you enjoy yourself, I guess that's all that matters...

Thing is that nobody will take you serious until you publish. So I would start doing that if I were you.

 

[DanTeplitskiy]

Dear Micromass,

Are you sure that I have been arguing over this thing all the time since then?

The thing of publishing is a complicated matter, you know ))

The goal was not to enjoy myself though. I wanted a discussion over my point...

Yours,

Dan

 

[Fredrik]

This thread is obviously not going anywhere, so I think it's time to end it.