Resolution of Russell's and Cantor's paradoxes

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In summary, the conversation discusses a paper that presents a resolution to the Russel's and Cantor's paradoxes within a non-axiomatic approach. The paper argues that the contradiction only arises due to the general principle in naive set theory that allows for the creation of sets with any formula, leading to paradoxes. The paper proposes a different approach using classical logic to avoid these paradoxes. The conversation also includes a discussion on the concept of a "replica" of the set of sets and the importance of consistent definitions in mathematics. The author of the paper asks for feedback and comments on the content.
  • #36
DanTeplitskiy said:
Dear Micromass,

If we can derive anything from a contradiction will the thing we derive this way be of any value for us? Will this thing be logically grounded if this could be anything?

Yes, it will be logically grounded. The truth table of the implication is logically sound. There are no problems with this.

For example, we can derive "he should be taken to the nearest hospital for legless people And he should not be taken to the nearest hospital for legless people" from any contradiction (including the above- "Dan is completely legless" AND "His right ankle is bleeding"). Right?

Yes, from a false hypothesis, we can derive everything. In latin: "ex falso sequitur quodlibet". This is not a flaw in logic, however it might be confusing to some. For example, the following is also true

"Dan is completely legless and his right ankle is bleeding" THEN "1+1=3"

This does not make 1+1=3 true. It only makes the implication true.
 
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  • #37
micromass said:
Yes, it will be logically grounded. The truth table of the implication is logically sound. There are no problems with this.



Yes, from a false hypothesis, we can derive everything. In latin: "ex falso sequitur quodlibet". This is not a flaw in logic, however it might be confusing to some. For example, the following is also true

"Dan is completely legless and his right ankle is bleeding" THEN "1+1=3"

This does not make 1+1=3 true. It only makes the implication true.

Dear Micromass,

I completely agree that the implication is true. My point is if we get something this way can this something (not the implication but what we get from it) be considered to be logically grounded?

Yours,

Dan
 
  • #38
DanTeplitskiy said:
Dear Micromass,

I completely agree that the implication is true. My point is if we get something this way can this something (not the implication but what we get from it) be considered to be logically grounded?

Yours,

Dan

Well, everything we get from following logical inference rules will be logically valid. So yes.
 
  • #39
micromass said:
Well, everything we get from following logical inference rules will be logically valid. So yes.

Dear Micromass,

Do you mean if we get "1+1=3" from some contradiction, "1+1=3" by itself is logically grounded?

As far as I understand if something can be derived only from some contradiction it is not logically valid at all.

Yours,

Dan
 
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  • #40
DanTeplitskiy said:
Dear Micromass,

Do you mean if we get "1+1=3" from some contradiction "1+1=3" by itself is logically grounded?

Yours,

Dan

Yes, provided that the contradiction is logically grounded.
 
  • #41
micromass said:
Yes, provided that the contradiction is logically grounded.

Dear Micromass,

If contradiction is just given to you, like, say, the following way:
If (this forum exists and it does not exist) Then 1+1=3, will you consider "1+1=3" logically grounded?

Yours,

Dan
 
  • #42
DanTeplitskiy said:
Dear Micromass,

If contradiction is just given to you, like, say, the following way:
If (this forum exists and it does not exist) Then 1+1=3, will you consider "1+1=3" logically grounded?

Yours,

Dan

No, since the premise is false.
 
  • #43
Dear Micromass,

That is my point!

If we analyze the formula: R = {x: x∉x} ⇒ R ∈ R ↔ R ∉ R we can see that it is actually the following one: (R = {x: x∉x} And R ≠ {x: x∉x}) ⇒ R ∈ R ↔ R ∉ R.
That, in my opinion, makes "R ∈ R ↔ R ∉ R" logically ungrounded.

Yours,

Dan
 
  • #44
DanTeplitskiy said:
Dear Micromass,

That is my point!

If we analyze the formula: R = {x: x∉x} ⇒ R ∈ R ↔ R ∉ R we can see that it is actually the following one: (R = {x: x∉x} And R ≠ {x: x∉x}) ⇒ R ∈ R ↔ R ∉ R.
That, in my opinion, makes "R ∈ R ↔ R ∉ R" logically ungrounded.

Yours,

Dan

It's not logically ungrounded, since it can be logically proven. The premise is true, thus so must the conclusion.
 
  • #45
micromass said:
It's not logically ungrounded, since it can be logically proven. The premise is true, thus so must the conclusion.

Dear Micromass,

Sorry but how can this (R = {x: x∉x} And R ≠ {x: x∉x}) be a true premise?!

Yours,

Dan
 
  • #46
DanTeplitskiy said:
Dear Micromass,

Sorry but how can this (R = {x: x∉x} And R ≠ {x: x∉x}) be a true premise?!

Yours,

Dan

If it can be proven with inference rules then it is a true premise. And since [itex]R\in R~\leftrightarrow R\notin R[/itex] can be proven from the axioms and the inference rules, means that it is true.
 
  • #47
micromass said:
If it can be proven with inference rules then it is a true premise. And since [itex]R\in R~\leftrightarrow R\notin R[/itex] can be proven from the axioms and the inference rules, means that it is true.

Dear Micromass,

What I mean is "R ∈ R ↔ R ∉ R" can be derived only from (R = {x: x∉x} And R ≠ {x: x∉x}) .
Does not this alone make "R ∈ R ↔ R ∉ R" logically ungrounded?

Yours,

Dan
 
  • #48
DanTeplitskiy said:
Dear Micromass,

What I mean is "R ∈ R ↔ R ∉ R" can be derived only from (R = {x: x∉x} And R ≠ {x: x∉x}) .
Does not this alone make "R ∈ R ↔ R ∉ R" logically ungrounded?

Yours,

Dan

Why can it be derived only from that?? It can also be derived from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex](R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\})[/itex]. In fact, that is still true, but there's no need for it.
 
  • #49
micromass said:
Why can it be derived only from that?? It can also be derived from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex](R=\{x~\vert~x\notin x\}~\text{and}~R\neq \{x~\vert~x\notin x\})[/itex]. In fact, that is still true, but there's no need for it.

Dear Micromass,

If R ∈ R R ≠ {x: x∉x}. That is so because under the assumption “R ∈ R” R includes a member that is included in itself (R itself is such a member). Do you agree?

That is, when we write
R = {x: x∉x} And R ∈ R -> R ∉ R
it is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R -> R ∉ R which is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} ⇒ R ∈ R -> R ∉ R

Yours,

Dan
 
  • #50
DanTeplitskiy said:
Dear Micromass,

If R ∈ R R ≠ {x: x∉x}. That is so because under the assumption “R ∈ R” R includes a member that is included in itself (R itself is such a member). Do you agree?

That is, when we write
R = {x: x∉x} And R ∈ R -> R ∉ R
it is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R -> R ∉ R which is equivalent to
R = {x: x∉x} And R ≠ {x: x∉x} ⇒ R ∈ R -> R ∉ R

Yours,

Dan

Yes, I agree that [itex]R\in R~\rightarrow~R\notin R[/itex] can be derived from that. But why can it only be derived from that?
 
  • #51
micromass said:
Yes, I agree that [itex]R\in R~\rightarrow~R\notin R[/itex] can be derived from that. But why can it only be derived from that?

Dear Micromass,

That is so because:
1. The only way to derive R ∉ R is (R = {x: x∉x} And R ∈ R). Right?
2. (R = {x: x∉x} And R ∈ R) is equivalent to (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). Right?
3. From 1. and 2. you can see that when you derive R ∉ R from (R = {x: x∉x} And R ∈ R) you actually derive R ∉ R from (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). You can not avoid it.

Like if you know that 2+2 = 4 you know that every time you use "2+2" you actually use "4". You can not avoid it either.

Yours,

Dan
 
  • #52
DanTeplitskiy said:
Dear Micromass,

That is so because:
1. The only way to derive R ∉ R is (R = {x: x∉x} And R ∈ R). Right?
2. (R = {x: x∉x} And R ∈ R) is equivalent to (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). Right?
3. From 1. and 2. you can see that when you derive R ∉ R from (R = {x: x∉x} And R ∈ R) you actually derive R ∉ R from (R = {x: x∉x} And R ≠ {x: x∉x} And R ∈ R). You can not avoid it.

Like if you know that 2+2 = 4 you know that every time you use "2+2" you actually use "4". You can not avoid it either.

Yours,

Dan

Well no, what you have said in (1) is that we can derive [itex]R\in R~\leftrightarrow~R\notin R[/itex] from [itex]R=\{x~\vert~x\notin x\}[/itex]. There is no need for [itex]R\neq \{x~\vert~x\notin x\}[/itex]...
 
  • #53
Dear Micromass,

Thanks for reply!

Let me explain my point another way.

"Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number.
If N >= 2 then N+1> N that is N is not the greatest natural number."

Do you consider the above proof that there is no greatest natural number to be OK? (just in case - I do)

Yours,

Dan
 
  • #54
DanTeplitskiy said:
Dear Micromass,

Thanks for reply!

Let me explain my point another way.

"Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number.
If N >= 2 then N+1> N that is N is not the greatest natural number."

Do you consider the above proof that there is no greatest natural number to be OK? (just in case - I do)

Yours,

Dan

Yes, that looks OK.
 
  • #55
micromass said:
Yes, that looks OK.

Dear Micromass,

What if I take this: "Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number."

and change it to that: "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number."

Will the latter be the OK implication to you? (just in case: to me it definitely won't)

Yours,

Dan
 
  • #56
DanTeplitskiy said:
Dear Micromass,

What if I take this: "Let's denote the greatest natural number by N.
If N < 2 then N is not the greatest natural number."

and change it to that: "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number."

Will the latter be the OK implication to you? (just in case: to me it definitely won't)

Yours,

Dan

No, the latter is wrong. I don't see where you're taking me.
 
  • #57
micromass said:
No, the latter is wrong. I don't see where you're taking me.

Dear Micromass,

We are close to the point:smile:

Remember this? :wink: : Let p="Dan is completely legless and his right ankle is bleeding", then p is false. Thus the implication [itex]p\rightarrow q[/itex] holds true never the less.

What if we apply the same reasoning to the case we are discussing now:

Let p = "N - the greatest natural number And N < 2", then p is false.
Thus the implication [itex]p\rightarrow q[/itex] holds true never the less.

Does the implication "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number." seem to be OK now? (just in case: to me it still doesn't :smile:)

Yours,

Dan

P. S. Micromass, please do not take the above for any kind of personal attack. :shy: It is not that - you are a nice guy! I am just eager to explain my point to you now.
 
  • #58
DanTeplitskiy said:
Does the implication "Let's denote the greatest natural number by N.
If N < 2 then N is the greatest natural number." seem to be OK now? (just in case: to me it still doesn't :smile:)

Well, this is actually ok by me. Considering that there is no natural number, and thus considering that N<2 is actually false. Thus the implication

N<2 => N is the greatest natural number

is ok.
 
  • #59
micromass said:
Well, this is actually ok by me. Considering that there is no natural number, and thus considering that N<2 is actually false. Thus the implication

N<2 => N is the greatest natural number

is ok.

Dear Micromass,

Does it mean that you changed your mind and now you consider the reasoning
"Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number."
to be OK? :confused:

Yours,

Dan
 
  • #60
DanTeplitskiy said:
Dear Micromass,

Does it mean that you changed your mind and now you consider the reasoning
"Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number."
to be OK? :confused:

Yours,

Dan

I did not change my mind. Both

"If N<2, then N is the greatest natural number"

as

"If N<2, then N is not the greatest natural number"

are ok. This is of course a contradictory situation and this leads to the conclusion that N<2 is false.

I mean, if you define "N=the greatest natural number", then for every statement p holds that

If p, then N is the greatest natural number.

is always true! By definition.
 
  • #61
micromass said:
I did not change my mind. Both

"If N<2, then N is the greatest natural number"

as

"If N<2, then N is not the greatest natural number"

are ok. This is of course a contradictory situation and this leads to the conclusion that N<2 is false.

I mean, if you define "N=the greatest natural number", then for every statement p holds that

If p, then N is the greatest natural number.

is always true! By definition.

Dear Micromass,

Sorry, pal. That is not what I was asking you about.

You confirmed that "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." "...is wrong" - please see your own message of T 03:12 PM.

My question is:
Do you consider "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." to be the OK implication now?

Yours,

Dan
 
  • #62
DanTeplitskiy said:
Dear Micromass,

Sorry, pal. That is not what I was asking you about.

You confirmed that "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." "...is wrong" - please see your own message of T 03:12 PM.

My question is:
Do you consider "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." to be the OK implication now?

Yours,

Dan

OK, I was wrong in that message. If N is the greatest natural number, then

N<2 ==> N is the greatest natural number

is correct. So yes, I changed my mind :smile:
 
  • #63
micromass said:
OK, I was wrong in that message. If N is the greatest natural number, then

N<2 ==> N is the greatest natural number

is correct. So yes, I changed my mind :smile:

Dear Micromass,

No! You were not wrong in that message. You were quite right!

Just think of it. If things like "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." were considered as OK mathematical reasoning where would it take us to? To what "proofs"?

To undertand this better, do not try to take the thing apart into pieces and try to consider it as a whole reasoning. How could such a thing be a valid one? Your first reaction was quite right because it was based on a "common mathematical sense". Your "common mathematical sense" tells you that if "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." is OK reasoning then we can prove that N is the greatest natural number which is absurd!

Yours,

Dan
 
  • #64
DanTeplitskiy said:
Dear Micromass,

No! You were not wrong in that message. You were quite right!

I was wrong :smile:

Just think of it. If things like "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." were considered as OK mathematical reasoning where would it take us to? To what "proofs"?

It would clearly take us to absurd proofs. But this is of course because such a greatest natural number doesn't exist. In the same way, you can prove

"1+1=3 ==> 1+1=3"

It's absurd, but it's true.

To undertand this better, do not try to take the thing apart into pieces and try to consider it as a whole reasoning. How could such a thing be a valid one? Your first reaction was quite right because it was based on a "common mathematical sense". Your "common mathematical sense" tells you that if "Let's denote the greatest natural number by N. If N < 2 then N is the greatest natural number." is OK reasoning then we can prove that N is the greatest natural number which is absurd!

Common sense is not something that we should do mathematics with. Mathematics should be done with rigorous arguments. Common sense can be wrong, rigorous arguments can not.

And yes, from "Let's denote the greatest natural number by N", we can prove that "N is the greatest natural number". I see no problem in this.
 
  • #65
micromass said:
I was wrong :smile:



It would clearly take us to absurd proofs. But this is of course because such a greatest natural number doesn't exist. In the same way, you can prove

"1+1=3 ==> 1+1=3"

It's absurd, but it's true.



Common sense is not something that we should do mathematics with. Mathematics should be done with rigorous arguments. Common sense can be wrong, rigorous arguments can not.

And yes, from "Let's denote the greatest natural number by N", we can prove that "N is the greatest natural number". I see no problem in this.

Dear Micromass,

Do you mean that if we start our reasoning with, say, "Let m be the greatest prime..." we can prove that such a thing like the greatest prime exists??

Yours,

Dan
 
  • #66
DanTeplitskiy said:
Dear Micromass,

Do you mean that if we start our reasoning with, say, "Let m be the greatest prime..." we can prove that such a thing like the greatest prime exists??

Yours,

Dan

Yes. And if we start of with "Let m be the greatest prime", then we can also prove that 1+1=3. This is absurd, and thus the original statement about existing a greatest prime was wrong.
 
  • #67
micromass said:
Yes. And if we start of with "Let m be the greatest prime", then we can also prove that 1+1=3. This is absurd, and thus the original statement about existing a greatest prime was wrong.

Dear Micromass,

If we could have a proof that "there is a greatest prime" (by "proof" I mean the widely accepted one - otherwise it is not a "proof" for math) we would have it! But we do not and you know it quite well!

Why don't we have such a proof in mathematics, Micromass? :smile::wink:

Yours,

Dan
 
  • #68
This has nothing to do with common sense failing (logic is common sense!). A statement "A implies B" does not assert A nor B. The confusion may arise by that in casual talk we sometimes omit asserting A and say "If A, then B", and taking A as understood by context (there is nothing wrong with this). Coupled by the fact that we almost never say "If A, then B" if A is known to be false, it will sound weird.
 
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  • #69
disregardthat said:
This has nothing to do with common sense failing (logic is common sense!). A statement "A implies B" does not assert A nor B. The confusion may arise by that in casual talk we sometimes omit asserting A and say "If A, then B", and taking A as understood by context (there is nothing wrong with this).

Dear Disregardthat,

Sorry, pal. I put the question to Micromass. Does what you tell me have anything to do with this (just in case: with the question, not with the topic)? :smile:

Yours,

Dan
 
  • #70
DanTeplitskiy said:
Dear Disregardthat,

Sorry, pal. I put the question to Micromass. Does what you tell me have anything to do with this (just in case: with the question, not with the topic)? :smile:

Yours,

Dan

I was responding to

Common sense is not something that we should do mathematics with. Mathematics should be done with rigorous arguments. Common sense can be wrong, rigorous arguments can not.

and I think that was micromass' statement.
 
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