The overlooked part of Russell's paradox

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RJG
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Re: The overlooked part of Russell's paradox

Post by RJG »

Steve3007 wrote:X is a reference to that.
What is "that"? ..."X"? ...are you saying X is a reference to X?

Which X is the reference, and which X is the X?

No matter how you want to wordplay it, you will always have two different meanings for X.

X as X, and X as the reference, or
X as the reference, and X as the reference for the reference.

Having two meanings for X is the error.
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Thomyum2
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Re: The overlooked part of Russell's paradox

Post by Thomyum2 »

RJG wrote: June 11th, 2021, 7:55 am
Terrapin Station wrote:Again, that's thinking about them as "objects."
It doesn't matter if they are "objects" or "abstract ideas" or "X". X<X is still logically impossible (for whatever you want to assign as X). The logic still holds true.
Try this - I send my children to the grocery and I want my list back, so I tell them I want you to bring me back these three things:

1) an apple
2) an orange
3) this list

So when they get back from the store, they hand me these three things. The list contains itself. What's illogical about that?
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Steve3007
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Re: The overlooked part of Russell's paradox

Post by Steve3007 »

Steve3007 wrote:The list is {X, 1, 2}. X is a reference to that.
RJG wrote:What is "that"? ..."X"? ...are you saying X is a reference to X?
https://dictionary.cambridge.org/gramma ... ammar/that
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RJG
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Re: The overlooked part of Russell's paradox

Post by RJG »

Steve, it seems that you agree that "boxes" can't contain themselves, but yet you believe "lists" can contain themselves.

I assume you believe X<X is logically impossible. But yet if X represents a "list", then it is somehow possible??? ...and if X represents a "box", then it is not possible??

What's the difference? Logic is logic.


********
Thomyum2 wrote: The list contains itself.
Not so. The list contains the words "this list", not the actual list itself.
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Re: The overlooked part of Russell's paradox

Post by Thomyum2 »

RJG wrote: June 11th, 2021, 8:19 am Steve, it seems that you agree that "boxes" can't contain themselves, but yet you believe "lists" can contain themselves.

I assume you believe X<X is logically impossible. But yet if X represents a "list", then it is somehow possible??? ...and if X represents a "box", then it is not possible??

What's the difference? Logic is logic.


********
Thomyum2 wrote: The list contains itself.
Not so. The list contains the words "this list", not the actual list itself.
But that's just the point. The words "this list" is not the actual list. Neither is the piece of paper that the list is written on. The list is the items you designate as being contained in it. When you create a list, you define what is in that list. An if you make the list one of those items, then your list is containing itself.
“We have two ears and one mouth so that we can listen twice as much as we speak.”
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RJG
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Re: The overlooked part of Russell's paradox

Post by RJG »

Thomyum2 wrote:The list contains itself.
RJG wrote:Not so. The list contains the words "this list", not the actual list itself.
Thomyum2 wrote:When you create a list, you define what is in that list. The words "this list" is not the actual list. Neither is the piece of paper that the list is written on. The list is the items you designate as being contained in it.
Agreed. The list is whatever you define it as.

Thomyum2 wrote:And if you make the list one of those items, then your list is containing itself.
Not so. Thom, this is not logically possible. The list (whatever it is) can only contain a 'reference' to itself, and never itself because X<X is logically impossible. (X cannot be outside/inside itself, or before/after itself, or above/below itself, or less than/greater than itself, etc etc).

- A box cannot logically contain itself. But if this box contains a photograph called (labeled) "this box" then this box contains a reference-of-itself; ...it does not contain itself.

- And likewise, a list cannot logically contain itself. But if this list contains the words "this list", then this list contains a reference-of-itself, ...it does not contain itself.

It doesn't matter if X is a "box" or a "list" or something imaginary. X<X is still logically impossible.
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Re: The overlooked part of Russell's paradox

Post by Thomyum2 »

RJG wrote: June 11th, 2021, 2:22 pm
Thomyum2 wrote:When you create a list, you define what is in that list. The words "this list" is not the actual list. Neither is the piece of paper that the list is written on. The list is the items you designate as being contained in it.
Agreed. The list is whatever you define it as.
Thomyum2 wrote:And if you make the list one of those items, then your list is containing itself.
Not so. Thom, this is not logically possible. The list (whatever it is) can only contain a 'reference' to itself, and never itself because X<X is logically impossible. (X cannot be outside/inside itself, or before/after itself, or above/below itself, or less than/greater than itself, etc etc).

- A box cannot logically contain itself. But if this box contains a photograph called (labeled) "this box" then this box contains a reference-of-itself; ...it does not contain itself.

- And likewise, a list cannot logically contain itself. But if this list contains the words "this list", then this list contains a reference-of-itself, ...it does not contain itself.

It doesn't matter if X is a "box" or a "list" or something imaginary. X<X is still logically impossible.
If you agree that I can define my list without restrictions, then you've accept this as a premise. If you later tell me I can't make the list a member of itself, you're changing the premise. Logic only follows the premises - it doesn't dictate them. Does this result in a contradictions? Yes. Is it illogical? No. What you're doing to resolve this contradiction is to go back and put a restriction on the definition - you're adding a new premise by saying that a set cannot be defined in terms of itself as a member.

If I'm using the right terminology, (I'm not a trained mathematician or logician, so perhaps some of those on the forum here can help me out on this) I think this is referred to as 'formalizing' - creating a set of rules around the premises to ensure consistency and that no contradictions logically follow from the premises. As I understand it, this is what Russell and his contemporaries were trying to do (which we seem intent on reenacting here on this thread) - to come up with a complete and coherent set of premises that would be free from contradictions. And after 30 years of debates like this one, it's what Gödel ultimately proved cannot be done - all formal systems of logic are 'incomplete' and will always result in contradiction and paradoxes. I think it's one of the great insights of the time - that there are true statements that cannot be proved. There are limits to what can be proved by logic.
“We have two ears and one mouth so that we can listen twice as much as we speak.”
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RJG
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Re: The overlooked part of Russell's paradox

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Thomyum2 wrote:If you agree that I can define my list without restrictions, then you've accepted this as a premise.
Not all premises are true or possible. Yes, anybody can define a non-sensical premise. But that does not automatically make it true or possible.

For example, you can define the premise "Box X contains itself". But that does not mean that it is true or logically possible. Your premise is false (logically impossible) via the form X<X.

Or you can define the premise "List X contains itself". But that does not mean that it is true or logically possible. Your premise is false (logically impossible) via the form X<X.


*************
Also note: paradoxes are no more real than magic is. Paradoxes (and magic) exist ONLY out of ignorance of understanding.
philosopher19
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Re: The overlooked part of Russell's paradox

Post by philosopher19 »

Steve3007 wrote: June 11th, 2021, 4:43 am It depends if this list is in the house. You didn't say it was. If it is then:
That's not what it depends on. Consider again that I was forming a list of all lists that are in this house that list themselves. I was not forming a list of all list in this house.
Why would you think is optional?
See above and reflect carefully on the underlined.
If the list is in the house and we define it as "the list of all lists in this house that are members of themselves" then it must be a member of itself.

If the list is not in the house and we define it as "the list of all lists in this house that are members of themselves" then it must not be a member of itself.
Whether the list is in the house or not, only determines whether the list is in the house or not. It does not determine whether the list is a member of itself or not. So if I was forming a list of all lists in my house, then if the list is in the house, then it lists itself.

It makes no difference if I form this list (the list of all self-listing lists in my house) inside or outside of my house. If I form this list inside my house, then this list will be include in the list of all lists in my house. Whether or not it should be included in the list of all self-listing lists in my house, is another matter.
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Re: The overlooked part of Russell's paradox

Post by philosopher19 »

Steve3007 wrote: June 11th, 2021, 5:05 am I've shown how it lists itself. It's pretty straightforward. It doesn't amount to "ALL plus one". It amounts to a list in the house of all self-listing lists in the house.
Consider the following:

Call any set that is not a member of itself an X. Call any set that is a member of itself a Y.

Suppose you were trying to form a set of all Xs. So after having ALL Xs, you form a new set. This set cannot be an X because it would amount to saying ALL Xs plus one X. Whilst you can say all Xs in this room plus one X outside this room, you cannot say ALL Xs in an absolute sense, plus another X. That amount to your ALL not truly being ALL. Thus:

You cannot have an X as the set of all Xs, or a Y as the set of all Ys (unless there was only one Y and you took that one to be all). This is because the notion of absolutely all plus one is contradictory (as in there is no more after absolutely all is exhausted). You do, however, have a Y as the set of absolutely all Xs (the set of all sets is that Y).

Proof/support of the above:

Consider x = {x, y, z}. Here, it is contradictory to say x is not a member of itself (because x is in x). It is also contradictory to say x, y, and z are all members of themselves (because x is in x. Consistency with the aforementioned would yield the following: either only x is a member of itself, or x is a member of itself twice, whilst y and z are members of themselves once. The latter is contradictory, the former is consistent).
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Re: The overlooked part of Russell's paradox

Post by philosopher19 »

RJG wrote: June 11th, 2021, 6:46 am
RJG wrote:X cannot logically contain itself. It can only contain a reference to itself. Again, a "member of itself" defies simple logic; and is therefore logically impossible.
philosopher19 wrote:Yes, x < x is impossible precisely because x = x. In this thread, my focus was strictly on self-membership (self-referencing), versus membership of other than self (being referenced by other than the self). It was not focused on self-membership such that x < x.
1. If you accept X with a double meaning; e.g. X (the "list"), and X (the "reference-to-the-list") then there is no contradiction or paradox whatsoever.

2. And if you accept X with a single meaning only; e.g. X (the "list") then again there is no contradiction or paradox whatsoever.

3. The trick/deception of Russell's Paradox comes from mixing/confusing X (the list) with X (the reference). Equating one as the other leads to the seemingly paradoxical.
If you look at my reply to Steve (the one that is directly above this reply to you), I think the problem with Russell's paradox is in suggesting that since you cannot have an X as the set of all Xs whilst you can have a Y of all Ys, you cannot have a set of all sets. Whereas in truth, there's no such thing as a Y of all Ys, nor an X of all Xs. The set of all sets encompasses all Xs, as well as itself (making it the only Y when our reference is sets in an absolute/pure sense).
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Re: The overlooked part of Russell's paradox

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philosopher19 wrote:If you look at my reply to Steve (the one that is directly above this reply to you), I think the problem with Russell's paradox is in suggesting that since you cannot have an X as the set of all Xs whilst you can have a Y of all Ys, you cannot have a set of all sets. Whereas in truth, there's no such thing as a Y of all Ys, nor an X of all Xs. The set of all sets encompasses all Xs, as well as itself (making it the only Y when our reference is sets in an absolute/pure sense).
Phil, the real problem is that we falsely believe that "something can contain itself" in the first place. ...we falsely believe in something logically impossible in the first place.

X<X is logically impossible. It doesn't matter what X represents; whether it be real or abstract; whether it be boxes, lists, or sets, (or reference-to-anyone of these!), it still does not make it any less impossible.

Russell's Paradox is a 'paradox' (a parlor trick; deception) because of our 'ignorance' (i.e. our inability to think logically).
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Re: The overlooked part of Russell's paradox

Post by Steve3007 »

philosopher19 wrote:There are definitely three items on this list. Does this list include itself as an item? In other words, should this list list itself?
Steve3007 wrote:It depends if this list is in the house. You didn't say it was.
philosopher19 wrote:That's not what it depends on. Consider again that I was forming a list of all lists that are in this house that list themselves. I was not forming a list of all list in this house.
Eh??!? Of course it depends on that. Let's go no further before getting this straight. Here is how you defined that list:
=philosopher19 wrote:So now I form a new list: The list of all lists in this house that are members of themselves.
Look at what you've said above. If this new list you're talking about is in the house, it lists itself, because it is a list in this house that lists itself. If it is not in the house it doesn't list itself because it's not in the house. If it's in the house it references 4 lists. If it's not in the house it references 3. If we can't agree on that then we speak different languages!
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Re: The overlooked part of Russell's paradox

Post by Thomyum2 »

RJG wrote: June 11th, 2021, 6:00 pm Not all premises are true or possible. Yes, anybody can define a non-sensical premise. But that does not automatically make it true or possible.

For example, you can define the premise "Box X contains itself". But that does not mean that it is true or logically possible. Your premise is false (logically impossible) via the form X<X.

Or you can define the premise "List X contains itself". But that does not mean that it is true or logically possible. Your premise is false (logically impossible) via the form X<X.
Of course not all premises are true, but in order to show the truth or falsehood of any premise, you'd have to supply the premises or evidence on which that claim is based. Logic can't make a claim of the truth of anything without the accepted premises (and definitions) behind it - logic can only take us from one truth to another.

The claim of the impossibility of 'X<X' that you're making is only meaningful if you precede it with, or understand that there's an implied premise such as 'for any given number X', or 'for every that whole number X'. In order to 'X<X' to be meaningful at all, X must be defined as something that has the property that it can be greater or less than something else, i.e. it must be a number of some kind.

For example:
  • The claim '8 < 9' can be said to be true or false
  • The claim 'lightbulb < schoolbus' cannot be said to be either true or false.
A 'set' is not a number, so the falsehood of 'X<X' would not logically apply to it.
RJG wrote: June 11th, 2021, 6:00 pm Also note: paradoxes are no more real than magic is. Paradoxes (and magic) exist ONLY out of ignorance of understanding.
Not sure where this is coming from as I wasn't claiming that a paradox is real, whatever that might mean anyway. But I think you're being too hard on them - paradoxes are useful for understanding and revealing things about our logical systems. (In my experience, they can also fun and instructive thought exercises.) Here's a quote from Kierkegaard that I like:
But one must not think ill of the paradox, for the paradox is the passion of thought, and the thinker without the paradox is like the lover without passion: a mediocre fellow.
But I think this is all getting off track - I was mainly trying to point out how the Russell paradox fits into the larger picture of our evolving understanding of formal systems, which I think is a fascinating topic. But I really don't have much interest in whether or not there is such thing as a set that contains itself beyond that.
“We have two ears and one mouth so that we can listen twice as much as we speak.”
— Epictetus
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Re: The overlooked part of Russell's paradox

Post by philosopher19 »

Steve3007 wrote: June 12th, 2021, 8:57 am Look at what you've said above. If this new list you're talking about is in the house, it lists itself, because it is a list in this house that lists itself. If it is not in the house it doesn't list itself because it's not in the house. If it's in the house it references 4 lists. If it's not in the house it references 3. If we can't agree on that then we speak different languages!
Again, it's not just a matter of whether this list is in the house or not, it's also about whether this list lists itself or not. Compare:

A) I have 3 lists in this house that list themselves. I form a new list of all lists in this house. If this new list is in the house, then yes, this list lists itself.

B) I have 3 lists in this house that list themselves. I form a new list of all lists in this house that list themselves. It is not enough for this list to be in this house to list itself (as was the case with A). It must also satisfy the criteria of being a list that lists itself.

Tell me where there is a contradiction in the following:

I form a list of the three self-listing lists in this house. This list does not list itself, therefore, this list does not list itself because it does not satisfy the red part of B (nor could it without running into contradictions as highlighted in the OP and my last two or three replies to you).
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