The overlooked part of Russell's paradox

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Thomyum2
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Re: The overlooked part of Russell's paradox

Post by Thomyum2 »

Didn't Kurt Gödel resolve this dispute about 90 years ago?
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Re: The overlooked part of Russell's paradox

Post by Steve3007 »

RJG wrote:There is a difference here. Pi (as with any variable) we can get to stand by itself...
I don't really know what you mean by "stand by itself". We can give pi a label. We can call it "pi". Likewise we can give x a label. We can call it "x". But the only way we can define pi, numerically, is by using an infinite sequence. As a transcendental number it can't be represented as a finite length piece of algebra.
...but x = {x, 1, 2} or X<X cannot be resolved; we can't get X to stand by itself so we can know what it is (what it means).
If "stand by itself" means being able to describe what it means, then x can be described as "the set of all sets that are members of themselves". That's what it means. That meaning may refer to the abstract concept of sets, but "the ratio of the circumference of a circle to its diameter" refers to the abstract concept of a circle.
If you believe X<X is logically impossible, then why don't you believe x={x, 1, 2} is likewise impossible?
Remember, x represents a set here, not a number or a variable.

With x representing a set, x={x, 1, 2} doesn't contain any logical contradictions. It just means that x is a list which refers to, among other things, itself. There's no logical contradiction in creating a list and adding, as an item on that list, "this list".

As you've pointed out, it may not be much good for representing the relationships between physical boxes, but that doesn't mean it contains logical contradictions. It just means it has limited possible uses.
What's the difference? In both cases, x can never be resolved, and therefore impossible.
What do you mean by "resolved" in the context of sets of sets? If I have a list of items, and one of the items on that list is the list itself, in what sense is that list not resolved?
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Re: The overlooked part of Russell's paradox

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Terrapin Station wrote:I didn't catch what the thread title promised: what's the "overlooked" part of Russell's paradox?
Thomyum2 wrote:Didn't Kurt Gödel resolve this dispute about 90 years ago?
Hopefully philosopher19 will tell us more explicitly what he thinks he's added.
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Re: The overlooked part of Russell's paradox

Post by philosopher19 »

Terrapin Station wrote: June 8th, 2021, 7:23 am Aside from Steve3007's comment above, I didn't catch what the thread title promised: what's the "overlooked" part of Russell's paradox?
Everyone recognises that you cannot have a set of all sets that are not members of themselves. The overlooked part is that you cannot have a set of all sets that are members of themselves.
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Re: The overlooked part of Russell's paradox

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RJG wrote: June 8th, 2021, 8:46 am
philosopher19 wrote:If x, y, and z are sets that are members of themselves
Phil, this is not possible. X (or Y, Z etc) cannot logically be a "set that is a member of itself". To help understand, imagine a box (to represent a "set"), now put this box inside itself. It can't be done.

"Sets that are members of themselves" are logical impossibilities (form X<X).
I think I know where you're coming from. Your focus is purely metaphysical. Here I'm purely focused on logic. Instead of viewing it as physically containing itself, just view it as logically encompassing itself. For example:

Call the list of all lists L. L lists itself because it is a list. L also lists all lists other than itself. So of all the items that L lists, one is a member of itself (that item is L), whilst the others are not members of themselves because they are members of L and they are not L itself.
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Re: The overlooked part of Russell's paradox

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Steve3007 wrote: June 8th, 2021, 6:17 am You're wrong to state above that "p = {x, y, z}" represents x, y and z being members of themselves. It doesn't. The statement "x is a member of itself", using similar notation, could be represented as "x = {x, ...}". Likewise for y and z.
So assume I have three sets that are not members of themselves: x, y, and z. I then form a set of these three sets. I write this in the following manner: p = {x y z}. No problems here. If I write x = {x y z} then problems occur because it implies x is a member of itself.

So assume you have three sets that are not members of themselves: x, y, and z. You are then told to form a set of these three sets. How would you write this in a manner that is consistent with the above? And by this I mean the letters outside { } represents the set, and the letters inside the { } represent the elements of that set.
You can't have a set of all sets that are not members of themselves because then it could neither be a member of itself nor not a member of itself without contradiction. That's the point of the paradox.
Take V to be the set of all sets. V encompasses all sets that are not members of themselves and it is a member of itself because it is a set. Where is the contradiction in what I've just written? Whilst it is clearly contradictory to have a set of all sets that are not members of themselves that is itself not a member of itself, it is necessarily the case that the set of all sets encompasses all sets that are not members of themselves as well as itself (because it too is a set).

Rejecting the set of all sets is blatantly contradictory. Yet, from what I've heard, this is what's ongoing in mainstream maths, philosophy, and logic.
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Re: The overlooked part of Russell's paradox

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Thomyum2 wrote: June 8th, 2021, 11:43 am Didn't Kurt Gödel resolve this dispute about 90 years ago?
Not to my knowledge.
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Re: The overlooked part of Russell's paradox

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philosopher19 wrote:
Steve3007 wrote:You're wrong to state above that "p = {x, y, z}" represents x, y and z being members of themselves. It doesn't. The statement "x is a member of itself", using similar notation, could be represented as "x = {x, ...}". Likewise for y and z.
So assume I have three sets that are not members of themselves: x, y, and z. I then form a set of these three sets. I write this in the following manner: p = {x y z}. No problems here. If I write x = {x y z} then problems occur because it implies x is a member of itself.
Why is that a problem? You've simply changed your mind and decided to define x as being a member of itself. In the above, you've effectively written: "x is not a member of itself. No, actually, let's make x a member of itself after all."
So assume you have three sets that are not members of themselves: x, y, and z. You are then told to form a set of these three sets. How would you write this in a manner that is consistent with the above? And by this I mean the letters outside { } represents the set, and the letters inside the { } represent the elements of that set.
p = {x, y, z}
philosopher19 wrote:
Steve3007 wrote:You can't have a set of all sets that are not members of themselves because then it could neither be a member of itself nor not a member of itself without contradiction. That's the point of the paradox.
Take V to be the set of all sets. V encompasses all sets that are not members of themselves and it is a member of itself because it is a set. Where is the contradiction in what I've just written?
As I said, the contradiction in the concept of a set of all sets, including sets that are not members of themselves, is that it cannot with logical consistency either be a member of itself nor not be a member of itself, even though those are the only two logically possible options. As I said, that's the whole point of the paradox.

But remember, this is just a logical game. Sets aren't real. They're abstract concepts that we create either for fun or because they happen to have some use in describing some aspects of the world. Or both. (Frequently they start as the former and end up, perhaps unexpectedly, as the latter). If we discover a paradox like this in set theory then all we have to do is note that the self-contradictory concept "the set of all sets that are not members of themselves" is not much use for anything except philosophical discussions.
Whilst it is clearly contradictory to have a set of all sets that are not members of themselves that is itself not a member of itself,...
The point is that the term "the set of all sets..." implies self-membership. It's self-referential. It doesn't say "the set of all sets except this one..." or similar.
...it is necessarily the case that the set of all sets encompasses all sets that are not members of themselves as well as itself (because it too is a set).
Yes, that's the whole point of the paradox! That's why it's called a paradox.
Rejecting the set of all sets is blatantly contradictory. Yet, from what I've heard, this is what's ongoing in mainstream maths, philosophy, and logic.
I've no idea what you mean by "rejecting the set of all sets". As I said, this is a game of logic. If anybody wants to consider the concept of "the set of all sets" and its logical implications they're free to do so. Doing so does not, in itself, say anything about the nature of the real world. Applicability of abstract concepts as descriptions of aspects of the real world is handy, but it's not necessary.
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Re: The overlooked part of Russell's paradox

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Typo: "As I said, the contradiction in the concept of a set of all sets, including sets that are not members of themselves..." should read: "As I said, the contradiction in the concept of a set of all sets that are not members of themselves...".
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Re: The overlooked part of Russell's paradox

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Steve3007" wrote:..."the set of all sets that are members of themselves"...
This is logically impossible. This is as non-sensical as saying "itself is the number before itself".

RJG wrote:...x={x, 1, 2} is logically impossible...
Steve3007 wrote:Remember, x represents a set here, not a number or a variable.
Which x are you talking about? ...x the set?, ...or x the set+member?


***********
RJG wrote:"Sets that are members of themselves" are logical impossibilities (form X<X).
philosopher19 wrote:I think I know where you're coming from. Your focus is purely metaphysical. Here I'm purely focused on logic. Instead of viewing it as physically containing itself, just view it as logically encompassing itself.
Phil, logically something cannot "encompass itself"! X<X is logically impossible. Logically, something cannot exist outside itself to then contain itself.

philosopher19 wrote:For example: Call the list of all lists L. L lists itself because it is a list. L also lists all lists other than itself. So of all the items that L lists, one is a member of itself (that item is L), whilst the others are not members of themselves because they are members of L and they are not L itself.
This is not even imaginable. Try to imagine your scenario. Not only is this logically and physically impossible, but one can't even imagine the scenario that you are describing unless we imagine two different L's, one as a list and an item, and the other as just a list. So, when you are talking about "L", which one are you talking about?

A "member of itself" is impossible in ALL senses (logically, physically, and imaginably).
Last edited by RJG on June 9th, 2021, 7:41 am, edited 2 times in total.
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Re: The overlooked part of Russell's paradox

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philosopher19 wrote: June 8th, 2021, 9:59 pm
Terrapin Station wrote: June 8th, 2021, 7:23 am Aside from Steve3007's comment above, I didn't catch what the thread title promised: what's the "overlooked" part of Russell's paradox?
Everyone recognises that you cannot have a set of all sets that are not members of themselves. The overlooked part is that you cannot have a set of all sets that are members of themselves.
The point of the paradox isn't to list every contradiction that results from the unrestricted comprehension principle.

At any rate, when you say, " nothing can be a member of itself twice," how would you formalize (or at least semi-formalize) what you're claiming the contradiction is there?
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Re: The overlooked part of Russell's paradox

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RJG, do you accept the difference between a reference and an object? So, if I have a list, and on that list I write the four items: "apples, oranges, pears, this list", you can see that the list doesn't physically contain any fruit or lists, but contains references to those things? And you can see that it's possible for a list to contain a reference to itself? That's what a set is, at least as I'm using the term here. A set is an example of a way of thinking about things, just as a shopping list is a way of thinking about what I'm planning to buy.
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Re: The overlooked part of Russell's paradox

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RJG wrote:
Steve3007 wrote:..."the set of all sets that are members of themselves"...
This is logically impossible. This is as non-sensical as saying "itself is the number before itself".
It's unrelated to numbers.
RJG wrote:
Steve3007 wrote:Remember, x represents a set here, not a number or a variable.
Which x are you talking about? ...x the set?, ...or x the set+member?
X is a set and it is also a member of a set. Do you accept that a set can simultaneously be a set and be a member of a set? I assume so. Your problem is specifically with it being a member of itself, and that seems to simply be because you don't see the members of sets as references and don't seem to accept the concept of self-reference. See above post.
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Re: The overlooked part of Russell's paradox

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RJG wrote: June 8th, 2021, 11:01 am
Terrapin Station wrote:First off, when we're talking about how people are thinking about things--which is what we're talking about when we're talking about sets, what would make it true that a given variable is unique?
If we want to make sense then we first have to know what we are talking about. So my question is - What is X?

Does X=X? or
Does X={X, 1, 2}?

Which of these two different (unique) meanings/solutions are we talking about? In other words, which X are we talking about? We can't have two different meanings/solutions to the same variable and still make sense.

X={X, 1, 2} logically contradicts X=X. Therefore (and assuming we hold X=X as true, then), X={X, 1, 2} is as logically impossible as X<X.
???

If X = {X, 1, 2} than X = X is (X = {X, 1, 2}) = (X = {X, 1, 2})
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Re: The overlooked part of Russell's paradox

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RJG wrote:...x={x, 1, 2} is logically impossible…
Steve3007 wrote:Remember, x represents a set here, not a number or a variable.
RJG wrote:Which x are you talking about? ...x the set?, ...or x the set+member?
Steve3007 wrote:RJG, do you accept the difference between a reference and an object?
Yes, a reference and its object are not the same; they are two different things.

Steve3007 wrote:So, if I have a list, and on that list I write the four items: "apples, oranges, pears, this list", you can see that the list doesn't physically contain any fruit or lists, but contains references to those things?
Correct. The List is one thing, and the Items on the list are other things, and the Objects represented by the Items are another.

Steve3007 wrote:And you can see that it's possible for a list to contain a reference to itself?
Yes, a list can contain a reference-to-itself, ...but NEVER, ever, ever, ever itself.

Steve3007 wrote:That's what a set is, at least as I'm using the term here. A set is an example of a way of thinking about things, just as a shopping list is a way of thinking about what I'm planning to buy.
1. So then you agree that the x's represented in the set x={x, 1, 2), are two different x's? (i.e. one x is a list (object) and the other x is a reference to the list; two different x's!)? ...YES/NO?

2. And so then you further agree that it is logically impossible for listX to contain listX? (i.e. it can only contain a reference-to-listX, but never itself). ...YES/NO?


***********
RJG wrote:...x={x, 1, 2} is logically impossible…
Steve3007 wrote:Remember, x represents a set here, not a number or a variable.
RJG wrote:Which x are you talking about? ...x the set?, ...or x the set+member?
Steve3007 wrote:Do you accept that a set can simultaneously be a set and be a member of a set?
Yes. A set can be also be a member, but NEVER a member of itself. A "member of itself" is logically impossible. Only a reference-to-X can be a member of X. A reference-to-X is not the same as X; these are two different animals.

Steve3007 wrote:Your problem is specifically with it being a member of itself...
It is not my problem, it is logic's problem. A "member of itself" is logically impossible.
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