## The overlooked part of Russell's paradox

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philosopher19
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### Re: The overlooked part of Russell's paradox

RJG wrote: June 9th, 2021, 7:35 am Phil, logically something cannot "encompass itself"! X<X is logically impossible. Logically, something cannot exist outside itself to then contain itself.
I get where you're coming from, but I don't think you get where I'm coming from. I will try and clarify where I'm coming from:

Suppose you had three lists in your room and you decided to make a fourth list. This list, is the list of all lists in your room. Tell me which of following statements you disagree with (if any):

1) This list, lists four items
2) This list lists itself as an item
3) This list lists three items other than itself. This list lists three items that are not itself
4) Because this list lists itself, it is a member of itself. If this list didn't list itself, it wouldn't be a member of itself. For example, the list of items in my car, is not a member of itself (in other words, it does not encompass itself). But if the list was an item in my car, then it would have to list itself (thereby making it a member of itself, or encompassing itself).
Terrapin Station
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### Re: The overlooked part of Russell's paradox

RJG wrote: June 9th, 2021, 1:44 pm So I take it that you believe "both x's are the SAME, and one x is NOT different nor does NOT refer to the other x"?
Correct, x in all instances refers to the set {x, 1, 2}

So, from your belief, the set "x" is a never-ending infinite regress? ...that thusly contains an infinite number of members?
ONLY insofar as anyone thinks of that. This stuff only exists only insofar as anyone is thinking it.
RJG
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### Re: The overlooked part of Russell's paradox

RJG wrote:Phil, logically something cannot "encompass itself"! X<X is logically impossible. Logically, something cannot exist outside itself to then contain itself.
philosopher 19 wrote:I get where you're coming from, but I don't think you get where I'm coming from. I will try and clarify where I'm coming from:

Suppose you had three lists in your room and you decided to make a fourth list. This list, is the list of all lists in your room. Tell me which of following statements you disagree with (if any):

1) This list, lists four items
Good so far. This List # 4 has 4 items listed (printed) on it.

philospher19 wrote:2) This list lists itself as an item.
Yes, List # 4 has the printed words "List # 4" on it.

philosopher19 wrote:3) This list lists three items other than itself. This list lists three items that are not itself
Yes, List # 4 also has the words "List # 1", "List # 2", and "List # 3" printed on it.

philosopher19 wrote:4) Because this list lists itself, it is a member of itself.
Not so. This is the error. The printed text is not the list itself. The words "List # 4" printed on List # 4 is NOT a list, ...these are just printed words that refer to List # 4.

philosoper19 wrote:If this list didn't list itself…]
Listing (or writing the words) "List # 4" onto List # 4 does not make the words somehow transform into an actual list. They (the printed words) only 'refer' to a list. They are not the list itself.

philosopher19 wrote:...it wouldn't be a member of itself.
A "member of itself" is a logical impossibility.

philosopher19 wrote:For example, the list of items in my car, is not a member of itself (in other words, it does not encompass itself). But if the list was an item in my car, then it would have to list itself (thereby making it a member of itself, or encompassing itself).
There is only ONE list here. It is that piece of paper (list) that now sits in your car. Anything and everything written on this list (including the word "list") are just items on the list that reference other stuff.
philosopher19
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### Re: The overlooked part of Russell's paradox

RJG wrote: June 9th, 2021, 4:29 pm The printed text is not the list itself.
Yes, we agree on this.
Listing (or writing the words) "List # 4" onto List # 4 does not make the words somehow transform into an actual list. They (the printed words) only 'refer' to a list. They are not the list itself.
Again, we agree.
A "member of itself" is a logical impossibility.
But I'm not discussing metaphysical possibilities or impossibilities here. Again, consider the list of all lists in my room. There is a meaningful distinction to be had between the item that is in reference to the list itself, and the items that are in reference to other lists. The latter I describe as being "not members of themselves", and the former I describe as being "a member of itself".

Whether or not something can existentially or metaphysically contain itself (as opposed to just referentially), is another matter.
RJG
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### Re: The overlooked part of Russell's paradox

philosopher19 wrote:Whether or not something can existentially or metaphysically contain itself (as opposed to just referentially), is another matter.
I have no issue with, nor is there a problem with "something referentially containing itself". The problem (logical impossibility) is with "something containing itself" or a "member of itself". Simple Logic tells us that X<X is logically impossible.

Therefore:
Set X = {X1, 1, 2} is mathematically and logically valid
Set X = {X, 1, 2} is mathematically and logically impossible; invalid

RJG wrote:A "member of itself" is a logical impossibility.
philosopher19 wrote:But I'm not discussing metaphysical possibilities or impossibilities here.
Neither am I. I am talking about math/logic. Mathematically, and logically, you can't have 2 simultaneously different assignments for the same variable. X cannot represent both "something" and "a reference-to-something" simultaneously. Pick one and call it "X", and then call the other "Y", or "X1", or something other than "X", as the meaning for "X" has already been taken!

philosopher19 wrote:There is a meaningful distinction to be had between the item that is in reference to the list itself, and the items that are in reference to other lists.
The relevant meaningful distinction is between the "list" itself, and the "reference-to-the-list". These are NOT the same thing! One is a list, and one is a reference.

philosopher19 wrote:The latter I describe as being "not members of themselves", and the former I describe as being "a member of itself".
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
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### Re: The overlooked part of Russell's paradox

RJG wrote: June 10th, 2021, 12:10 pm 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.
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.
philosopher19
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### Re: The overlooked part of Russell's paradox

Steve3007 wrote: June 8th, 2021, 11:50 am
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.
There is something that I've noticed as a result of the discussion I was having with RJG with regards how I can the overlooked part of Russell's paradox better.

Suppose you have a house with 3 rooms. Each room contains three lists. Of the three lists that each room contains, one of these lists is a member of itself. So there are at least three lists in this house that are members of themselves (because they reference/list themselves).

So now I form a new list: The list of all lists in this house that are members of themselves.

There are definitely three items on this list. Does this list include itself as an item? In other words, should this list list itself?

You would think that this is entirely optional, but it isn't. As in you'd think, if we choose for this list not to list itself, then no contradictions occur, and if we choose for this list to list itself, then again, no contradictions occur. The former is not contradictory, whereas the latter is. Here's why:

You cannot take something as an ALL, and then add to it whilst keeping the context the same.

First you established ALL the self-listing lists in the house (of which there were 3), then you made a list of ALL the self-listing lists in the house (of which there were 3). So the ALL had already been exhausted. It is contradictory for this list to list itself because the list will not longer semantically qualify as ALL the self-listing lists in the house. You cannot have ALL plus one.

Now suppose you found a secret room in this house that also contained a list that lists itself. You'd just amend your list of all self-listing lists in this house to include this new self-listing list. But in doing so, you have acknowledged that your previous list was in fact inaccurate, precisely because it was not a list of ALL the self-listing lists in the house because it did not include the list in the secret room.

You can even form a new self-listing list in your house (like a list of all lists in this house), and then form a new list of ALL self-listing lists in this house to include this new list. What you cannot do, is form this new list, and then add itself to it as that would amount to ALL plus one. ALL plus one is semantically not the same as ALL.

A list of all self-listing lists in this house can never list itself because it would amount to ALL plus one (which is not the same as ALL plus nothing). This is why you can never have a set of ALL sets that are members of themselves that is itself a member of itself.
Steve3007
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### Re: The overlooked part of Russell's paradox

philosopher19 wrote:Suppose you have a house with 3 rooms. Each room contains three lists. Of the three lists that each room contains, one of these lists is a member of itself.
OK. Like this:

Room 1:
A = {X, ...}
B = {X, ...}
C = {C, ...}

Room 2:
D = {X, ...}
E = {X, ...}
F = {F, ...}

Room 3:
G = {X, ...}
H = {X, ...}
I = {I, ...}
So there are at least three lists in this house that are members of themselves (because they reference/list themselves).
In the situation as you've stated it there are exactly three of the lists in the house are members of themselves.
So now I form a new list: The list of all lists in this house that are members of themselves.
OK.

J = {C, F, I}
There are definitely three items on this list. Does this list include itself as an item? In other words, should this list list itself?
It depends if this list is in the house. You didn't say it was. If it is then:

J = {C, F, I, J}
You would think that this is entirely optional, but it isn't.
Why would you think is optional? If list J is in the house and is "the list of all lists in this house that are members of themselves", as you've stated, then it contains itself. If it isn't in the house then it doesn't contain itself. Where's the option?
As in you'd think, if we choose for this list not to list itself, then no contradictions occur, and if we choose for this list to list itself, then again, no contradictions occur. The former is not contradictory, whereas the latter is.
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.

No choice.
Steve3007
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### Re: The overlooked part of Russell's paradox

Continuing from the previous post:
You cannot take something as an ALL, and then add to it whilst keeping the context the same.
You've lost me. I don't yet see the sense in which we're "taking something as an ALL and then adding to it". We're making a list of all self-member lists in the house. As I said, the question of whether this list is a self-member depends simply on whether this list is in the house.
First you established ALL the self-listing lists in the house (of which there were 3), then you made a list of ALL the self-listing lists in the house (of which there were 3). So the ALL had already been exhausted. It is contradictory for this list to list itself because the list will not longer semantically qualify as ALL the self-listing lists in the house. You cannot have ALL plus one.
If J is in the house, and we define J as "the list of all lists in this house that are members of themselves" then:

J = {C, F, I, J}

No contradiction. You possibly seem to be talking about the fact that before list J was created there were only 3 self-listing lists in the house. That's true, but it doesn't result in any contradictions any more than the definitions of C, F and I result in contradictions. Any list that lists itself obviously doesn't list itself before it's been created. If you wanted to you could define J as "the list of all lists in this house that were members of themselves before this list was created" or some such.
Now suppose you found a secret room in this house that also contained a list that lists itself.
OK

Secret Room:
K = {K, ...}
You'd just amend your list of all self-listing lists in this house to include this new self-listing list.
If you wanted the description "the list of all lists in this house that are members of themselves" to remain accurate (and to include secret rooms) yes, you would.
But in doing so, you have acknowledged that your previous list was in fact inaccurate, precisely because it was not a list of ALL the self-listing lists in the house because it did not include the list in the secret room.
It was accurate according to the information available at the time and turned out, when new information came to light, to be innacurate.
You can even form a new self-listing list in your house (like a list of all lists in this house), ...
OK (If this goes on we'll be in danger of running out of letters! )

L = {C, F, I, J, K, L}
...and then form a new list of ALL self-listing lists in this house to include this new list. What you cannot do, is form this new list, and then add itself to it as that would amount to ALL plus one. ALL plus one is semantically not the same as ALL.
It wouldn't amount to "All plus one". It would amount to a list of all self-listing lists in the house.
A list of all self-listing lists in this house can never list itself because it would amount to ALL plus one (which is not the same as ALL plus nothing). This is why you can never have a set of ALL sets that are members of themselves that is itself a member of itself.
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.
RJG
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### Re: The overlooked part of Russell's paradox

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.
RJG
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### Re: The overlooked part of Russell's paradox

Wikipedia wrote:If you have a list of lists that do not list themselves, then that list must list itself, because it doesn't contain itself. However, if it lists itself, it then contains itself, meaning it cannot list itself. This makes logical usages of lists of lists that don't contain themselves somewhat difficult. It was developed by Bertrand Russell.
To help better understand and visualize the nonsense (logical impossibility) here, just replace the word "list" with "box"

First statement:
1A. "If you have a list of lists that do not list themselves…"
1B. If you have a box of boxes that do not box (contain) themselves...

Can boxes box (contain) themselves? - NO, not logically possible.
Can lists list (contain) themselves? - NO, not logically possible.
Important Note: Lists don't actually list lists, they only list names-of-lists.

This so-called paradox starts by having a gullible reader believe and accept a logical impossibility as a possibility.

Second statement:
2A. "...then that list must list itself, because it doesn't contain itself"
2B. "...then that box must box itself, because it doesn't box itself"
-
Firstly, since it is logically impossible for a box to box itself, then ALL boxes meet this criteria, therefore (according to the first part of the statement) ALL boxes must do the impossible and box themselves. This is just more nonsense, that we, the gullible reader, falsely believe as possible.

Third statement:
3A. "However, if it lists itself, it then contains itself, meaning it cannot list itself."
3B. However, if it boxes itself, it then contains itself, meaning it cannot box itself.

Hence the so-called "paradox", created by deceiving the gullible reader into believing that logical impossibilities are possible!
Steve3007
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### Re: The overlooked part of Russell's paradox

Steve3007 wrote:Can we take that to mean we now both agree that lists and sets reference things and that sets aren't boxes?
I should have known that was too much to hope for.
Terrapin Station
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### Re: The overlooked part of Russell's paradox

RJG wrote: June 11th, 2021, 7:41 am To help better understand and visualize the nonsense (logical impossibility) here, just replace the word "list" with "box"

First statement:
1A. "If you have a list of lists that do not list themselves…"
1B. If you have a box of boxes that do not box (contain) themselves...

Can boxes box (contain) themselves? - NO, not logically possible.
Can lists list (contain) themselves? - NO, not logically possible.
Important Note: Lists don't actually list lists, they only list names-of-lists.
Again, that's thinking about them as "objects."
RJG
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### Re: The overlooked part of Russell's paradox

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.
Steve3007
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### Re: The overlooked part of Russell's paradox

RJG wrote: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.
X is, in all cases, a reference to a list. The list is {X, 1, 2}. X is a reference to that. (It goes without saying, this has been said previously numerous times.)
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.
No, the paradox doesn't come from that mixing. No such mixing takes place. The paradox comes from considering whether a list containing only references to lists which don't contain references to themselves contains a reference to itself. (I've explicitly added "reference" there in the hope that it'll make you happy and stop fixating on a matter of terminology that's not relevant to the paradox, but since symbols like 'x' are always references, it shouldn't be necessary.)

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