I challenge interpretation placed on Russell's Paradox and claim it is an excellent description of conditional null sets. The condition being statement of what elements or sets are permitted to belong to a set. Basis a set begins with one of these { then ends with one of these } and has rule(s) about what can belong in the set I call this the set shopping list. Set shopping list is an a priori event (or posteriori) Conditional null set is a posteriori event (or priori) Of course this is a symmetric relation as as shopping list may be defined by what it includes and excludes. And: construction of predicate may be True/False according to its rules processing of machine logic may be True/False Thinks: how many times have we constructed a predicate that looks good but is not good? How many times does machine logic produce a good result or a problematic result. I propose pragmatic approach and let the awkward results define a conditional null set Plus! Universal is a very big thing and I claim we cannot know everything but we can know specified things hence conditional universal set and conditional null set. We still have paradox than any thing in conditional null set is also a thing in compliment of conditional null set. Easy side step: philosophy and logic are fractal in nature?
A wavy hand reply about Universal Set existences. A Conditional Universal Set exists and exists before any sets can be defined based on whatever is used on shopping list criteria, sets, elements and subsets in the Conditional Universal Set. A Universal Set is unlikely to exist as it would need to contain all information needed to define itself which is likely to be impossible simply by saying the Universal set consists of all elementary particles in universe enumerated and doubled. Under my proposed set theory adjustments this would generate a TF FT loop which would be infinitely increasing and have infinitely increasing cardinality. Also there would not be any steady universal null set consisting of infinite union of all subsets and, of course, as a consequence the Universal set would never be in a stead state long enough to exist
@@terryendicott2939 No, everything is a set, because ZFC literally doesn't allow you to have anything else. If you say x is an element of X, then x is a set because it can't be anything else, what else is there besides sets?
@@terryendicott2939There're things called indexed families, but they are also formalised using zfc and thus sets. A class, however, is not a thing in zfc. But there're other axiomatic systems that have classes which are, indeed, not sets
Russells paradox worked on a differnet version of set theory. Here the axiom of spesification makes it so that this paradox cannot be constructed in this schema. Nevertheless, Gödels statement G can still be constructed, if we allow enough complexity to produce arithmetic.
Well no, it is an inevitability, we can avoid it using Axiomatic Set Theory but that's not good enough. We still use the set of all sets or arbitrarily large sets in Set Theory. The way we avoid problems is by calling them something else. So we call it the Class of All Sets. Pretty neat workaround. We still think of it like the Set of All Sets in practice but the language modification to avoid self referencing prevents the Russell's Paradox.
@@larianton1008in practice we still use the set of all sets. We just call it something else. The change from set of all sets to class of all sets solves the problem.
Are you using zf axioms? Isn't the existence of a set of all sets precluded directly by the axiom of regularity, since it implies that no set can contain itself?
Why have I never seen this? Lovely little proof. I've see the Axiom of Specification to delimit comprehension to avoid Russell's paradox, but never seen it actually used to prove the non-existence of the set of all sets!
I love your videos: they're always incredibly insightful and well-explained. I must admit that this proof baffled me somewhat. I'm vaguely familiar with Russell's paradox, but I always took as the conclusion, "therefore, we just can't define a set that contains itself without breaking set theory." But the contradiction in your proof appears to rely on having built such a set B. That set is empty, as far as I can tell, but provided that the empty set itself is not an element of A, the idea that it is not contained in A goes through. The contradiction might as well just be, "this set B just doesn't make sense / isn't well-defined."
A = { empty }, so A has 1 element, the empty set By the definition, B = A, since the empty set is in A and not in itself. B is not empty, it has 1 element.
For the ones who ask " Then what is the "collection" of all sets, is there any object that includes all sets. Check what is "Class" in math. "Collection" of all sets is not a set but a class
Lovely! I claim that B as defined is contained in A and is the null set of A. (actually the conditional null set of A) The set A containing set B satisfying x ∈ B AND x ∉ B allows set B to define null set of A. Further B as conditional null set of A denoted ∅ₐ - a definition With property that compliment of conditional null set of A is A itself ∅ₐᶜ = A It is complicated but thanks for your video 🙂 */ Hence Universal set exists and is equal to compliment of "infinite chained union of all conditional null sets" Because Universal null set exists if and only if Universal set exists */ EDIT: */ to */ revised to: Hence a Conditional Universal set exists and is equal to compliment of "infinite chained union of all conditional null sets" Because Conditional Universal null set exists if and only if Conditional Universal set exists So I think this makes more sense as it means a "universal set" exists but is conditional on information used to define sets and elements in that universe. Whether a 'truly' universal set exists depends on whether there are truly universe of elements and truly universe of sets Probably the best approach is to take that in Probability and Stats and extend the finite concatenated union into an imagined infinite concatenated union. But even then that is just countable infinity rather than uncountable infinity no?
Okay I was wrong about universal set existence and here is why ... A wavy hand reply about Universal Set existences. A Conditional Universal Set exists and exists before any sets can be defined based on whatever is used on shopping list criteria, sets, elements and subsets in the Conditional Universal Set. Surreal number people might challenge me for saying things have to exist before sets can exist. Maybe they existed together? A Universal Set is unlikely to exist as it would need to contain all information needed to define itself which is likely to be impossible simply by saying the Universal set consists of all elementary particles in universe enumerated and doubled. Under my proposed set theory adjustments this would generate a TF FT loop which would be infinitely increasing and have infinitely increasing cardinality. Also there would not be any steady universal null set consisting of infinite union of all subsets and, of course, as a consequence the Universal set would never be in a stead state long enough to exist
10 second argument: Let S be the set of all sets. Its cardinality is strictly smaller than that of its power set. However, its power set must be contained in S, so the cardinality of S must be at least as great as that of its power set. Contradiction.
@@thegoldenratioandbeyond232 how to prove that P(S) in S imply cardinality of S > cardinality of P(S) ? Since P(S) can be inside S but still satisfy cantor’s theorem. e.g. cardinality of set of Natural numbers is |N|. But the set X = {N} has cardinality of 1. ie, I don’t see how S = {a,b,c,…,P(S)} implies that |S| > |P(S)|. Another e.g. Y = set of reals = R By cantor’s theorem |R| > |N| Z = {R,1,2,3,4,…}. R is inside Z but cardinality of Z = |N|. So |Z| < |R| even though R is contained in Z Edit: I used the notation |X| to refer to the cardinality of X
@@CutleryChips You are confusing subsets with elements of a set. Let's take your second example: R would be an element of Z:= {R,1,2,3,4,…}, but not a subset of Z. What I am saying is that by definition of S, the powerset of S is a subset of S, i.e., any element of the powerset of S is also an element of S, which leads to a contradiction. An elementary property of cardinality is that if A is a subset of B, then |A|
So if the work done has achieved a conclusion or rather a result, and this statement is true. Then for case two have you not shown that as a result of computation that B is in fact an element of B. My question is, why are you checking it twice? Didn’t you just make your own problem?
By definition of the set B, B = {x in A : x not in x}. In particular, to say B is in B is equivalent to saying B is in A and B is not in B. Given statement P means B is in B Given statement Q means B is in A Then we have P (~P ^ Q). Given this as a premise, we can conclude ~Q (B is not in A) using the argument in the proof. It really just comes down to natural deduction, and can be checked using the proof checker at openlogicproject if you desire
Elements can be sets, who says they can't? The set {0,{0}} is the number 2, and it clearly contains a set as an element. And thus you can both have that {0} is an element AND a subset of 2.
@ If we look at the set A={0,1}, we would say 0 is an element of A, but {0} is a subset of A. Or if A={{0},{1}}, we would say {0} is an element of A, but {{0}} is a subset of A.
@ Yes, in general an element of a set is a single thing inside a set (it could be a set, a number, a vector, a blue cow), and subsets are sets made from elements (also the empty set is a subset of any set). The video takes an arbitrary set A and then makes a set B where the members are sets in A that don't contain themselves as elements (this is prohibited by the axiom of regularity in ZF but axiom of regularity also immediately gives that a set of all sets is not a set). Then you can clearly see that if B does not contain itself as an element then it does (or A is not the set of all sets), and if B does contain itself as an element then it should not (by the construction of B). At no point are we taking subsets, we are talking about elements and element inclusion, the elements just happen to be sets.
@@fushifushi9441 in Russel's paradox, we define a set R such that R contains every set S with the property that S is not an element of S. In other words, R is a set of all sets that don't contain themselves So R is not quite the set of all sets
It really means for any value x that exists in any set A, for the set B to be the set of all sets it must contain the set that doesnt contain itself, thus x must not contain itself, i believe at least, if I am wrong please correct me
For reference, if anyone is interested, this proof is given in Enderton's "Elements of Set Theory", as Theorem 2A
I challenge interpretation placed on Russell's Paradox and claim it is an excellent description of conditional null sets.
The condition being statement of what elements or sets are permitted to belong to a set.
Basis a set begins with one of these { then ends with one of these } and has rule(s) about what can belong in the set I call this the set shopping list.
Set shopping list is an a priori event (or posteriori)
Conditional null set is a posteriori event (or priori)
Of course this is a symmetric relation as as shopping list may be defined by what it includes and excludes.
And: construction of predicate may be True/False according to its rules processing of machine logic may be True/False
Thinks: how many times have we constructed a predicate that looks good but is not good?
How many times does machine logic produce a good result or a problematic result.
I propose pragmatic approach and let the awkward results define a conditional null set
Plus! Universal is a very big thing and I claim we cannot know everything but we can know specified things hence conditional universal set and conditional null set. We still have paradox than any thing in conditional null set is also a thing in compliment of conditional null set.
Easy side step: philosophy and logic are fractal in nature?
A wavy hand reply about Universal Set existences.
A Conditional Universal Set exists and exists before any sets can be defined based on whatever is used on shopping list criteria, sets, elements and subsets in the Conditional Universal Set.
A Universal Set is unlikely to exist as it would need to contain all information needed to define itself which is likely to be impossible simply by saying the Universal set consists of all elementary particles in universe enumerated and doubled.
Under my proposed set theory adjustments this would generate a TF FT loop which would be infinitely increasing and have infinitely increasing cardinality.
Also there would not be any steady universal null set consisting of infinite union of all subsets and, of course, as a consequence the Universal set would never be in a stead state long enough to exist
“Ah yes, to B to not to B.
That is the contradiction.”
-William Shakespeare.
Carl Friedrich Shakespeare
{all sets}
there you go.
damn....
Contradiction because the set containing all sets is not contained in the set containing all sets🤨
@@Mrlonely345 its a joke dawg
@@Mrlonely345sure it is, it says right there, "all sets", not "all sets except this one" 😁
x meaning any arbitrary set really throws me off.
yeh - x is switching from an "element" to a "set" and that is cool
@@Alan-zf2ttliterally everything is a set, bro
@@lukandrate9866 Not everything is a set. That is what the proof was all about.
@@terryendicott2939 No, everything is a set, because ZFC literally doesn't allow you to have anything else. If you say x is an element of X, then x is a set because it can't be anything else, what else is there besides sets?
@@terryendicott2939There're things called indexed families, but they are also formalised using zfc and thus sets. A class, however, is not a thing in zfc. But there're other axiomatic systems that have classes which are, indeed, not sets
"So, is Russell's paradox fake?"
Russells paradox worked on a differnet version of set theory. Here the axiom of spesification makes it so that this paradox cannot be constructed in this schema. Nevertheless, Gödels statement G can still be constructed, if we allow enough complexity to produce arithmetic.
Well no, it is an inevitability, we can avoid it using Axiomatic Set Theory but that's not good enough. We still use the set of all sets or arbitrarily large sets in Set Theory. The way we avoid problems is by calling them something else. So we call it the Class of All Sets. Pretty neat workaround. We still think of it like the Set of All Sets in practice but the language modification to avoid self referencing prevents the Russell's Paradox.
@@larianton1008in practice we still use the set of all sets. We just call it something else. The change from set of all sets to class of all sets solves the problem.
Are you using zf axioms?
Isn't the existence of a set of all sets precluded directly by the axiom of regularity, since it implies that no set can contain itself?
Yep you’re exactly right. This is just using the OTHER axioms. ❤
There's no set of all sets in a very small fragment of ZF. May as well do the non-trivial version :P
This prooves that the very concept of totality is inconsistent
Why have I never seen this? Lovely little proof. I've see the Axiom of Specification to delimit comprehension to avoid Russell's paradox, but never seen it actually used to prove the non-existence of the set of all sets!
I love your videos: they're always incredibly insightful and well-explained.
I must admit that this proof baffled me somewhat. I'm vaguely familiar with Russell's paradox, but I always took as the conclusion, "therefore, we just can't define a set that contains itself without breaking set theory." But the contradiction in your proof appears to rely on having built such a set B. That set is empty, as far as I can tell, but provided that the empty set itself is not an element of A, the idea that it is not contained in A goes through.
The contradiction might as well just be, "this set B just doesn't make sense / isn't well-defined."
you're mixing up the subsets and elements of a set
A = { empty }, so A has 1 element, the empty set
By the definition, B = A, since the empty set is in A and not in itself. B is not empty, it has 1 element.
If I remember correctly, A is the number 0 in a possible construction of the natural numbers from set theory.
There is a pretty much similar proof of that "There is no surjection of a set A to its power set P(A)".
That is used to proove that you can always build a higher order infinite
that is why we use classes and not sets 😂
I can’t comprehend the notion of B ∈ B. Doesn’t that just nest forever?
I suppose it does, but is that a problem?
For the ones who ask " Then what is the "collection" of all sets, is there any object that includes all sets. Check what is "Class" in math. "Collection" of all sets is not a set but a class
That is the classical answer.
for some clarity, sets are classes, but classes are not necessarily sets.
@ Yeah and enough to make someone curious about classes
yo i thought this was math, wheres the numbers??
He has two cases: 1 and 2
B must be B. Mindblowing.
it really is
Lovely! I claim that B as defined is contained in A and is the null set of A. (actually the conditional null set of A)
The set A containing set B satisfying x ∈ B AND x ∉ B allows set B to define null set of A.
Further B as conditional null set of A denoted ∅ₐ - a definition
With property that compliment of conditional null set of A is A itself ∅ₐᶜ = A
It is complicated but thanks for your video 🙂
*/ Hence Universal set exists and is equal to compliment of "infinite chained union of all conditional null sets"
Because Universal null set exists if and only if Universal set exists */
EDIT: */ to */ revised to:
Hence a Conditional Universal set exists and is equal to compliment of "infinite chained union of all conditional null sets"
Because Conditional Universal null set exists if and only if Conditional Universal set exists
So I think this makes more sense as it means a "universal set" exists but is conditional on information used to define sets and elements in that universe.
Whether a 'truly' universal set exists depends on whether there are truly universe of elements and truly universe of sets
Probably the best approach is to take that in Probability and Stats and extend the finite concatenated union into an imagined infinite concatenated union.
But even then that is just countable infinity rather than uncountable infinity no?
Okay I was wrong about universal set existence and here is why ...
A wavy hand reply about Universal Set existences.
A Conditional Universal Set exists and exists before any sets can be defined based on whatever is used on shopping list criteria, sets, elements and subsets in the Conditional Universal Set. Surreal number people might challenge me for saying things have to exist before sets can exist. Maybe they existed together?
A Universal Set is unlikely to exist as it would need to contain all information needed to define itself which is likely to be impossible simply by saying the Universal set consists of all elementary particles in universe enumerated and doubled.
Under my proposed set theory adjustments this would generate a TF FT loop which would be infinitely increasing and have infinitely increasing cardinality.
Also there would not be any steady universal null set consisting of infinite union of all subsets and, of course, as a consequence the Universal set would never be in a stead state long enough to exist
you showed that for any set A there is a set that doesn't contain set A. you have not proven that there is no set that contains all sets.
Don't we need A to consist of only sets as elements? Otherwise the definition of B is nonsensical.
10 second argument:
Let S be the set of all sets. Its cardinality is strictly smaller than that of its power set.
However, its power set must be contained in S, so the cardinality of S must be at least as great as that of its power set.
Contradiction.
Then where’s the rest proof that a set’s cardinality is strictly smaller than the power set’s?
@@CutleryChipsCantor's theorem
@@lukandrate9866 ok 👍
@@thegoldenratioandbeyond232 how to prove that P(S) in S imply cardinality of S > cardinality of P(S) ?
Since P(S) can be inside S but still satisfy cantor’s theorem.
e.g. cardinality of set of Natural numbers is |N|. But the set X = {N} has cardinality of 1.
ie, I don’t see how S = {a,b,c,…,P(S)} implies that |S| > |P(S)|.
Another e.g.
Y = set of reals = R
By cantor’s theorem |R| > |N|
Z = {R,1,2,3,4,…}. R is inside Z but cardinality of Z = |N|. So |Z| < |R| even though R is contained in Z
Edit: I used the notation |X| to refer to the cardinality of X
@@CutleryChips You are confusing subsets with elements of a set. Let's take your second example: R would be an element of Z:= {R,1,2,3,4,…}, but not a subset of Z.
What I am saying is that by definition of S, the powerset of S is a subset of S, i.e., any element of the powerset of S is also an element of S, which leads to a contradiction.
An elementary property of cardinality is that if A is a subset of B, then |A|
Please proof that B exists.
So if the work done has achieved a conclusion or rather a result, and this statement is true. Then for case two have you not shown that as a result of computation that B is in fact an element of B. My question is, why are you checking it twice? Didn’t you just make your own problem?
By definition of the set B, B = {x in A : x not in x}. In particular, to say B is in B is equivalent to saying B is in A and B is not in B.
Given statement P means B is in B
Given statement Q means B is in A
Then we have P (~P ^ Q). Given this as a premise, we can conclude ~Q (B is not in A) using the argument in the proof. It really just comes down to natural deduction, and can be checked using the proof checker at openlogicproject if you desire
This argument obfuscates things when it does not clearly distinguish between elements vs sets, or “is an element of” vs set inclusion.
what is an element?
Elements can be sets, who says they can't? The set {0,{0}} is the number 2, and it clearly contains a set as an element. And thus you can both have that {0} is an element AND a subset of 2.
@ I mean elements of the set, or members of the set.
@ If we look at the set A={0,1}, we would say 0 is an element of A, but {0} is a subset of A. Or if A={{0},{1}}, we would say {0} is an element of A, but {{0}} is a subset of A.
@ Yes, in general an element of a set is a single thing inside a set (it could be a set, a number, a vector, a blue cow), and subsets are sets made from elements (also the empty set is a subset of any set).
The video takes an arbitrary set A and then makes a set B where the members are sets in A that don't contain themselves as elements (this is prohibited by the axiom of regularity in ZF but axiom of regularity also immediately gives that a set of all sets is not a set). Then you can clearly see that if B does not contain itself as an element then it does (or A is not the set of all sets), and if B does contain itself as an element then it should not (by the construction of B). At no point are we taking subsets, we are talking about elements and element inclusion, the elements just happen to be sets.
You gotta eat more
Are you a professional dietician?
@@thegoldenratioandbeyond232 The guy is super thin, I bet he'll feel amazing with a little more mass on. 😊
He is eating but the brain took all the nutrition
Russell's paradox?
Not quite
@@methatis3013What do you mean: “Not quite”? 😂 Please elaborate.
@@fushifushi9441 in Russel's paradox, we define a set R such that R contains every set S with the property that S is not an element of S.
In other words, R is a set of all sets that don't contain themselves
So R is not quite the set of all sets
what u mean by X is not in X, how can a set in itself huh??
given any sets A and B, "A is in B" is a well-formed statement, and its negation is too
It really means for any value x that exists in any set A, for the set B to be the set of all sets it must contain the set that doesnt contain itself, thus x must not contain itself, i believe at least, if I am wrong please correct me
For example A={1,A} is a normal set, so it can be true