Thank you very much very clear explanation. I am currently taking an introduction course to modern mathematics in which 1/4 of the content are related to set theory, which takes me lots of time to understand it.
property of the elements of A is the set of all sets that dont contain themselves A = {B,C} Sinse A doesnt contain itself, A is now an element of the set of sets that dont contain themselves, wich is A A ={A,B,C} but now A contains itself so by its property i gotta remove it A = {B,C} but now A doesnt contain itself so is a member of A A ={A,B,C} but now A contains itself so by its property i gotta remove it A = {B,C} but now A doesnt contain itself so is a member of A . . . Is this correct?
Is the restriction of this axiom what forced us to create the union axiom? To me it looks like that with the unrestricted version of the axiom we could create unions with it and the pairing axiom, without needing the union axiom
3:00 I think a better description of the set of all sets that do not contain themselves is "regular" , i believe that's the reason we use the capital letter R. Most regular or normal sets do not contain themselves. Thus it is misleading to call R pathological. The problem is that it is not a well defined set, since we cannot answer if R ϵ R or R ∉ R.
+xoppa09 It's true that most sets don't contain themselves but this is a set which is of all those sets that do not contain themselves. But it is nonetheless pathological (this is known to me but not the viewer at this point). I believe R is used in referring to Bertrand Russell.
Thanks for making this supplemental video on Russell's paradox. Prior to coming across your lecture series, just learning about the Barber paradox was not satisfying because I wanted to see the actual math, the notations and definitions and such. Now that I have gone through all your videos, I can understand the paradox in the notation form and that makes it much much clearer. A professor in another video, in talking about Russell's barber paradox and Turing's halting problem, said that there has to be the element of "not-ness." Now I see that in Naive Set Theory, the problem occurs with R, the set of sets that do NOT include themselves. I also now understand what the "naive" is in the Naive Set Theory. It's the condition of being unrestricted in the Axiom of comprehension, which you explained so clearly. I was a little confused about the set A in the restricted comprehension at first. What is this other set that is assumed to exist? I was thinking that an actual example problem would be helpful here, as you did with the set of red things and the set of not-motorcycle things at the beginning of your video. But then I figured out that the set A is the set of all the elements of R, from which R is excluded. So, as you said, ZFC dodges the paradox by saying that R can't exist. So if we think of this in terms of the Barber paradox, it's saying that the barber is restricted from entertaining the paradoxical question "Do I shave myself?" because he is not allowed to think of himself as belonging to the set of those that do not shave themselves. So the word "dodge" is the right word because the Axiom of subsets/Axiom of schema of separation is simply saying that we are just not going to have a set of all sets that do not list themselves. So I guess, for the barber, the village council has to come up with an amendment that says that the barber is the only one that can be bearded or something like that. Hey, thanks again for making all this clear!
John Ok Was that a "computerphile" video by any chance? I also saw that one when I was first getting interested in Turing's work on the halting problem. Also, that set, A, just has to be some set that we're sure exists, for the purpose of creating the new set, R, which is going to be a subset of A. Another interpretation of the subset axiom would be that if you grant me the existence of A, then I can form whatever subsets I like by applying some condition (the technical term is "predicate" or "formula") to form some new set, R. The predicate could always be false about all the members of A, in which case R would just be the empty set, which makes sense because the empty set is a subset of all sets. With the Barber paradox, I always feel like there's some linguistic way one could dodge the paradox by saying that the barber is from a different town, or is an alien, or some other bizarre answer, but the whole point of the Barber paradox is to give a verbal metaphor for the set theory paradox, so these answers aren't satisfying. Perhaps we might say that the barber's definition of shaving all and only those who don't shave themselves is incoherent. After all, that's what's going on in naive set theory, the assumption that if we can imagine any formula, there must be a way to list out all the objects making the formula true. But, we get into problems when we list out the people shaved by the barber.
mdphdguy1 Yes, it was a Computerphile/Numberphile video. Their videos are like tiny morsels that give you appetite for more. Right now, I am viewing a lecture by Prof. Raymond Flood at Gresham College, titled "Cantor's Infinities." I first watched "Infinity is bigger than you think" by Numberphile, and that got me want to find out some more.
"The barber has to simultaneously shave himself and not shave himself" says the presenter. The heart and solution of the paradox lies with "simultaneously". The paradox isn't about barbers or sets but features any verb or verb phrase - shave, paint, adore, share a pizza with, include in a set - that can take a subject, object and tense. But here we're concerned with logical tense. Unlike grammatical tense, logical tense is not necessarily signified linguistically by, say, the suffix "-ed", by auxiliaries like "will" or "going to", or words like "previously", "now", "tomorrow". Instead it depends on the position of the sentence or proposition the verb is in relative to other sentences in the argument which contain that same verb. It follows that each successively repeated verb (and the clause it's in) has a different logical tense, refers to a different time in the stream of utterances. A painter paints all and only those who don't paint themselves. Making the logical tenses explicit, this can become: She will paint tomorrow all and only those who didn't paint themselves yesterday". So if she didn't paint herself yesterday, she will tomorrow. And if she did, then she won't. Simple. Logical. If Russell's set didn't include itself, it does now. But if it did, then it doesn't. This is not an entirely unfamiliar idea. In arithmetic and algebra we use brackets to indicate operational tense, the order which operations are to be performed, to avoid a contradictory result. Think of set inclusion as an operation, like painting and shaving or multiplying and adding.
Lets make it more verbose and less prone to grammar-Nazi attack. Who shaved the barber, given that: (1) Everyone was shaved yesterday (nobody had beard more than 24 hours long yesterday evening). (2) Everyone shaved today (3) Everyone can shave themselves, but only barber can also shave others (4) Barber doesn't shave a person if person has less than a week worth of beard. (5) Barber is shaved today. Java language have an answer for this one. The answer is "NullPointerException" - no such barber. If you allow someone else to shave this barber - problem, is solved. But under all these constraints the problem has no solution.
+Aphrontic Alchemist Perhaps one could work the Russell paradox down into a statement that looks like _this statement is false_ in natural language, but I wouldn't say the Russell paradox is convoluted. The statement in natural language most similar to the Russell paradox is the barber paradox
"This statement is false" paradox is like a saying "some theories are incorrect". Well, sure. So what and how is that my responsibility? Russell's paradox is more like "Cantor, YOUR theory is broken - you'd better fix it". This is a directed message and Set theory was fixed to address the problem.
I think you need tp assume or prove nonexistence of set of all sets too, to avoid Russel's paradox in ZFC. Otherwise one can take the universal set for A and R always belongs to A How exactly is the Barbers paradox solved? In that we are taking subsets of the set of all people in the village
* If the set of all sets exist, then the set of all sets x for which the statement x∉x is true also exists. (By the axiom of separation.) * Such a set does not exist; the assumption that it does leads to a contradiction. * Therefore, the set of all sets does not exist, either. If you want to prove that the statement "there is a set containing every set" is not provable in ZFC, then this is going to be very difficult; in particular, Gödel has proven that no sufficiently strong theory can prove its own consistency (unless the theory is itself inconsistent). But on the other hand, it can be easily shown that the negation of the above statement, i.e. "there doesn't exist a set containing every set" (or equivalently, "given any set X, there is some set Y such that Y is not in X") is provable in ZFC. (From the axiom of regularity it can be proven that no set can contain itself, from which it immediately follows that the set of all sets can't exist. But proving that there is no set of all sets does not actually require the axiom of regularity. Remember: if a theory is inconsistent, then adding an axiom to it does not make it consistent.)
Thank you. Great video. With these videos learning tough math theories is fun.
i am hoping that you can made some video about abstract algebra or analysis, your video is so nice!
Thank you very much very clear explanation. I am currently taking an introduction course to modern mathematics in which 1/4 of the content are related to set theory, which takes me lots of time to understand it.
property of the elements of A is the set of all sets that dont contain themselves
A = {B,C}
Sinse A doesnt contain itself, A is now an element of the set of sets that dont contain themselves, wich is A
A ={A,B,C} but now A contains itself so by its property i gotta remove it
A = {B,C} but now A doesnt contain itself so is a member of A
A ={A,B,C} but now A contains itself so by its property i gotta remove it
A = {B,C} but now A doesnt contain itself so is a member of A
.
.
.
Is this correct?
Very good explanation of how Russel's paradox is killed off in ZFC.
Is the restriction of this axiom what forced us to create the union axiom? To me it looks like that with the unrestricted version of the axiom we could create unions with it and the pairing axiom, without needing the union axiom
3:00 I think a better description of the set of all sets that do not contain themselves is "regular" , i believe that's the reason we use the capital letter R. Most regular or normal sets do not contain themselves. Thus it is misleading to call R pathological. The problem is that it is not a well defined set, since we cannot answer if R ϵ R or R ∉ R.
+xoppa09
It's true that most sets don't contain themselves but this is a set which is of all those sets that do not contain themselves. But it is nonetheless pathological (this is known to me but not the viewer at this point). I believe R is used in referring to Bertrand Russell.
Thanks for making this supplemental video on Russell's paradox. Prior to coming across your lecture series, just learning about the Barber paradox was not satisfying because I wanted to see the actual math, the notations and definitions and such. Now that I have gone through all your videos, I can understand the paradox in the notation form and that makes it much much clearer.
A professor in another video, in talking about Russell's barber paradox and Turing's halting problem, said that there has to be the element of "not-ness." Now I see that in Naive Set Theory, the problem occurs with R, the set of sets that do NOT include themselves. I also now understand what the "naive" is in the Naive Set Theory. It's the condition of being unrestricted in the Axiom of comprehension, which you explained so clearly.
I was a little confused about the set A in the restricted comprehension at first. What is this other set that is assumed to exist? I was thinking that an actual example problem would be helpful here, as you did with the set of red things and the set of not-motorcycle things at the beginning of your video. But then I figured out that the set A is the set of all the elements of R, from which R is excluded. So, as you said, ZFC dodges the paradox by saying that R can't exist. So if we think of this in terms of the Barber paradox, it's saying that the barber is restricted from entertaining the paradoxical question "Do I shave myself?" because he is not allowed to think of himself as belonging to the set of those that do not shave themselves. So the word "dodge" is the right word because the Axiom of subsets/Axiom of schema of separation is simply saying that we are just not going to have a set of all sets that do not list themselves. So I guess, for the barber, the village council has to come up with an amendment that says that the barber is the only one that can be bearded or something like that.
Hey, thanks again for making all this clear!
John Ok Was that a "computerphile" video by any chance? I also saw that one when I was first getting interested in Turing's work on the halting problem. Also, that set, A, just has to be some set that we're sure exists, for the purpose of creating the new set, R, which is going to be a subset of A. Another interpretation of the subset axiom would be that if you grant me the existence of A, then I can form whatever subsets I like by applying some condition (the technical term is "predicate" or "formula") to form some new set, R. The predicate could always be false about all the members of A, in which case R would just be the empty set, which makes sense because the empty set is a subset of all sets. With the Barber paradox, I always feel like there's some linguistic way one could dodge the paradox by saying that the barber is from a different town, or is an alien, or some other bizarre answer, but the whole point of the Barber paradox is to give a verbal metaphor for the set theory paradox, so these answers aren't satisfying. Perhaps we might say that the barber's definition of shaving all and only those who don't shave themselves is incoherent. After all, that's what's going on in naive set theory, the assumption that if we can imagine any formula, there must be a way to list out all the objects making the formula true. But, we get into problems when we list out the people shaved by the barber.
mdphdguy1 Yes, it was a Computerphile/Numberphile video. Their videos are like tiny morsels that give you appetite for more. Right now, I am viewing a lecture by Prof. Raymond Flood at Gresham College, titled "Cantor's Infinities." I first watched "Infinity is bigger than you think" by Numberphile, and that got me want to find out some more.
"The barber has to simultaneously shave himself and not shave himself" says the presenter. The heart and solution of the paradox lies with "simultaneously". The paradox isn't about barbers or sets but features any verb or verb phrase - shave, paint, adore, share a pizza with, include in a set - that can take a subject, object and tense. But here we're concerned with logical tense. Unlike grammatical tense, logical tense is not necessarily signified linguistically by, say, the suffix "-ed", by auxiliaries like "will" or "going to", or words like "previously", "now", "tomorrow". Instead it depends on the position of the sentence or proposition the verb is in relative to other sentences in the argument which contain that same verb. It follows that each successively repeated verb (and the clause it's in) has a different logical tense, refers to a different time in the stream of utterances. A painter paints all and only those who don't paint themselves. Making the logical tenses explicit, this can become: She will paint tomorrow all and only those who didn't paint themselves yesterday". So if she didn't paint herself yesterday, she will tomorrow. And if she did, then she won't. Simple. Logical. If Russell's set didn't include itself, it does now. But if it did, then it doesn't. This is not an entirely unfamiliar idea. In arithmetic and algebra we use brackets to indicate operational tense, the order which operations are to be performed, to avoid a contradictory result. Think of set inclusion as an operation, like painting and shaving or multiplying and adding.
Lets make it more verbose and less prone to grammar-Nazi attack. Who shaved the barber, given that:
(1) Everyone was shaved yesterday (nobody had beard more than 24 hours long yesterday evening). (2) Everyone shaved today (3) Everyone can shave themselves, but only barber can also shave others (4) Barber doesn't shave a person if person has less than a week worth of beard. (5) Barber is shaved today.
Java language have an answer for this one. The answer is "NullPointerException" - no such barber. If you allow someone else to shave this barber - problem, is solved. But under all these constraints the problem has no solution.
r is not in r is a subset of r?? lovely videos
So basically the Russel paradox is a more convoluted way of saying the "this statement is false" paradox
+Aphrontic Alchemist
Perhaps one could work the Russell paradox down into a statement that looks like _this statement is false_ in natural language, but I wouldn't say the Russell paradox is convoluted. The statement in natural language most similar to the Russell paradox is the barber paradox
I meant "technical"
"This statement is false" paradox is like a saying "some theories are incorrect". Well, sure. So what and how is that my responsibility?
Russell's paradox is more like "Cantor, YOUR theory is broken - you'd better fix it". This is a directed message and Set theory was fixed to address the problem.
I think you need tp assume or prove nonexistence of set of all sets too, to avoid Russel's paradox in ZFC. Otherwise one can take the universal set for A and R always belongs to A
How exactly is the Barbers paradox solved? In that we are taking subsets of the set of all people in the village
* If the set of all sets exist, then the set of all sets x for which the statement x∉x is true also exists. (By the axiom of separation.)
* Such a set does not exist; the assumption that it does leads to a contradiction.
* Therefore, the set of all sets does not exist, either.
If you want to prove that the statement "there is a set containing every set" is not provable in ZFC, then this is going to be very difficult; in particular, Gödel has proven that no sufficiently strong theory can prove its own consistency (unless the theory is itself inconsistent). But on the other hand, it can be easily shown that the negation of the above statement, i.e. "there doesn't exist a set containing every set" (or equivalently, "given any set X, there is some set Y such that Y is not in X") is provable in ZFC. (From the axiom of regularity it can be proven that no set can contain itself, from which it immediately follows that the set of all sets can't exist. But proving that there is no set of all sets does not actually require the axiom of regularity. Remember: if a theory is inconsistent, then adding an axiom to it does not make it consistent.)
I have always wondered why more logic is not required course work for pure mathematics major