Zermelo-Fraenkel Set Theory

Поділитися
Вставка
  • Опубліковано 16 гру 2024

КОМЕНТАРІ • 52

  • @ptmad7910
    @ptmad7910 4 роки тому +11

    I am a fresher in COMP+MATH major and feeling so lucky there's someone who really explains things that got me stuck. Thank you Eliott! Surely I will try to watch every video you make as well.

  • @simba_1765
    @simba_1765 Рік тому +1

    Hello, I'm writing from Russia, your video is great. I've watched a lot of videos on this topic, but only you have an emphasis on the little things. Thanks! Amazing content!

  • @klammer75
    @klammer75 4 роки тому +35

    This gives a whole new meaning to being pro-choice☺️

    • @Dadarotas303
      @Dadarotas303 3 роки тому

      Pro choice has no meaning, The meaning however has always been in the choice itself.

  • @axiomateorema6355
    @axiomateorema6355 3 місяці тому +1

    Excelent video, as always, Mr. Nicholson.
    Best regards from Barcelona-Catalonia.

  • @rtg_onefourtwoeightfiveseven
    @rtg_onefourtwoeightfiveseven 4 роки тому +5

    I would watch every single one of your videos if only I had the time.

    • @Dr.Cassio_Esteves
      @Dr.Cassio_Esteves 4 роки тому

      I will (eventually) watch every single one of his videos.

  • @notablycurious6261
    @notablycurious6261 4 роки тому +4

    Thank you very much for your work, your videos are wonderful. Saved a lot of your playlists, I'll finish all of them someday.

  • @jimmyb998
    @jimmyb998 3 роки тому +5

    30:40 Yes, "Russel's Set" defined according to the axioms is equal to N: R=N, correct.
    Now according to the power set axiom, I can define P(N), of which N is an element. Next, using the axiom of subsets, I can construct {N} as {x e P(N) | x=N }. Since N=R, the first axiom implies {N} = {R}, which therefore also exists. Finally now, using the union axiom, I can construct {R} U N = {R,0,1,2,3,4,...}. So your claim (at 32:02) that this set cannot be constructed is wrong...
    Notice that the set {R,0,1,2,3,4,...} is NOT R. Call it Q instead: Q={R,0,1,2,3,4,...}. The question whether Q e Q can be answered unambiguously: it is false. So no contradiction.
    I think what you were trying to do 31:03, was to consider whether a DIFFERENT set, say R', could be defined as R'={R',0,1,2,3,...}. This is indeed impossible, and it would lead again to Russel's paradox.

  • @vladmarianciuc7574
    @vladmarianciuc7574 2 роки тому

    This is great. I was struggling understanding these concepts but you explain them very clear. Thank you for your work!

  • @nadavshemesh1231
    @nadavshemesh1231 3 роки тому +1

    Clear and Elegant explanation, thank you! much much appreciated

  • @jamesobrien8267
    @jamesobrien8267 Рік тому +2

    Self-Referential Consistency Theorem with Constructive Sets
    For fun I thought let's combine Zermelo-Fraenkel set theory with parts of Russell's paradox and Cantor's paradise theorems to create a completely new theorem.
    Example: "Self-Referential Consistency Theorem with Constructive Sets"
    Introduction: In this new theorem, we will explore the relationship between self-reference, consistency, and constructive sets within the framework of Zermelo-Fraenkel set theory. We will consider a set S that contains both its own complement and its power set.
    Self-Reference and Paradox: The new rules prompt us to examine the self-referential nature of set S. If we include S in itself, we create a potential paradox, as S would contain its own complement and its power set, leading to contradictions similar to Russell's paradox.
    Constructive Sets: To avoid paradoxes and inconsistencies, the new rules emphasize constructive approaches. We introduce a new set R, which is the set of all sets that do not contain themselves. In symbolic terms, R = {x | x ∉ x}.
    Resolving Paradox: By considering set R instead of S, we prevent self-reference and paradoxes. Set R contains all constructive sets that are free from contradictions. We also define a new set T, which contains the power set of R.
    Defining the New Theorem: Using constructive sets and avoiding self-reference, we now state the "Self-Referential Consistency Theorem with Constructive Sets": For any set R that contains all sets not containing themselves, the power set of R, denoted as T, is free from contradictions and inconsistencies arising from self-reference.
    Implications: The new theorem sheds light on the interplay between self-reference, consistency, and constructive sets within Zermelo-Fraenkel set theory. It provides a framework for defining sets without encountering paradoxes. The theorem opens up new avenues for exploring constructive set theory and its implications for the foundations of mathematics.
    Applications: The "Self-Referential Consistency Theorem with Constructive Sets" has potential applications in formal logic, set theory, and automated reasoning systems. It can inspire new research on the limitations of self-reference in mathematical systems and the importance of constructive approaches in avoiding paradoxes.
    In conclusion, by combining Zermelo-Fraenkel set theory with elements from Russell's paradox and Cantor's paradise theorems, we have created the "Self-Referential Consistency Theorem with Constructive Sets." This theorem highlights the significance of constructive sets in ensuring the consistency of self-referential systems and extends our understanding of the boundaries of provability and consistency in mathematical reasoning.

    • @dng88
      @dng88 21 день тому

      Very strange why the refutation of this construction by indeterminate theorem in 192x?

  • @solanita222
    @solanita222 3 роки тому +1

    This video truly saved me.

  • @warwolt
    @warwolt 4 місяці тому

    13:26 minor nit pick, the empty set symbol is null, not phi

  • @BleachWizz
    @BleachWizz 3 роки тому +2

    Axiom of choice is only obvious to me if the intersection between the elements of the set is 0, then you can choose any representative for the elements of the set that are inside the set.
    So axiom of choice basically says you can make {{even}, {odd}} => {0,1}

    • @MikeRosoftJH
      @MikeRosoftJH 2 роки тому

      That's two different (but equivalent) formulations of the axiom of choice. One way is: given any collection of disjoint non-empty sets, there exists a set which contains a single element from each set in the collection. (And here the 'disjoint' condition is important; if the collection were {{1,2}, {2,3}, {1,3}}, then no set can have exactly one element from each set.) The other formulation is: given any collection of non-empty sets, there exists a function which maps each set in the collection to its element. (For collection {{1,2}, {2,3}, {1,3}} the choice function obviously exists; for example: {1,2}->1, {2,3}->2, {1,3}->3.
      For finite sets the axiom of choice is trivial; it's really relevant when the original collection contains infinitely many sets. (And even axiom of choice for infinitely many finite sets isn't obvious. One mathematician has put it: Axiom of choice isn't needed to pick one of each from infinitely many pairs of shoes - you can always pick the left shoe; but it's needed to pick one of each from infinitely many pairs of socks. That is: it's consistent without axiom of choice that there exists some countably infinite collection of two-element sets, which doesn't have a choice set. It also follows: it's not necessarily the case without axiom of choice that a union of countably many finite sets is countable.)

  • @thedoctordowho2022
    @thedoctordowho2022 3 роки тому +1

    Year man!!!Aweosome video or yours.Really intresting.

  • @CardiganBear
    @CardiganBear 3 роки тому +1

    Really excellent video - thanks. Leaves me wanting more... How about those other 3 axioms?

  • @markovwallenstein9357
    @markovwallenstein9357 7 місяців тому

    Doesn't contain the axiom of replacement?

  • @erikdellidlpuntocom4632
    @erikdellidlpuntocom4632 3 роки тому

    Lovely explanation!

  • @manuelocana8074
    @manuelocana8074 3 роки тому +1

    Are there more examples of sets like Rusell's?

    • @manuelocana8074
      @manuelocana8074 3 роки тому

      Um, and how do you know the axioms in zfc do not leave contradictions?

    • @guilhermefurquim8179
      @guilhermefurquim8179 3 роки тому

      @@manuelocana8074 you don't know

    • @tim40gabby25
      @tim40gabby25 3 роки тому

      Including 10 could be called "fragmentation set theory", old duffer here, enjoying the ride :)

    • @manuelocana8074
      @manuelocana8074 3 роки тому

      @@tim40gabby25 i don't understand what you say

  • @dng88
    @dng88 21 день тому

    But that Soln does not work because of another theorem in 192x …

  • @mariaritacorreia9340
    @mariaritacorreia9340 Рік тому

    I'm a fan! 🤩

  • @badtimes5247
    @badtimes5247 3 роки тому

    how did you do it can you share with me , thank you

  • @ROForeverMan
    @ROForeverMan Рік тому +2

    Russell's paradox is fixed in a totally different way, namely by understanding consciousness. To see how this is done, see my paper "The Self-Referential Aspect of Consciousness". And for people that might read my message in the future, check for the paper that I'm currently working on where I go even deeper into the subject, which will be named "How to Build a World". Author: Cosmin Visan.

  • @AlexandrePorto
    @AlexandrePorto 3 роки тому +2

    I CAN'T watch this, the audio ia too low even at maximum volume

    • @johnandrheyabordo644
      @johnandrheyabordo644 7 місяців тому

      I think the issue isn't the video. I can precisely hear the speaker from my end.

  • @ronaldrussel1158
    @ronaldrussel1158 4 роки тому +1

    Great, thank you!!!

  • @fallonhendrie4350
    @fallonhendrie4350 Рік тому

    This is awesome

  • @jongraham7362
    @jongraham7362 2 роки тому

    It is not entirely clear to me why one needs the pairing axiom if one already has the empty set axiom and the power set axiom. Don't the two together imply that one can build sets like {Ø, {Ø}} ... I'm not clear on the point.

    • @jongraham7362
      @jongraham7362 2 роки тому

      I suppose it is useful in order to create this: {{{{{ }}}}} as something meaningful. Which leads to this: {{{{ Ø }}}} = 4.

    • @MikeRosoftJH
      @MikeRosoftJH 2 роки тому +2

      @@jongraham7362 Axiom of pairing says that for any elements (sets) a, b, there exists the set {a,b} (the set P, such that x∈P if and only if x=a or x=b); when a=b, this yields the set {a,a}={a}. Sure, you can build specific sets using axiom of powerset on the empty set; but the axiom says that the common set exists for every pair of sets. (Though I believe that you can derive axiom of pairing from other axioms, such as from the axiom of powerset and axiom of replacement.)
      Sure, you can define 0 as empty set, 1={0}={{}}, 2={1}={{{}}}, and so on. But it's more convenient to use the von Neumann definition of natural numbers, where every number is a set of all numbers less than itself: 0 is the empty set, 1 is {0}, 2 is {0,1}, 3 is {0,1,2}, and so on. (Observe that any natural number n is a set with exactly n elements; and given natural number n, the number n+1 is obtained by taking the set n, and adding n itself to it as an additional element: n+1=n∪{n}.) This naturally extends to the infinite case (ordinal numbers).

  • @cooterhead_jones
    @cooterhead_jones 7 місяців тому

    Who is the man?

  • @sergeivesna4091
    @sergeivesna4091 3 роки тому

    0 1st 6th 10th and what about 7th 8th 9th ?

  • @chico6854
    @chico6854 4 роки тому

    Thanks !!!

  • @imaxus1128
    @imaxus1128 3 роки тому +2

    Who are you? You seem to know so much about so many fields of knowledge...

  • @BirdsRntReal
    @BirdsRntReal Рік тому

    Maths or mathematics? So by taking away free choice you have the zermelo set theory. So as long as there is a perfect restriction mathematics is valid.

  • @jps17183
    @jps17183 2 роки тому +1

    Axiom 0 is already a philosophical absurd. It's like saying Nothing is something (in this case a set). But, if you consider the semantics of it the preposition is already problematic in the moment you say "Nothing IS"... "Is" is already stating an existence of x (i.e. not nothing).

    • @هيلة-ع8م
      @هيلة-ع8م 2 роки тому

      How would you define nothingness?

    • @jps17183
      @jps17183 2 роки тому

      ​@@هيلة-ع8م You can't. It's a transcendental idea

    • @هيلة-ع8م
      @هيلة-ع8م 2 роки тому

      @@jps17183 ._. Hmm excuse my ignorance, but would it not be “something” the moment we think about it? I mean the idea already has a name (=nothingness)

    • @jps17183
      @jps17183 2 роки тому

      @@هيلة-ع8م Very well observed. Exactly and that's why is transcendental . The name is just that... A name to refer to some intuition we all share, however understanding an idea is different that compreending it. You can understand infinity (otherwise you couldn't know the truth value of the statment: "is allways bigger of that you are thinking now") but you can't compreend it (i.e. you cannot circusncribe/capture it it to concept) . Another example is the term consciousness. We all know what that is because weshare that otherwise you couldn't make sense of that word. It's like trying to explain the redness of red to a blind man. Those names you give to ideas are not relevant, are just symbolic means through witch you can communicate them. The ideas however are transcendental and trying to define them will led to logical paradoxes.

  • @lincolnuland5443
    @lincolnuland5443 5 місяців тому

    ok so there's this thing called the empty set and it's got nothing in it. And we take that empty set and put another empty set inside it and that's not an empty set anymore...blah blah blah and that's the integers!!! ta da!...... Absolutely absurdly ridiculous nonlogic that has nothing to do with anything. How TF did a label / symbol used to represent no quantity become a "number"? wtf...

    • @warwolt
      @warwolt 4 місяці тому +1

      Set theory provides common mathematical objects to _represent_ all other mathematical objects, e.g. we can represent the natural numbers using sets. It's not meant as a _reduction_ , i.e. we don't need to the consider the number 2 to actually _be_ the set {0,1}, it's just a representation. It allows any arbitrarily complex mathematical object to be ultimately built out of the same stuff, so all mathematicians have a common language

  • @ThinkTank255
    @ThinkTank255 2 роки тому

    How can you prove that no natural number is red? Indeed (I'm not kidding here), I think there are natural numbers that are red. In fact, I think that is exactly what "red" is, just some (probably) gigantic natural number. So, then, you have asserted a fallacy, from which anything can follow. IMO, non-constructive mathematics is nonsense. Asserting an object exists using pseudo-mathematical notation doesn't make it exist. You have to actually demonstrate an object exists mathematically and non-constructive mathematics cannot do that (it can only assert). Therefore, 20th century (and now, sadly, 21st century) mathematics is on VERY shaky ground. It doesn't matter how clever you try to be, non-constructive set theory ultimately fails. You did a good job at pointing out the controversies around the Axiom of Choice, however, the Axiom of Infinity is also rather controversial in my opinion. At first, it seems rather benign. However, what you ahve actually done is assert an object exists without actually demonstrating its existence. What you have actually demonstrated is that you can construct {0}, {0,1}, {0,1,2}, etc.... Each of those sets will always have the sucessor of the greatest element missing. You can, instead, talk about an equivalence class of sets defined by {0: f(x)} and f(x) is a set generating function such that f({0}) = {0, 1 : f(x)}, for example, or anything else. In some sense, it is the opposite of the Axiom of Restricted Comprehension, however, we are defining an equivalence class and not a set. Note, there is never a complete set {1,2,3,...}. Only the sets {1:f(x)}, {1,2:f(x)}, {1,2,3:f(x)}, etc....