The Axiom of Choice

Поделиться
HTML-код
  • Опубликовано: 4 дек 2024

Комментарии • 79

  • @chair547
    @chair547 4 года назад +76

    The set of my friends = ø

    • @cookiecrumbzi
      @cookiecrumbzi 2 года назад +10

      Just put that set inside another and u’ll have one

    • @cristianm7097
      @cristianm7097 11 месяцев назад +3

      ​@@cookiecrumbzi True, the empty set is my friend.

    • @mostrat
      @mostrat 8 месяцев назад

      it’s phi not that ugly thing

  • @gabe5865
    @gabe5865 Год назад +4

    Thank you so much! I am a senior undergraduate math student doing a paper on the Banch-Tarski paradox and this literally helped me understand it so much better than any video could!!!

  • @albertkang9886
    @albertkang9886 5 лет назад +29

    This is the best intuitive introduction to AoC I've ever seen. Thank you!

    • @maxpercer7119
      @maxpercer7119 2 года назад

      no, it's not, its full of conceptual errors and is misleading...
      but if you have a mathematical background you can kinda get the gist of it
      also it's a great philosophical-esque video ;o

  • @vector8310
    @vector8310 Год назад +5

    It was only after you related the axiom of choice to well ordering and to the Banach-Tarski paradox did the the concept of the axiom gel for me. Thank you for this intuitive exposition.

  • @MikeRosoftJH
    @MikeRosoftJH 5 лет назад +41

    The description of well-ordering is inaccurate. A set is well-ordered by order relation

    • @SP-qi8ur
      @SP-qi8ur 4 года назад +2

      What if we considered the rationals in the open interval from 0 to 1? Would these have a minimum for every subset?

    • @MikeRosoftJH
      @MikeRosoftJH 4 года назад +3

      @@SP-qi8ur Not under the usual order; but because rationals are countably infinite, there exists a different order on the same set which has this property. (In fact, axiom of choice implies - and is equivalent to - the proposition that a well-ordering relation exists on every set.)

    • @SP-qi8ur
      @SP-qi8ur 4 года назад +2

      @@MikeRosoftJH I see. So countable sets can always be well-ordered regardless of the axiom of choice. It is with uncontably infinite sets that choice is necessary and sufficient for well-ordering. Thanks for replying.
      P.D. Is there a known well-ordering of the reals? Does the axiom of choice merely state that there is one, or can it be used to define it?

    • @MikeRosoftJH
      @MikeRosoftJH 4 года назад +8

      @@SP-qi8ur There are many uncountable sets which can be well-ordered. In fact, Hatogs theorem says that for every well-ordered set there exists another well-ordered set with strictly greater cardinality.
      As for "known" well-ordering of reals: while axiom of choice implies that real numbers (and any other set) can be well-ordered, no formula can be proven to define a well-ordering of reals (without additional axioms). There does exist a specific definable relation for which it's consistent that this is a well-ordering of reals (more precisely: the relation is a well-ordering of some subset of reals, but it can't be proven that it covers all reals).
      As I have said, axiom of choice is equivalent to the proposition that all sets can be well-ordered. So if axiom of choice is false, then there is some set which can't be well-ordered - it could be the reals, it could be the set of all functions on reals, it could be some other set. (Of course, if reals can't be well-ordered, then nor can functions on reals.) Another proposition which is equivalent to axiom of choice is that the cardinality of any two sets is comparable (either one can be mapped one-to-one with a subset of another, or vice versa): if every set can be well-ordered, then cardinality of all sets is comparable; and if some set can't be well-ordered, then by Hartogs theorem there exists a set whose cardinality is incomparable with it.

    • @SP-qi8ur
      @SP-qi8ur 4 года назад +3

      @@MikeRosoftJH Thank you for your insight

  • @wargreymon2024
    @wargreymon2024 Месяц назад

    It's straight to the point when it comes to intuition, well-done!

  • @TomTom-rh5gk
    @TomTom-rh5gk 3 года назад +6

    Finally someone explains how one sphere can be made into two identical spheres. This is the only discussion of the Axiom of Choice that tried to make sense. The rest suck.

  • @georgelaing2578
    @georgelaing2578 2 года назад +4

    Beautiful and enlightening!!

  • @WaubObWobble
    @WaubObWobble Год назад

    A Christian college in Michigan just told me to hit that bell icon so I can be notified of any new videos

  • @pikksburgh420
    @pikksburgh420 4 года назад +6

    I was intrigued when hearing about this "Axiom of choice" on the TV show Chance. However I'm totally lost lol.

  • @ExistenceUniversity
    @ExistenceUniversity Год назад

    7:40 How many oranges did you cut to make the orange peel shells? 1 orange or 2?

  • @onecommunistboi
    @onecommunistboi 3 года назад +5

    I think I misunderstand the meaning of the Well Ordering Theorem. If I can order the reals, doesnt that mean that I have a bijective function that maps the naturals to the reals? But by the definition I learned that would mean that the reals are only countably infinite. So I guess my question is, doesn't the Well Ordering Theorem contradict the existence of an uncountably infinite set?

    • @dekippiesip
      @dekippiesip 3 года назад +1

      I don't see a paradox here. The set is well ordered in the sense that their exists a relation < such that for any 2 distinct numbers x and y either x < y or y < x. For the reals it just happens to be the case that their are an infinite number of numbers t such that x < t < y, that's why you can't use this ordering to construct a bijection with the naturals.

    • @onecommunistboi
      @onecommunistboi 3 года назад

      @@dekippiesip Thank you! :)

    • @MikeRosoftJH
      @MikeRosoftJH 3 года назад +3

      ​@@dekippiesip That's not what a well-ordering relation means. A set is well-ordered by ordering relation

    • @bruh-pj3kq
      @bruh-pj3kq Месяц назад

      @onecommunistboi the person that responded to you was wrong; they had no idea what they’re talking about.
      The reason we can well-order R and have that not be isomorphic to N is because, unlike N, the ordering on R will have limit points.
      That is to say, when you well-order a set, it makes sense to talk about a successor.
      In N, the successor of 1 is 2, of 2 is 3, and so on.
      The key difference here is that every natural number can eventually be obtained by taking the successor of previous elements. I.e for any integer, let’s say 10 for example, we can write a chain given by 1 < 2 < 3 < 4 < 5 < 6 < 7 < 8 < 9 < 10.
      However, in the well-ordering for R, this is not the case. When you well-order R, you will obtain a first element, say a. But if you try an define a bijection by 1 -> a and 2 -> s(a) and so on, you will not get every element in R; there will be elements in R that follow that infinite sequence.

    • @MikeRosoftJH
      @MikeRosoftJH 8 дней назад

      @@bruh-pj3kq I'd say that this is backwards; it's not immediately obvious why a well-ordering on real numbers needs to have limit points. Conversely, a well-ordering on natural numbers can have limit points; a trivial example is an order where 1

  • @arekkrolak6320
    @arekkrolak6320 3 года назад +4

    You can leave exercise to your dog

  • @alireza2974
    @alireza2974 3 года назад +1

    that was good explanation thank you.

  • @davidwright8432
    @davidwright8432 6 месяцев назад

    In the Z-example, ordering {0, 1 -1, ...} why is 0 the least ? Unless the ordering involved is not the usual one of the integers, but simply defines the set as starting with 0? -1 (and all other negative integers) are smaller, in the usual ordering, than 0.

    • @MikeRosoftJH
      @MikeRosoftJH 6 месяцев назад

      Well, if you want to, you can put it to the millionth position. The whole point is to come up with a non-standard ordering under which the set of integers has the same order type as the set of natural numbers under the usual order; in other words, to find an order such that every integer appears at some finite position.

  • @syed9576
    @syed9576 4 года назад +5

    I think it's interesting how he keeps writing "Axiom of Choice" but WOT for well ordered theorem... lol

    • @anticorncob6
      @anticorncob6 3 года назад +1

      I think we know why he did that. :)
      Although he could've just written AC for it, and a lot of mathematicians even several decades ago wrote AC for axiom of choice.

  • @liuzhen2008
    @liuzhen2008 4 года назад +1

    So.. how would one order real numbers between 0 and 1 given WOT.?

    • @MikeRosoftJH
      @MikeRosoftJH 4 года назад +2

      Well, axiom of choice tells you that the real numbers - and any other set - can be well-ordered; but it doesn't tell you how to do that. In fact, no formula can be proven to define a well-ordering of reals. (Without axiom of choice, it's consistent - assuming set theory itself is consistent - that real numbers can't be well-ordered.) There's one thing you *can* do: if you take the axiom of constructibility, then there is a specific (highly complicated) formula which defines a well-ordering not just on reals, but on the class of all sets. So going back, there is a formula for which it's consistent that it's a well-ordering of reals. (And before you ask: that formula has nothing to do with the numerical value of the reals, but depends on how the real numbers are realized as sets in set theory.) To be precise: this formula defines a well-ordering on some set of reals, but - without assuming the axiom of constructibility - you can't prove that this set includes all reals.

  • @divermike8943
    @divermike8943 4 дня назад

    5:11 F is a collection of sets. What the heck is uF ?
    I got completely lost here.

    • @divermike8943
      @divermike8943 4 дня назад

      Okay. I Googled it. UF is the union of all the elements. If F is a collection of sets, UF is a set that contains all the elements of all those sets.
      Still lost at this point in the video.

    • @divermike8943
      @divermike8943 4 дня назад

      I fear I will never understand the axiom of choice. Thank goodness I don't need to.

    • @MikeRosoftJH
      @MikeRosoftJH 13 часов назад

      @@divermike8943 Axiom of choice is the following proposition: Given any collection of disjoint non-empty sets, there exists a set containing a single element from each. This is equivalent to the following formulation: Given any collection of non-empty sets, there exists a function which maps every set in this collection to an element of that set. (Note that in set theory - ZF or ZFC - all elements are themselves sets. A function is therefore also a set - a set of ordered pairs; if pair [a,b] is an element of f, then we write f(a)=b. And an ordered pair also needs to be realized as a set; the conventional realization is {{a}, {a,b}} - it can be shown that it's indeed true that [a,b]=[c,d] if and only if a=c and b=d.)
      This uses the second formulation. So let F be some set - collection of sets. We posit that there exists a function f whose domain is F, and for every A∈F is f(A)∈A. Therefore, this is a function from F to the union of all sets in F. We write this as ∪F; or, more verbosely, as ∪x: x∈F (the "x∈F" clause can be written under the union sign). That the set ∪F exists is a consequence of the axiom of union.

  • @maxpercer7119
    @maxpercer7119 2 года назад

    I would add this video to a 'youtube university - philosophy and math' section/playlist.

  • @AlexSav
    @AlexSav 7 месяцев назад +1

    I think Christ did something similar with bread. People then called it a miracle, not paradox.

    • @MikeRosoftJH
      @MikeRosoftJH 6 месяцев назад +1

      It follows that Jesus is pro-choice, because Banach-Tarski theorem is a consequence of the axiom of choice.

  • @jcool2199
    @jcool2199 Год назад +1

    This helped me a lot thank you so much

  • @shantanukulkarni6237
    @shantanukulkarni6237 3 года назад +5

    Can't believe I was assigned a proof as an exercise by a youtuber while I was trying to relax away from my studies. Math you got me again. Aaargh.
    I'm joking of course. Great video. 😁

  • @CjqNslXUcM
    @CjqNslXUcM 2 года назад

    thx

  • @thbeam2044
    @thbeam2044 3 года назад +1

    3:56 surely the set of absolute-values of integers isn't well-ordered, since e.g. |1| = |-1|? well-ordering requires total order :/

    • @MisterAppleEsq
      @MisterAppleEsq 3 года назад +3

      "With negative numbers coming earlier in the list than their positive counterparts" was specified.

    • @thbeam2044
      @thbeam2044 3 года назад +3

      @@MisterAppleEsq Whoops, missed that part. thanks!

  • @hearteyedgirl
    @hearteyedgirl 7 месяцев назад

    came in for Skittles stay for

  • @trillionman2105
    @trillionman2105 3 года назад +2

    Good video but some of your curly brackets are simply disturbing

  • @CristianGambino-qs2ey
    @CristianGambino-qs2ey 10 месяцев назад

    A really nice and useful video, considering that it explain two- three theorem that i have never been able to understand very well. If i understand correctly, one could say that the Banach-Tarski theorem is the equivalent, on the three-dimesional plan, of the set theory and Hilbert Hotel

  • @pashute12
    @pashute12 2 года назад

    His name was pronounced Avraham Frankl. not frank-Ell. The emphasis is on the Frank.
    His grandchildren are quite famous in Israel, one of them was my brother's boyhood best friend.

  • @Darrida
    @Darrida 4 месяца назад

    Babylonian school (physics) against Greek School (mathematics)

  • @tl1326
    @tl1326 2 года назад

    when you wrote axiom
    i read that as A x 10m

  • @jimbronson687
    @jimbronson687 2 года назад

    Although I very much like abstract and non abstract thus a set I see no practical application for AoC? Yes No..

  • @markwrede8878
    @markwrede8878 7 месяцев назад

    Choice is not rigorously defined in that it must assume sets and members of sets as within its warrant. Choice must acknowledge the distinction in its selected set between set elements and member elements of other sets.

  • @JTient
    @JTient 2 года назад

    Who is watching chance?

  • @HalcyonSerenade
    @HalcyonSerenade Год назад

    **sniff**

  • @benjaminshiels1824
    @benjaminshiels1824 Год назад

    Almost sounds like a form of quantum entanglement meets fractions.

  • @Kishiru324
    @Kishiru324 Год назад +1

    The axiom of choice. "Well, my sphere in my head can be split and put together to be the same!" Math will one day realize its the same as spiritual energy.

  • @nicosilva4750
    @nicosilva4750 4 года назад +4

    Once again, mathematicians murdering the definition of an axiom. For the record, this is not AT ALL what Euclid thought an axiom was (something self-evidently true)--which is why he doesn't use the term. The "self-evidently true" idea came from the scholastics with respect logic, which is not what Aristotle described as an axiom--the originator of the term. The consequence is that axioms have simply become arbitrary agreed upon notions--ugh. No wonder there are so many paradoxes.

    • @justinsankar1164
      @justinsankar1164 3 года назад

      😐

    • @carlosgaspar8447
      @carlosgaspar8447 3 года назад +2

      axioms being self-evidently true makes no sense. they are assumed true and consistent, otherwise anything could be proven.

    • @nicosilva4750
      @nicosilva4750 3 года назад +1

      @@carlosgaspar8447 Wrong. This is just modern practice. The original term was used for propositions that can't be denied without contradiction (i.e. by denying the proposition, you have to make use of the very proposition you are denying). Aristotle used the Law of Non-contradiction as an example. Modern tradition is an attempt to use Aristotle's Posterior Analytics guide for science, whereby there are undefined terms, and then primitive propositions. He never uses the idea of axioms there, nor does he suggest its use for mathematics.

    • @interwebff
      @interwebff 3 года назад +1

      so what is an axiom nerd?

    • @nicosilva4750
      @nicosilva4750 3 года назад

      @ཀཱYes. that's true, and important to understand. So don't call it an axiom, or call it a "local" axiom, or a math primitive etc. (which is what Aristotle did--common notion). Most importantly though, there needs to be a rule for determining why something is local (hence NEEDED). In math it usually has to do with magnitude preservation of some type (the ability to compare magnitudes after a transformation or operation). See Rodin's book "Axiomatic Method and Category Theory." The first few chapters outline the difference between Euclid and Arisitotle's understanding, with today's modern understanding.

  • @tranngochungdevwannabe
    @tranngochungdevwannabe 6 месяцев назад

    so this axiom can be used for infinity huh? I remember something like inf+x=inf
    but this only means that if u have 1 set of infinity points, then u can choose so that u create 2 sets each equal to that first set, from that first set. That is all
    you can't from that come to the conclusion that u can create 2 spheres identical to the first sphere. too many words like " create, identical" that need definition.
    in math: I mean, imagine u have 1 set of infinity points, how the fuk can you "create" a sphere with that in math? just by saying u "create" it? what does that even mean?
    I feel like u can only say that a set from all points of a sphere is equal to another set from all points of another sphere. Adding stuff like "break down a sphere", and "create a new sphere" is too out of math. those work first of all is not math, second, it make ppl think about phys, hence the so-called "paradox". but it is not a paradox if they use the correct words for it.

  • @anthonym2499
    @anthonym2499 2 года назад

    But 0 is not the least element. 0 was chosen arbitrarily as a least element because no least element exists. An arbitrary ordering is not a well ordered anything, as it does not define, nonarbitrarily, a "smallest" unit element. The inclusion of a "zero" element indicates that a set is incomplete - that something is missing from the set; namely, the ability to nonarbitrarily define the smallest unit element.

    • @anthonym2499
      @anthonym2499 2 года назад +1

      Infinity is unmetricable. It is the condition of no measurement is possible.
      Suppose we have a fundamental measurable element: {a} what is the fundamental measure of this element? It is; itself. Since there is no fundamental element to oppose it, therefore the single element alone has no properties that are measurable in this condition.
      What can we do with this element? Well, no other fundamental measurable element exists. So if we want more, we duplicate it - but we dont know what a number is yet, so its duplication is infinite.
      We create a space, in which we can define the measurable properties that emerge from a collection.
      { aaa....}
      This is a new measurable element, - infinite space. It is its own self, that depends on the existence of another. It is one, composed of many.
      This new measurable element, is divisible by its fundamental nonarbitrary unit element.
      space 1: {a,a,a,a,a,a,a,a,...}
      space 2: { aa,aa,aa,aa,aa,aa,aa,aa,...}
      space 3: { aaa,aaa,aaa,aaa,aaa,aaa,aaa,aaa,...}
      ...
      space n: { na,na,na,na,na,na,na,na,...}
      ....
      space infinite: { aaaaaaaaaaaaaaaa...}
      ....
      space random: {random segmentations ... , a, aa, aaaaaaa, aa, aaaaa, aa, aaa, aa, aaaaa, aaaaaaaaa, aa, ........}
      Now we are able to define properties when there is a difference by degree.
      Create a new space, in which we can view the emergent properties of an arbitrary collection of these elements.
      We can describe a measure of 1s, 2s, 3s, ns, all, or random. We cannot describe a measure of "zero" because a space needs to have a measurable element in order to need to exist.
      The number of ways we can segment our space element, is countable, a nonarbitrary unit element exists. It is complete because we have included all the ways we can segment the section up - 0 is not invented yet; but soon will be.
      If we arbitrarily remove an element from the space; then we get the sense of incompleteness .... something is missing; but we just dont know which one, or how many, it was.
      "zero" is the inability to choose a nonarbitrary unit element - because "at least 1" choice is missing.
      So what happens if we include 2 spaces? What we remove from one, we can put into the other ... that should be able to keep track of them. So how do we nonarbitrarily choose an element to remove? There is no order value placed on our collection; so we are unable to nonarbitrarily pick the most important, or the least important. We can count the number of elements in the set: infinite. But we lack a fundamental property of order.
      If order can even hope to be a measurable we must be able to nonarbitrarily define importance. Well, then the segmentation that we choose has to mean something.
      Z = {0,-1,1,-2,2, ...} is an arbitrary order. We imposed the order by choosing any value possible as a least value; that is just a random ordering. Well ordered implies a degree of importance. We see that the direction of the integers is unimportant, otherwise we wouldnt have been able to mix them all up into this configuration. Distance from 0 is not a choice of importance, it is a standardization.
      We could just have arbitrarily chosen to order by Z + n, add n to each value and we get an equivalent order.
      Z+5 = {5, 4, 6, 3, 7, 2, 8, 1, 9, 0, 10, -1, 11, -2, 12, -3, 13, ...}; any set which can be standardized is closed. It will never produce anything of a nonarbitrary value; a value that anything else doesnt already provide.
      If it is possible to standardize a collection of sets; then no set is more important than another.