I hadn't understood AC before this because I didn't see why an element could not be selected randomly - thank you for clarifying.
@VisualMath Жыл бұрын
Thanks for the feedback, I am very glad that the video was helpful. I hope you enjoyed it as well!
@rebusd3 жыл бұрын
Thank you very very much for this explanation.
@VisualMath3 жыл бұрын
Glad it was helpful! That's why I make these videos.
@realbart2 ай бұрын
About Russels paradox: nobody starts describing it starts off with the hidden assumption that every statement must be true or false (the Law of the Excluded Middle). You could also adapt an alternative logic system that does not use the Law of the Excluded Middle. If you allow for statements to be undetermined, the paradox is no problem. (However, many parts of logic won't work than)
@VisualMath2 ай бұрын
Well there is good reason why the real world is almost always based on some form of fuzzy logic 🤣
@saxxie-dev2 ай бұрын
Intuitionistic type theory, which does not require (or in some variants even contradicts) LEM still suffers from a variant of Russel's paradox called Girard's paradox. It ends up a lot more convoluted and subtle but the end result is pretty much the same - no type of all types
@VisualMath2 ай бұрын
@@saxxie-dev Thanks for sharing: I was not aware of this, and I will have a look 😀
@samanosvasilias7 ай бұрын
I love your videos, I love the format, and your humor, I hope this channel grows!
@VisualMath7 ай бұрын
Thank you so much, your feedback is very much appreciated. But I am not sure whether I have humor to offer 😂
@hassanalihusseini17174 ай бұрын
I like this video! Explaining very well. I would like to see some examples of the problems with f(x+y) = f(x) + f(y) with and without the AC.
@VisualMath4 ай бұрын
Thanks for the comment, I am glad that you liked the video ☺ Haha, I am not sure how an “example” is supposed to look like (as you require AoC) 😂 But here we go: Consider R as a Q vector space and choose a basis B, using AoC. The any choice B→R extends to a linear map R→R. Most such maps are not continuous. I hope that helps!
@hassanalihusseini17174 ай бұрын
@@VisualMath Thank you!
@VisualMath4 ай бұрын
@@hassanalihusseini1717 Welcome!
@NathanielHellerstein6 ай бұрын
Accepting the Axiom of Choice or not is... _your choice!_
@VisualMath6 ай бұрын
Haha, maybe that is the real axiom of choice 🤣
@drefplinth6362 Жыл бұрын
@10:29 in the video when one considers the socks to be indistinguishable (. . . sorry if what follows is a sequence of very naive questions!), is one still allowed to consider all those sets of pairs of indistinguishable socks as sets having two elements each? Or . . . should one see all of those "sets of socks" as singleton sets due to the idea of extensionality? So, in the sock example, wouldn't the axiom of choice simply be choosing the only element in each of those presumably singleton sets? Further, would it be appropriate to consider a collection such as C = {{a}, {a, a}, {a, a, a}, {a, a, a, a}, ...} to be an "equivalence class" with its simplest element {a} taken as its representative, and should such notation (again by extensionality) collapse via C = {{a}, {a}, {a}, {a}, ...} to something like C = {{a}} (i.e., a singleton set with a singleton set as member)? Also, is "class" in the term "equivalence class" intended to mean the same thing as "class" in the term "the Russell class", . . . or does "class" in the term "equivalence class" mean no more than what "grouping" and "collection" mean, respectively, in these artificial paraphrasings "equivalence grouping" or "equivalence collection"? Lastly, thank you very much for producing and posting all of your videos on mathematics: they all look very intriguing!
@VisualMath Жыл бұрын
Let me see whether I can help you: - No, the axiom of choice for singleton sets is boring. We are talking about sets {a_i,b_i}. The analogy with the socks is saying that one (=whoever tries to get one sock out of each set) cannot distinguish them. If one has extra knowledge, then, for example, “take the left sock” is a choice function whose existence we do not need to assume. In general one does not have extra knowledge, so one needs the axiom of choice. - Note that {a}={a,a}={a,a,a}=… as sets, so C={{a}} without any equivalence. - The history of the name is explained here: mathoverflow.net/questions/135347/who-introduced-the-terms-equivalence-relation-and-equivalence-class It seems like it could have ended up being anything, like equivalence grouping as you suggest. I would read too much into names in math - most names are pretty random ;-) - Thanks ;-) I hope you will enjoy some of the videos! I hope that helps.
@drefplinth6362 Жыл бұрын
@@VisualMath Thank you!
@VisualMath Жыл бұрын
@@drefplinth6362 Welcome!
@antoniusnies-komponistpian21723 ай бұрын
Have people tried to formulate a weaker version of the axiom of choice that might work for the cases we want it to work for but doesn't produce the weird results? Or is that proven to be impossible?
@VisualMath3 ай бұрын
Good question. There is indeed a whole zoo of weaker versions of AoC 🙂 See here for example: archive.org/details/LectureNotesInMathematics1876HorstHerrlichauth.AxiomOfChoiceSpringerVerlagBerlinHeidelberg2006/page/n29/mode/2up
@antoniusnies-komponistpian21723 ай бұрын
@@VisualMath Cool, interesting
@VisualMath3 ай бұрын
@@antoniusnies-komponistpian2172 Yes, I like it too 😁
@NathanielHellerstein6 ай бұрын
Koenig's Lemma is a weak version of the Axiom of Choice. I like Koenig's Lemma, and its consequent Compactness Theorem, because they validate Nonstandard Analysis, and I like infinitesimals. What disasters arise from Koenig's Lemma? Or from denying it?
@VisualMath6 ай бұрын
That is a good question. I have never seen anyone discussing disasters that follow from Kőnig's lemma. Since the lemma is still outside ZF (unless you only allow countable graphs), I would expect there are several disasters 😁
@tim-701cca Жыл бұрын
Base on 11:03, I have a question about when we don't need AC. I am thinking whether "The Infinite Cartesian product of N is nonempty" require AC. Since we can always pick a natural number from N, say 0, we can define a choice function. So I think we don't need AC here. Am I right?
@VisualMath Жыл бұрын
Exactly 👍: as soon as your sets in the infinite product are well-ordered (or have some other way to find a preferred element like minimum), then no AoC is needed. The AoC is for sets with "not distinguishable"🤔 elements.
@tim-701cca Жыл бұрын
Thank you for your explanation. You explain this ‘shoe sock example’ very clear to me.😊
@VisualMath Жыл бұрын
@@tim-701cca 🤗
@lukevideckis22602 ай бұрын
why is ZF used then if there are so many problems with it?
@lukevideckis22602 ай бұрын
I'm sure the other ways to avoid russels paradox also have a bunch of problems with them too
@VisualMath2 ай бұрын
@@lukevideckis2260 It is not used in the same way the Peano axioms are not used (very few people really work with them). So one can easily ignore problems 😂 Maybe the better answer is what you already observed: its not we have many alternatives 🤔
@xbylina26413 ай бұрын
A little mistake: at 2:06 in purple box (definition of the Russel set), there should be "x NOT IN x" (instead of "x IN x")...
@xbylina26413 ай бұрын
Oh, silly me, you cerrected yourself at 4:01 :)
@VisualMath3 ай бұрын
@@xbylina2641 Haha, still: good catch 👍 Thanks for the comment and for watching ☺
@TheMorhaGroup Жыл бұрын
"This sentence is false" is only a paradox, when assuming a sentences claim of a truth value, makes it that value. If we are to observe the non-self referential sentence. "that claim is false" we can see. there is no reason to take the self-proposed truth value prima facie, as such a claim can be articulated, whether it is true, or not. A sentence can not determine its own truth value
@VisualMath Жыл бұрын
Yes, different axioms different outcomes. And that is why one need to be careful when one translates sentences into formal logic 😁
@edwardmacnab3546 ай бұрын
"this sentence is false" is a true sentence in that it "IS" a sentence . What the sentence means is debateable but it could be considered as a joke or a brain twister , a type of sentence that has NO PLACE in mathematics
@TheMorhaGroup6 ай бұрын
@@edwardmacnab354 yes it is a sentence, but so is "all cats can fly". Can you see my point?
@edwardmacnab3546 ай бұрын
@@TheMorhaGroup 'all cats can fly' is a true sentence , it's meaning is debateable . Perhaps it is a deliberate attempt at cynicism. How about--"all sentences are mathematical constructs"
@TheMorhaGroup6 ай бұрын
@@edwardmacnab354 it's true on what basis?
@SteveThePster9 ай бұрын
Do you have an example of a family of sets for which it is not obvious that a choice function exists? I.e. a possible counter example to the Axiom of Choice
@VisualMath9 ай бұрын
My favorite party pooper are the real numbers 🤣 Take the set of all subsets of real numbers. How could one define a choice function that picks an element from every subset? You cannot do it “step-by-step” as there are too many such subsets. You cannot say “Take the minimum” since subsets often have no such minimum. There is no “easy” choice function, so one needs an axiom to ensure that one can proceed.
@ゾカリクゾ2 жыл бұрын
great video
@VisualMath2 жыл бұрын
Thank you for the feedback. Glad that you liked the video, and I hope it was helpful.
@Bethos1247-Arne6 ай бұрын
when I heard about this axiom I thought it is obviously true. In some sense I still think it is but I am not so sure about things which follow. Perhaps this axiom should not be used. Perhaps well-ordering any set is not possible.
@VisualMath6 ай бұрын
And it some sense it doesn't matter whether we believe in it or not: there are disasters either way 🤣
@apokalyptiktoxic27383 ай бұрын
I dont think that a disaster of Choice is the existente of Sets without Volume. If you haven‘t Choice your may Not be able to Proof if there is a measure or not. If you could Proof wirhout Choice that there is a measure for R^n then Choice couldnt be Independent from the other axioms
@VisualMath3 ай бұрын
It depends what you call 'a disaster', but having non-measurable sets was unexpected at the time of discovery 🙂
@antoniusnies-komponistpian21723 ай бұрын
Couldn't we actually say, the probability of "This sentence is false" is 1/2? Because that is not self-contradictory.
@VisualMath3 ай бұрын
Absolutely. The sentence is only nonsense in binary logic, but if you allow a "Maybe" value, then this is fine 😀
@NukeCloudstalker3 ай бұрын
It's funny that that sentence isn't just treated as an indeterminate limit expression. The "this" should be infinitely expanded in this form: "This sentence is(this sentence is(this sentence is())". It effectively is the same as the infinite alternating series 1-1+1-1[...]. Recursive statememts can really only be made sense of concretely if they termimate, or there exists a limit of it (not the same).
@VisualMath3 ай бұрын
@@NukeCloudstalker Interesting. Maybe one can model all infinite sums on "honest" sentences 🤣
@pauselab55698 ай бұрын
It’s too useful to abandon. Pretty much all of linear algebra and field theory relies on it
@VisualMath8 ай бұрын
Well, the argument against it would say that most of “honest” linear algebra and field theory does not need it. For example, if you work with finite dimensional vector spaces. So it really depends how one interprets “honest” 😂
@ralphinoful6 ай бұрын
Why does the axiom of choice require a choice "function"? Why can't we just say that this choice "thing" is the existence of an arbitrary mapping? I feel like that should solve some problems.
@VisualMath6 ай бұрын
I am not sure whether I am following. The AoC is an existence statement of some map, this is probably easiest to see in the equivalent statement that “A Cartesian product of a collection of non-empty sets is non-empty”
@ralphinoful6 ай бұрын
@@VisualMath As I was trying to explain what I was trying to say, I think I answered my own question. But for the sake of trying to explain my thought process. A function is the existence of a specific set, of pairs, of inputs and their outputs. So f(x) = 2x is the set {{1,2}, {2,4}, {3,6}, ... {x,f(x)}}. And I was thinking of some mapping that is in a way "not well defined", but also able to give unique outputs at the same time. But this concept is sort of self-contradicting. I was thinking about a mapping that would essentially act as a random mapping, where it "can" map to multiple elements, but when you call the function it collapses to a single element. Sort of similar to how an electron might exist in a field, and then collapses into a single position when observed. That my mapping is taking a single element, and mapping it to an arbitrarily large set, but it then "collapses" when you call it. But I guess you would be able to just call the mapping, have everything collapse to something random, and then retroactively define that mapping as a bona fide function. I wanted to get away from the fact that the AoC defines a sort of "procedural" way to assign our choice.
@samuelsanchez40204 ай бұрын
@ralphinoful what you are describing is equivalent to aoc: Given a relation R there exists a subset F of R such that F is a function and Dom(F)=Dom(R). You can prove that this statement follows from aoc and that it also implies aoc.
@arigato93403 жыл бұрын
Please make an intro to tensor analysis
@VisualMath3 жыл бұрын
Thank you for the idea. Do you have anything specific in mind?
@victormd11002 жыл бұрын
The controversy of AC nowadays is basically null and virtually all set theorists believe in it, so i dont agree with you putting it in analogy with niels bohr sentence. The axioms of set theory were meant to capture what the von neumann universe satisfies, which is basically everythinf you get from beginning with just the empty set and keep on powersetting it to potential infinity. Given this, it has been largely argued that AC is obviously true ( on V ), i mean, thats what is meant by powerset and subsets, subsets of a given set are supposed to be every single combination of some part of its elements. Just like the powerset of {a,b,c} is supposed to contain the empty set, {a},{b},{c},{a,b},{a,c},{b,c},{a,b,c} the subsets of N are supposed to contain every single part of N, or may i say, every single combination of different elements of N, pick any random elements of N to countable infinity and you'll have a subset of it. If you have an arbitrary collection of nonemptysets Ai, of course you'll also have a set which contains exactly {a1,a2,a3,...,aw,...,aw_w...} again, that is obviously a "part" of the union of all Ai and so, must exist. The reason why AC was controversial at the beginning was because it wasnt agreed which "universe" the axioms are supposed to model. Nowadays it is viewed as simply true ( in fact, some other axioms are largely viewed as true such as cons(ZFC), cons(ZFC+cons(ZFC)), some large cardinal axioms,...
@VisualMath2 жыл бұрын
Of course not, but I am told it works even if you don’t believe in it - Niels Bohr Hmm, maybe I was unclear: My take is that it doesn't matter whether you believe in it or not. It doesn't even matter whether its true or not. I feel that is what most people think as well; people work with what works.