What are some other restrictions we could place on f, X, and Y, to guarantee a right inverse without needing the axiom of choice? (Essentially, we need to impose enough "information" on those things for a choice function to be formed.)
We could let X and Y be such that f is injective, and that there exists some other injection from Y to X. This forces a bijection, hence a right inverse, by the Schröder-Bernstein theorem.
Again, since creating a right inverse (from a surjective map) without the axiom of choice would mean that we need to find some distinguished member of f^-1({y}) for each y in Y. This would be easy in some set with additional structure, the extra bit of 'information'. For instance if X is the tangent bundle of Y, then each fiber f^-1({y}) has the structure of a vector space, hence has a distinct element, the additive identity of the abelian group underlying the vector space. Also, such a vector is unique, which forces a right inverse.
It still amazes you that you only have 20k subs. Your standard of quality is easily comparable to the biggest math channels, like 3b1b and Mathologer. By far the most underrated math channel.
I think his quality and knowledge is easily better than those channels. Mathologer has the par on knowledge but not in the nice visual representation and cool background lol. This channel is by far better than most in terms of the topic choices and his knowledge.
I may not be super advanced in math, but math is what I want to study when I am older. And mathematicians like you provide in-depth explanations of super specific topics that make learning way easier when a textbook kinda fails. Even if I may not fully understand this yet, or I may not need to even be somewhat familiar, thanks a lot 😊
Here is a list of topics. Get familiar with graphs and functions (I would personally skip inequalities but it’s super easy). Algebra 1 and algebra 2. Simple geometry. I dont think there is any need in studying stuff like angle theorems inside a circle (there are so many angle theorems for a circle) and there are also in geometry “corresponding angles” “alternating angles” “vertically opposite angles” all of that felt pretty useless in school and not needed for more advanced math so thats why you shouldnt stress out on geometry it should only take a day or 2 to brush up on. Learn the formulas for volumes and areas of shapes like area of circle and get familiar with graphing the equations of circles, straight lines, all of that which should be in algebra 1 and 2. After that do trigonometry and parametrisation. Parametrisation isn’t something you need to stress on much. Trigonometry however is super important learn about the unit circle and all of that. Now you are ready for calculus! :) start with learning how to do differentiation and what it represents. You will understand this better when you practice it with questions so its okay if things dont click whilst learning the basics. It can click better after practice problems. From there on out the road and map should be more clear and easy
Just a videography tip here: Wear something “contrastive.” Your T-shirt and heard blend with the background. You're basically a part of the wall. 😂 I love the quality tho.
Here's a nice AoC equivalent: Every infinity set A is in bijection with the product A x A. I personally don't think choice is something that should be added but I also consider it a bonus if your system is weak enough to just make it true. That said, you'll never convince the Zorn users to give it up. They build their gingerbread house a hundred years ago and it will keep being inherited until their lineage dies out.
I see no reason why we should expect Cartesian products to have larger cardinalities, simply because that is true for finite sets. It's somewhat analogous in my view to being surprised the naturals and the rationals have the same cardinality, despite one containing the other... Also, I do not think constructivism is anywhere near being on the rise; if anything, set theory itself is dying as formalisms with greater predictive power such as HoTT and category theory attempt to replace them as mathematical foundations.
@@duncanw9901 I guess a big part of the confusion and "paradox" in infinite sets is that people expect them to behave the same way as finite sets. Once you let go of that assumption, they actually make loads of sense and feel consistent.
Dude I love your style, I didn’t understand everything in the video because I haven’t studied more about those topics, but it’s always nice to see your videos because of the way you present things, it’s super fun. And I think I learned the important aspects of what AC means , thanks :)
I find your illustration of the logic link between the axiom of choice and the fact all sets are ordered most helpful ! We have 2 equivalent ways to get to the same conclusions ! Thanks for the video :)
The Axiom of Choice! Did you know there's a band by that name? It's a southern California based world music group of Iranian émigrés who perform a modernized fusion style rooted in Persian classical music with inspiration from other classical Middle Eastern and Eastern paradigms.
Interesting to note the following facts. Let A and B be two sets. If there is an injection from A to B and an injection from B to A, then there exists a bijection between A and B (Cantor-Bernstein theorem). The proof is not explicit, but does NOT require AC. But if you replace injection by surjection, then the conclusion is the same and does require AC. The axiom of choice also implies that, given two sets, there is alwyas an injection from one to the other (comparability of sets)
And the proof is as follows: if every set can be well-ordered, then cardinality of any two well-ordered sets is comparable. And if there exists some set which can't be well-ordered, then by Hartogs theorem (see the Wikipedia article 'Hartogs number') there exists a well-ordered set whose cardinality is incomparable with it.
I've got to say man, you keep surprising me with your quality. I was afraid this video might be slightly more sketchy/handwavey than usual, but you really make sure you know what you talk about and are not afraid to include it. Nice! I'd share my own preferred take on including the axiom of choice, that I don't hear often: It is not so much a limitation of ZF, that we need to include C. It is more so that we formulate ZF in first-order logic, which inherently describes finitistic reasoning. This is also why the finite version works fine: one just uses existential quantification elimination some finite number of times. But trying that same proof for infinite sets proofs impossible within first-order logic; but the mathematician wants to do so! And here we should keep in mind logic has always had the role of giving an explenation about the reasoning of a mathematician. We logicians merely point out to the mathematicians that *if* they want to have this reasoning, and *if* they prefer a foundation in first order logic, *then* they are forced to inclide AoC. There is not necessarily a bad thing.
Holy crap dude I can't believe I just found your channel- I'm an undergrad and I love set theory and wanna keep studying mathematical logic- this video was epic haha. Would you consider making some more like this, maybe on ordinals, cardinals, transfinite induction etc?
Great video! Here's a quick question : When we have a set S of k linearly independant vectors in an finite dimensional vector space V of dimension n, with k
No it doesn’t! You just have a finite amount of steps to make in order to end up with a base of V: you’ll have to find successively n-k vectors, so the axiom of choice is not needed (only a finite product of nonempty sets)
5:30 how can you use the axiom of choice to put a well order on a set? I see how the inverse is possible but I’m not sure about this direction to prove the well ordering principle and axiom of choice are equivalent
The very broad intuition is that you pick some element to be your first element. Then you remove that element from your set and pick some element from your shrunken set to be your second element. Then you remove that element from your set and pick some element your further shrunken set to be your third element, etc. The "picking" process comes from the Axiom of Choice. To be slightly more rigorous, let S be a nonempty set. Then, by the Axiom of Choice, there is a "choice function" (let's call it f) defined on P(S)\{∅}. In other words, for every nonempty subset A of S, f(A) is an element of A. The idea is that you take your first element in S to be f(S). So we define x_0 = f(S). Then you look at S\{x_0}, which if nonempty, we can define x_1 = f(S\{x_0}). Then you look at S\{x_0,x_1}, which if nonempty, we can define x_2 = f(S\{x_0,x_1}). etc. Essentially, your choice function "chooses" an element you haven't yet picked. So, you're "constructing" the well-ordering by transfinite induction. (This is not circular since the ordinal numbers can be defined without any appeal to the well-ordering theorem.) Even more formally, suppose S is a nonempty set, and let f be a choice function on P(S)\{∅}. Define x_0 = f(S). Then, for all ordinal numbers α > 0, if S\{x_β | β < α} is nonempty, define x_α = f(S\{x_β | β < α}). Note: Since S is a set, there must exist a smallest ordinal number γ so that S\{x_β | β < γ} = ∅. (If not, S would have a subset in "one-to-one correspondence" with the ordinal numbers, and hence, would be too big to be a set.) To give yet another intuitive view of what's going on: the choice function f allows you to define a bijection from some ordinal number γ to S. Since γ is, itself, a well-ordered set, you can transfer the order-structure of γ through the bijection to S. --- Although, I don't think it is necessary to state, I will also give an even more rigorous approach. Take some element ★ not in S, say for instance ★ = {S}. Then take a choice function f define on P(S)\{∅}. You then take the following definition: For all ordinal numbers α, define x_α in S∪{★} to be f(S\{x_β | β < α}) if S\{x_β | β < α} is nonempty or x_α = ★ otherwise. Then there exists a least ordinal γ such that x_γ = ★. Next, define g : γ → S by g(α) = x_α. By how we have defined all of the x_α, g is a bijection. Then define an ordering ≤ on S by x ≤ y if and only if g⁻¹(x) ≤ g⁻¹(y). You then have that ≤ is a well-ordering on S.
Fun consequence of the axiom of choice (it might even be equivalent to it - we don't know yet): A partition of a set cannot be larger than the set itself. If you assume things like "all sets are measurable" - which contradicts the axiom of choice - then you can, in laymen's terms, sort every real number into a bucket in such a way that you end up with more buckets than you had real numbers.
IIRC, this is the longest-standing open problem in set theory. Someone recently posted an article on the arXiv which made a claim that they had shown the Partition Principle not to be equivalent to Choice. However, after more analysis it seems like this isn't a solid claim. karagila.org/2020/flowing-to-conclusions/
I’ve been trying to understand how AoC and the well ordering theorem are equivalent, and I’m not sure if I totally understand it. It is just that you choose and arbitrary real number to be the lowest, then choose a different number to be the next lowest, and so on? I’m not exactly sure what would make this work with an uncountable set.
The proof is usually done indirectly: axiom of choice implies Zorn's lemma, Zorn's lemma implies well-ordering theorem, and well-ordering theorem implies axiom of choice. (And, as Jerry Lloyd Bona has said: "axiom of choice is obviously true, well-ordering theorem is obviously false, and who can tell anything about Zorn's lemma?")
I think I’m going to need to study set theory some more to fully understand it I guess, I don’t understand what zorns lemma is and I don’t know a ton about ordinals, which it looked like we’re used in the proofs
Consider all subsets of R. For every subset A, use the axiom of choice to select an element of it. We can think of this element as F(A) where A ranges over the subsets of R. The well ordering will be F(R), F(R\F(R)), F(R\ (F(R) U F(R\F(R)))).... in other words at each step to get the next element you're selecting from R \ _every element you've selected so far._ The problem is in proving that this is a well ordering which doesn't leave anything out. Take a look at chapter 9 in _Schaums outline of theory and problems of set theory and related topics_ by Seymour Lipschutz.
I think he's actually left handed and he has made himself mirror-imaged in the scene where he writes the proof. You'll see he's holding his pen in his left hand in the other scenes, and his hair is brushed the other way.
Hey John @Epic Math Time - a question. How would you show using mapping algebra or formal logic that transcendental and complex valued functions are not well ordered? (say for example a logarithm or complex number function). You cannot perform modular arithmetic on logs and definitely not complex numbers (latter because you cannot definitely compare them). Does that contradict the formulation of AOC that all sets can be well ordered or no?
I'm not sure what you're asking. Axiom of choice implies that every set can be well-ordered (and, in fact, it's equivalent to this proposition). So for example, assuming axiom of choice real numbers (or functions on reals, and so on) can be well-ordered. But this ordering has nothing to do with the usual numeric order (obviously, the numeric order is not a well-ordering relation on reals - for example, the set of all numbers strictly between 0 and 1 doesn't have a minimum). And in fact, this order can't be defined in the usual sense: no formula can be proven to define a well-ordering of reals. (That is: there is no formula for which the following can be proven: 1) there exists exactly one object satisfying this formula, and 2) this object is a well-ordering relation on real numbers.)
It can be easily shown that there are infinitely many primes. Let p1, p2, p3, ..., pn be any finite set of primes. Let x=p1*p2*p3*...*pn. Consider the number x+1. By construction, x+1 is not divisible by any of the primes p1, p2, ..., pn. So it must be either itself a prime, or divisible by some prime not in the set. (This follows from the theorem that every positive integer has a unique factorization into primes.) Therefore, any finite set of primes misses some prime number and so it cannot be complete; it follows that there are infinitely many primes.
the axiom of choice itself seems to be intuitively right. But some of its implications appear absurd. I am not sure if standard math, used as tool which maps to realitiy as closely as possible, should include it.
"Axiom of choice is obviously true, well-ordering principle is obviously false, and who can tell about the Zorn's lemma?" (Of course, all three propositions are equivalent in ZF set theory.)
Once you reach a certain point in maths, the appearance of a single number is surprising, and it's typically just a index. I think once you reach that point, it's more about general functions and sets and stuff rather than specific elements of the sets. Also, you soon realise that numbers are nothing but equivalence classes of abstract sets.
Or: 1. He writes with his left hand, stands in the viewer's position (in your place as you watch this video) and is filmed from the background side, to appear right-handed. This recording is superimposed to the next one. 2. The surface he writes on is "filmed" (recorded) from within itself only (your position as you watch this). Note: this surface must be totally transparent from the background side but the writing must be visible from your point of view (like a prompter screen ?)
What are some other restrictions we could place on f, X, and Y, to guarantee a right inverse without needing the axiom of choice?
(Essentially, we need to impose enough "information" on those things for a choice function to be formed.)
We could let X and Y be such that f is injective, and that there exists some other injection from Y to X. This forces a bijection, hence a right inverse, by the Schröder-Bernstein theorem.
Again, since creating a right inverse (from a surjective map) without the axiom of choice would mean that we need to find some distinguished member of f^-1({y}) for each y in Y. This would be easy in some set with additional structure, the extra bit of 'information'. For instance if X is the tangent bundle of Y, then each fiber f^-1({y}) has the structure of a vector space, hence has a distinct element, the additive identity of the abelian group underlying the vector space. Also, such a vector is unique, which forces a right inverse.
@@AlphaCurveMath That implies the existence of a bijection, yes. But it doesn't guarantee that it's the original function
@@pleaseenteraname4824 true
I'm thinking X = R, f is continuous surjective, and Y is Hausdorff. Can we do it then?
Skills they don't tell you you'll learn in math:
I can recognize a Springer-Verlag spine from a mile away.
It still amazes you that you only have 20k subs. Your standard of quality is easily comparable to the biggest math channels, like 3b1b and Mathologer. By far the most underrated math channel.
Exactly!
Thank you 🙂
I think his quality and knowledge is easily better than those channels. Mathologer has the par on knowledge but not in the nice visual representation and cool background lol.
This channel is by far better than most in terms of the topic choices and his knowledge.
I may not be super advanced in math, but math is what I want to study when I am older. And mathematicians like you provide in-depth explanations of super specific topics that make learning way easier when a textbook kinda fails. Even if I may not fully understand this yet, or I may not need to even be somewhat familiar, thanks a lot 😊
Here is a list of topics. Get familiar with graphs and functions (I would personally skip inequalities but it’s super easy). Algebra 1 and algebra 2. Simple geometry. I dont think there is any need in studying stuff like angle theorems inside a circle (there are so many angle theorems for a circle) and there are also in geometry “corresponding angles” “alternating angles” “vertically opposite angles” all of that felt pretty useless in school and not needed for more advanced math so thats why you shouldnt stress out on geometry it should only take a day or 2 to brush up on. Learn the formulas for volumes and areas of shapes like area of circle and get familiar with graphing the equations of circles, straight lines, all of that which should be in algebra 1 and 2.
After that do trigonometry and parametrisation. Parametrisation isn’t something you need to stress on much. Trigonometry however is super important learn about the unit circle and all of that.
Now you are ready for calculus! :) start with learning how to do differentiation and what it represents. You will understand this better when you practice it with questions so its okay if things dont click whilst learning the basics. It can click better after practice problems. From there on out the road and map should be more clear and easy
Just a videography tip here:
Wear something “contrastive.” Your T-shirt and heard blend with the background. You're basically a part of the wall. 😂
I love the quality tho.
“Reach into the cosmos and grab” is a fantastic summarization of what the Axion of Choice does.
Here's a nice AoC equivalent: Every infinity set A is in bijection with the product A x A. I personally don't think choice is something that should be added but I also consider it a bonus if your system is weak enough to just make it true. That said, you'll never convince the Zorn users to give it up. They build their gingerbread house a hundred years ago and it will keep being inherited until their lineage dies out.
I see no reason why we should expect Cartesian products to have larger cardinalities, simply because that is true for finite sets.
It's somewhat analogous in my view to being surprised the naturals and the rationals have the same cardinality, despite one containing the other...
Also, I do not think constructivism is anywhere near being on the rise; if anything, set theory itself is dying as formalisms with greater predictive power such as HoTT and category theory attempt to replace them as mathematical foundations.
@@duncanw9901 I guess a big part of the confusion and "paradox" in infinite sets is that people expect them to behave the same way as finite sets. Once you let go of that assumption, they actually make loads of sense and feel consistent.
Dude I love your style, I didn’t understand everything in the video because I haven’t studied more about those topics, but it’s always nice to see your videos because of the way you present things, it’s super fun. And I think I learned the important aspects of what AC means , thanks :)
I find your illustration of the logic link between the axiom of choice and the fact all sets are ordered most helpful !
We have 2 equivalent ways to get to the same conclusions !
Thanks for the video :)
This is a really nice video, especially how you go into the vibes of invoking the axiom.
AC (and ZFC, GCH, etc.): my favorite frenemy.
Has anyone ever told you that you resemble Dr. Disrespect?
@@EpicMathTime Had to Google him just now, but having done a quick check there... damn, there really is an uncanny resemblance! 😲
Loving the TOOL background music
best explanation I found on TH-cam, good job
Legendary, as always!
Very good! I've never understood the AoC as well as this. Thank you
Me: Looking for intuition of AoC...
Epicmathtime: Reach into the cosmos and grab it.
Yup, that's what I'm looking for ✨
what does
@@aashsyed1277 It means we are allies in the battle against mathematical anomalies.
@@aashsyed1277 to me it looks like a heart, placed sideway though.
@@aashsyed1277 “
@@aashsyed1277 π
Great video! = Graduate Texts in Background + Tool + Math
I saw your channel after reading a comment you left on Leo's channel, and love it.
Appreciated!
The Axiom of Choice! Did you know there's a band by that name? It's a southern California based world music group of Iranian émigrés who perform a modernized fusion style rooted in Persian classical music with inspiration from other classical Middle Eastern and Eastern paradigms.
Actually a really nice explanation. Thank you!
How haven't I met your channel before? Amazing work
Simply Superb... tons of thanks bro.
Interesting to note the following facts. Let A and B be two sets. If there is an injection from A to B and an injection from B to A, then there exists a bijection between A and B (Cantor-Bernstein theorem). The proof is not explicit, but does NOT require AC. But if you replace injection by surjection, then the conclusion is the same and does require AC. The axiom of choice also implies that, given two sets, there is alwyas an injection from one to the other (comparability of sets)
And the proof is as follows: if every set can be well-ordered, then cardinality of any two well-ordered sets is comparable. And if there exists some set which can't be well-ordered, then by Hartogs theorem (see the Wikipedia article 'Hartogs number') there exists a well-ordered set whose cardinality is incomparable with it.
I loooove your videos! thank you very much :D
What about Zorn's lemma? Might be cool to see a proof of the equivalence of the two
Tool in the background, nice.
I've got to say man, you keep surprising me with your quality. I was afraid this video might be slightly more sketchy/handwavey than usual, but you really make sure you know what you talk about and are not afraid to include it. Nice!
I'd share my own preferred take on including the axiom of choice, that I don't hear often:
It is not so much a limitation of ZF, that we need to include C. It is more so that we formulate ZF in first-order logic, which inherently describes finitistic reasoning. This is also why the finite version works fine: one just uses existential quantification elimination some finite number of times. But trying that same proof for infinite sets proofs impossible within first-order logic; but the mathematician wants to do so!
And here we should keep in mind logic has always had the role of giving an explenation about the reasoning of a mathematician. We logicians merely point out to the mathematicians that *if* they want to have this reasoning, and *if* they prefer a foundation in first order logic, *then* they are forced to inclide AoC. There is not necessarily a bad thing.
thanks so much for making me understand so well!
Awesome my bro! 👍
and i thought it was so complicated, now it feels very natural
I relearned how based this video truly is, after having gone through various studies concerning similar topics
"Every relation contains a function " is also an application of the AC.
That was a \mathbb{R}ad video.
Awesome
Holy crap dude I can't believe I just found your channel- I'm an undergrad and I love set theory and wanna keep studying mathematical logic- this video was epic haha. Would you consider making some more like this, maybe on ordinals, cardinals, transfinite induction etc?
These sound like fine additions to my mission.
@@EpicMathTime 😀😍
Every choice is a new iteration.
please upload more
Thank you, I was wondering what that was.
Great video! Here's a quick question :
When we have a set S of k linearly independant vectors in an finite dimensional vector space V of dimension n, with k
No it doesn’t! You just have a finite amount of steps to make in order to end up with a base of V: you’ll have to find successively n-k vectors, so the axiom of choice is not needed (only a finite product of nonempty sets)
Love it!
5:30 how can you use the axiom of choice to put a well order on a set? I see how the inverse is possible but I’m not sure about this direction to prove the well ordering principle and axiom of choice are equivalent
The very broad intuition is that you pick some element to be your first element. Then you remove that element from your set and pick some element from your shrunken set to be your second element. Then you remove that element from your set and pick some element your further shrunken set to be your third element, etc. The "picking" process comes from the Axiom of Choice.
To be slightly more rigorous, let S be a nonempty set. Then, by the Axiom of Choice, there is a "choice function" (let's call it f) defined on P(S)\{∅}. In other words, for every nonempty subset A of S, f(A) is an element of A.
The idea is that you take your first element in S to be f(S). So we define x_0 = f(S).
Then you look at S\{x_0}, which if nonempty, we can define x_1 = f(S\{x_0}).
Then you look at S\{x_0,x_1}, which if nonempty, we can define x_2 = f(S\{x_0,x_1}).
etc.
Essentially, your choice function "chooses" an element you haven't yet picked.
So, you're "constructing" the well-ordering by transfinite induction. (This is not circular since the ordinal numbers can be defined without any appeal to the well-ordering theorem.)
Even more formally, suppose S is a nonempty set, and let f be a choice function on P(S)\{∅}.
Define x_0 = f(S). Then, for all ordinal numbers α > 0, if S\{x_β | β < α} is nonempty, define x_α = f(S\{x_β | β < α}).
Note: Since S is a set, there must exist a smallest ordinal number γ so that S\{x_β | β < γ} = ∅. (If not, S would have a subset in "one-to-one correspondence" with the ordinal numbers, and hence, would be too big to be a set.)
To give yet another intuitive view of what's going on: the choice function f allows you to define a bijection from some ordinal number γ to S. Since γ is, itself, a well-ordered set, you can transfer the order-structure of γ through the bijection to S.
---
Although, I don't think it is necessary to state, I will also give an even more rigorous approach. Take some element ★ not in S, say for instance ★ = {S}. Then take a choice function f define on P(S)\{∅}. You then take the following definition:
For all ordinal numbers α, define x_α in S∪{★} to be f(S\{x_β | β < α}) if S\{x_β | β < α} is nonempty or x_α = ★ otherwise.
Then there exists a least ordinal γ such that x_γ = ★.
Next, define g : γ → S by g(α) = x_α.
By how we have defined all of the x_α, g is a bijection.
Then define an ordering ≤ on S by x ≤ y if and only if g⁻¹(x) ≤ g⁻¹(y).
You then have that ≤ is a well-ordering on S.
Where did he come from where did he go?
You're an awesome person. Your style and fun presentation helps math be digestible. Thanks for what u do
Really sad that you need that kind of stuff in order to “digest” math.
Fucking Zorns Lemma. Imagine Commutative Algebra without Choice lol
Liked for Tool.
I think the emphasis on needing AoC to avoid affirmatively constructing the choice filled in a missing piece for me.
Fun consequence of the axiom of choice (it might even be equivalent to it - we don't know yet): A partition of a set cannot be larger than the set itself.
If you assume things like "all sets are measurable" - which contradicts the axiom of choice - then you can, in laymen's terms, sort every real number into a bucket in such a way that you end up with more buckets than you had real numbers.
IIRC, this is the longest-standing open problem in set theory. Someone recently posted an article on the arXiv which made a claim that they had shown the Partition Principle not to be equivalent to Choice. However, after more analysis it seems like this isn't a solid claim.
karagila.org/2020/flowing-to-conclusions/
I’ve been trying to understand how AoC and the well ordering theorem are equivalent, and I’m not sure if I totally understand it. It is just that you choose and arbitrary real number to be the lowest, then choose a different number to be the next lowest, and so on? I’m not exactly sure what would make this work with an uncountable set.
The proof is usually done indirectly: axiom of choice implies Zorn's lemma, Zorn's lemma implies well-ordering theorem, and well-ordering theorem implies axiom of choice. (And, as Jerry Lloyd Bona has said: "axiom of choice is obviously true, well-ordering theorem is obviously false, and who can tell anything about Zorn's lemma?")
I think I’m going to need to study set theory some more to fully understand it I guess, I don’t understand what zorns lemma is and I don’t know a ton about ordinals, which it looked like we’re used in the proofs
Consider all subsets of R. For every subset A, use the axiom of choice to select an element of it. We can think of this element as F(A) where A ranges over the subsets of R. The well ordering will be F(R), F(R\F(R)), F(R\ (F(R) U F(R\F(R)))).... in other words at each step to get the next element you're selecting from R \ _every element you've selected so far._ The problem is in proving that this is a well ordering which doesn't leave anything out. Take a look at chapter 9 in _Schaums outline of theory and problems of set theory and related topics_ by Seymour Lipschutz.
① Great-looking video!
② Hoping that adding this comment boosts the algorithm a tad... you truly deserve more subs.
sets don't have inverses with regards to cartesian products
Fantastic video lol
How can you write left-to-right so well? 🙂
I think he's actually left handed and he has made himself mirror-imaged in the scene where he writes the proof. You'll see he's holding his pen in his left hand in the other scenes, and his hair is brushed the other way.
Hey John @Epic Math Time - a question. How would you show using mapping algebra or formal logic that transcendental and complex valued functions are not well ordered? (say for example a logarithm or complex number function). You cannot perform modular arithmetic on logs and definitely not complex numbers (latter because you cannot definitely compare them). Does that contradict the formulation of AOC that all sets can be well ordered or no?
I'm not sure what you're asking. Axiom of choice implies that every set can be well-ordered (and, in fact, it's equivalent to this proposition). So for example, assuming axiom of choice real numbers (or functions on reals, and so on) can be well-ordered. But this ordering has nothing to do with the usual numeric order (obviously, the numeric order is not a well-ordering relation on reals - for example, the set of all numbers strictly between 0 and 1 doesn't have a minimum). And in fact, this order can't be defined in the usual sense: no formula can be proven to define a well-ordering of reals. (That is: there is no formula for which the following can be proven: 1) there exists exactly one object satisfying this formula, and 2) this object is a well-ordering relation on real numbers.)
i saw your comment on a maths video about infinite primes.
could you show us how you would do it? im genuinely interested. please
It can be easily shown that there are infinitely many primes. Let p1, p2, p3, ..., pn be any finite set of primes. Let x=p1*p2*p3*...*pn. Consider the number x+1. By construction, x+1 is not divisible by any of the primes p1, p2, ..., pn. So it must be either itself a prime, or divisible by some prime not in the set. (This follows from the theorem that every positive integer has a unique factorization into primes.) Therefore, any finite set of primes misses some prime number and so it cannot be complete; it follows that there are infinitely many primes.
@@MikeRosoftJH riiight, thank you
You def have some editing skills
So I guess the intuitionists reject the axiom of choice?
Yes, because the axiom of choice implies the law of excluded middle, and therefore must be purged by the intuitionist flames
And um ... can you tell us about the Axiom of Determinacy? The anti-choice axiom, in some way?
the axiom of choice itself seems to be intuitively right. But some of its implications appear absurd. I am not sure if standard math, used as tool which maps to realitiy as closely as possible, should include it.
"Axiom of choice is obviously true, well-ordering principle is obviously false, and who can tell about the Zorn's lemma?" (Of course, all three propositions are equivalent in ZF set theory.)
In type theory it's theorem of choice
It’ s amazing I did a video about this axiom just 3 days ago- do youtubers think the same ?)
Why does it feel like this guy is related to the guy from Langfocus?
They look similar.
@@anticorncob6 Similar hairstyle too
I love you
The music is distracting.
The speaker's gesture really hinders me to concentrate the contents.
I prefer to take set teory without AoC, and add GCH instead. AoC is only a theorem then... 😁
Yeh, but ..
An example would be good. And words like surjection, injection, bijection, and right inverse, could do with a reminder.
Taking a break from HW to look at a math video... and that is why ppl have called me a masochist.
This is math channel right? Where's the number? 🤕
math ain't about numbers!
Once you reach a certain point in maths, the appearance of a single number is surprising, and it's typically just a index.
I think once you reach that point, it's more about general functions and sets and stuff rather than specific elements of the sets. Also, you soon realise that numbers are nothing but equivalence classes of abstract sets.
How does he write backwards, he just learned that skill?!?!?! Or all the topology I know fails me and it's some kind of visual trick...
Or:
1. He writes with his left hand, stands in the viewer's position (in your place as you watch this video) and is filmed from the background side, to appear right-handed. This recording is superimposed to the next one.
2. The surface he writes on is "filmed" (recorded) from within itself only (your position as you watch this). Note: this surface must be totally transparent from the background side but the writing must be visible from your point of view (like a prompter screen ?)
@@georgecazacu6118 he's writing on a sheet of plexiglass, it's not super complicated
@@applessuace But the writing (from his perspective) is backwards, as you see it in a mirror.
@@georgecazacu6118 He just horizontally flips the video in editing.
Nice! I understood nothing!
Set theory is so ugly. Half the axioms are just there for the infinite case.
"Half the axioms are just there for the infinite case."
???
@@MuffinsAPlenty Absolutely. E.g. the power set axiom. Totally superfluous for finite sets.
I dont like it