The symbol for the empty set is not a phi, it was introduced by the Bourbaki group inspired by the Danish-Norwegian Ø (which sounds a little like the i in bird). The axiom of choice is only necessary when dealing with infinite collections of sets. If you have a finite set of sets, you can just do exactly what you said - pick an element from each set in order. Similarly, finite-dimensional vector spaces have bases without the axiom of choice. Moreover, there is an axiom of countable choice that is strictly weaker than the axiom of choice, and it is in a way more intuitive. Separable Hilbert spaces have Hilbert space bases using just the axiom of countable choice. Bases not that often used for other types of infinite-dimensional vector spaces, so that is not the main issue. A bigger concern to functional analysts may be trying to develop functional analysis without the Hahn-Banach theorem, which relies on Zorn's Lemma. However this can be done e.g. in constructive theories like Homotopic Type Theory, or Nik Weaver's constructive theories.

@gauravdoley1527

2 жыл бұрын

I came to this video cuz I had an argument with my friend over ZFC icarus set and absolute infinity, he said since ZFC icarus set resolves cantor's paradox along with many, ZFC icarus set is actually bigger than absolute infinity itself, but from what i read absolute infinite should be the epitome of all the numbers, it breaks the notion of size itself, but I can't find anything on the net for confirmation, so I decided to study into ZFC and came here, but I'm really new to infinite series and cardinality, so would you pls clear my doubt on which is bigger, ZFC icarus or absolute infinite? Also if you can give reasonings on how, that would be really helpful, so do you know something on this topic? Would really help me out :)

@TheMaginor

2 жыл бұрын

@@gauravdoley1527 I don't really know anything about the Icarus sets, sorry!

@gauravdoley1527

2 жыл бұрын

@@TheMaginor it's alright mate dont sweat it, thanks for replying either way!!

@kevalan1042

Жыл бұрын

@Gaurav Doley thanks, that's helpful

@CamEron-nj5qy

Жыл бұрын

My first thought

A of C is trivial for finite sets. Infinite sets give independence.

@eyelockedpro3203

3 жыл бұрын

Yes

@peorakef

3 жыл бұрын

wdym independence?

@ccdavis94303

3 жыл бұрын

@@peorakef with infinity it can be proved (through forcing) that A of C can't be derived.

@peorakef

3 жыл бұрын

@@ccdavis94303 so AoC isnt trivial for finite sets, it is not needed in the first place? only necessary for infinite sets?

@ccdavis94303

3 жыл бұрын

@@peorakef yes trivial in the sense that you can derive it.

Thanks for your video! I have several objections to its content, but I'll just point out what seems to me the most important, because it can easily lead to a misunderstanding. As you present it, it could be interpreted as saying that the Principle of Well-Ordering (PWO) is clearly contradictory with our intuitions, because "open intervals don't have smallest elements." But this has nothing to do with the PWO. What the PWO says about the real numbers is that there is an order, a way of putting the reals one after the other, such that, according to this order, every nonempty subset of real numbers has a least element. This order will then be very different from the standard order. This cannot contradict our perception of the reals, because we don't have any idea how this order would look like and no analysis of the standard order will help us to understand this alternative way of ordering the reals.

@tonaxysam

3 жыл бұрын

Yeah, that's the point of being contradictory; because such an order would imply that, no matter how weird or diverse your subset S of the reals R, there is always going to be some smalest element... if you think carefully about this that maybe could imply that R is a countable infinity set

@jgilferez

3 жыл бұрын

@@tonaxysam No, there is no contradiction there either. There are all sorts of well ordered sets, some of them are finite, some of them are countably infinite, and some of them are uncountably infinite. What the PWO says about the real numbers is that there is a well ordered set with as many elements as real numbers there are.

@tonaxysam

3 жыл бұрын

@@jgilferez Yeah, but if some well ordered set with the same amount of elements as the reals exists; there must be an isomorphism with some order that preserves the order of that set...

@StewartMcGinnis

3 жыл бұрын

@@tonaxysam For infinite sets you can get two sets with the same cardinality but different well order structures For example, consider the order 1 where I've added a single element w that is bigger than every natural number. It is a well ordered set of cardinality the same as the natural numbers. However, it's not order equivalent to the natural numbers because the natural numbers don't have a maximal element, but my new set has the maximal element w

@tonaxysam

3 жыл бұрын

@@StewartMcGinnis ​ @Stewart McGinnis Oh, so I guess that if a ordered the real number like: -1 That order would be a well order? Well that would actually happen right? I mean, every single not-empty subset of R will have a least element, right? But like, the least element has to be on the set?

2:10 not exactly. If the family of sets is finite, the choice can be proven by induction. It cannot be proven for infinite families of sets. The axiom of choice is about infinity.

@TheAcer4666

3 жыл бұрын

Yes exactly. The axiom of choice is about all collections of sets. This is made abundantly clear in the video, your ranting about finite families are not relevant to either the axiom of choice or this video.

@learpcss9569

Жыл бұрын

sorry for necrocommenting, but I still don't get it. If we can prove that it holds true for finite sets, what's the problem to take some finite subset of infinite set? I mean then we can prove this statements by two lemmas: if we can pick element from non-empty subset then we can pick it from initial set, since initial set contains the subset that contains "pickable" elements. Then by subset axiom we can pick subset from initial set and it is done? I don't claim this to be right since I'm not good at set theory, but still I wanted to know if there are some problems with this proof

@Miaumiau3333

9 ай бұрын

@@learpcss9569 I think you are confusing the size of the family with the size of the individual members of the family. When the family is finite, for example imagine a family comprised of 3 nonempty sets (say, A, B, and C), then it is possible to prove choice without using the axiom of choice, you can simply do induction. However, you cannot do this if the family is infinite (A, B, C, D, ...), you need an extra assumption such as the axiom of choice. Notice that we did not specify the sizes of the individual members A, B, etc., it does not matter if they are finite or infinite, they just have to be nonempty.

I believe this missed the point. I consider that the axiom of choice can be better described intuitively as "generalize to infinity whatever intuitive facts we know about the finite". (This is expressed by the "for every" in the formula.) Just as a note, the axiom of choice is also surprisingly equivalent to the fact that any surjective function has a right inverse, but only in classical logic. Intuitionistically the latter is weaker.

@pmcate2

3 жыл бұрын

​@Trevor Chase AOD seems to involve more complicated concepts in it's statement than AOC. I guess I see this as better reason to believe AOC. However, I am saying this as someone who has never heard of AOD before, and I do not know much about topological games.

@drewduncan5774

3 жыл бұрын

@Trevor Chase He didn't claim it's a justification of AC, just a description.

@santerisatama5409

13 күн бұрын

The intuitive fact we know about finite, is that a finite "forall" cannot be generalized to infinite "forall".

@tricanico

17 сағат бұрын

@@santerisatama5409 Well, then we are agreeing.

{a,b} is very different from (a,b). {a,b}={b,a} is unordered and (a,b)≠(b,a) in general, unless a=b.

@seneca983

4 ай бұрын

If you're talking about what is shown at 4:00 you seem to have misunderstood. Here (a,b) doesn't refer to an ordered pair of a and b. It refers to the open interval between them, i.e. the set of real numbers x for which a

@sufronausea

4 ай бұрын

​@@seneca983wasnt talking about that

@seneca983

4 ай бұрын

@@sufronausea OK, what were you referring to then?

@sufronausea

4 ай бұрын

@@seneca983 0:28 functions are defined such that their members are ordered pairs, not unordered

@seneca983

4 ай бұрын

@@sufronausea Ah, you were referring to that. That's true, though a definition using unordered pairs would still work in the case were the function is one set to another such that the sets are disjoint.

In my experience, the Barnard-Tarsky Theorem and other weird things that happen with the AoC reveal that all the counterintuitive consequences of AoC depend on the existence of pathological functions that break something in our intuitive understanding of those systems, and reveal how limited our intuitions are. In B-T, the sets are not measurable - that is, there's no way to assign to each set a volume that makes sense. Because the sets don't have a sensible way to assign volume, there is an intermediate state where the sets don't have a total volume that is well-defined, so of course you can end up with a case where you start with some volume V and end up with volume 2V. If you disallow such sets, demanding that each component set to still be measurable, then the B-T theorem collapses. Similarly with the W-O theorem; the ordering function is not the usual ordering function, and would have to be pretty gonzo in order for the W-O theorem to work. Order, after all, is additional structure on the real numbers, and a different order will have different properties. Our usual ordering of the reals is out choice out of the infinity of ordering functions available, and the AoC reveals that some of them are strange beasts indeed.

@santhoshs-vr3un

2 жыл бұрын

These points reflect my thoughts

@Smooth_Manifold

6 күн бұрын

“Banach-Tarski” are the correct names

@wyrmofvt

5 күн бұрын

@@Smooth_Manifold Yep. Derped on the name, as well as a few typos.

i'd like to first point out that when u said functions are tuples, u wrote a set of sets instead of a set of tuples

@andrasfogarasi5014

3 жыл бұрын

Do you really want to get into an argument about notation and whether 2-element sets can be considered tuples?

@CainGantt

3 жыл бұрын

@@andrasfogarasi5014 two-element sets aren't necessarily equivalent to 2-tuples, since ordering doesn't matter for the former. {1,2} = {2,1} but (1,2) =/= (2,1) following convention. the wiki article on ordered pairs says the modern set-theoretic definition is from Kuratowski. (a,b) := {{a},{a,b}}, and also offers a few equivalent definitions. this lets us get around the degenerate case with a=b, where we find {a,b}={a,a}={a}. without the additional structure from ordered pairs (in general, n-tuples). in this case, ordered pairs are very different from 2-element sets!

@pecfexfextus4437

3 жыл бұрын

@@andrasfogarasi5014 yes because order is not defined on arbitrary sets. and as such the function can fail to be a function since we can make it such that its not right-unique

@miguelandrade4439

3 жыл бұрын

Plus a function is a triple (A,B,g) where A is the domain, C is the codomain and g is the graph. He only showed us the graph. In his definition, an inclusion map would be the same as the identity...

@ARVash

3 жыл бұрын

@@pecfexfextus4437 It's an interesting anecdote but also it's understood implicitly by anyone who cares. As Andras said, it's a notation argument. It's implicitly understood to be an ordered set because otherwise it would make no sense. You might not choose to write it this way for a proof but for everyday conversation it is far clearer to assume the sets are ordered. He could have said that, but honestly how obtuse do you have to be to not be able to understand you can't swap the input/output of a function?

Consider an infinity shelf with containing pairs of shoes. If you want one of each kind, you simply choose the set of all leftys, and you're done even for an uncountable set. That's equivalent to choosing the set of all posivie reals out of the reals (and that get's messy already when go into the complex plane as there are no "positive" imaginary numbers...). But consider a shelf filled with pairs of socks. There is not way of universally distinguing between one sock and the other in each pair and at this point you need the AoC to declare wether the task is doable for an infinite abount of socks or not.

@imengaginginclown-to-clown9363

2 жыл бұрын

🧦

@bsatyam

2 жыл бұрын

I heard this example in a Frederic Schuller lecture iirc.

@MrLikon7

8 ай бұрын

This well known analogy is due to Bertrand Russel. Yes, Schuller mentioned it.@@bsatyam

4:14 confuses the natural order with wellorder. It would also apply to an open interval of rationals wich is trivially wellorderable as it's countable. A wellorder of any uncountable set get's quite "long" in a sense and is thus hard to imagine.

What you're saying at 4:14 is basically just "the usual ordering of R is not a well-ordering". The well-ordering principle states that there is _some_ ordering of R (which will look nothing like the usual one) that is a well-ordering.

@danmerget

3 жыл бұрын

Agreed. If my reasoning is correct, then one such ordering is to first sort by the fractional part, then the integer part. To be more precise: define two functions I(r) and F(r) so that I(r) is the "integer part" and F(r) is the "fractional part": * I(r) = floor(r) * F(r) = r - floor(r) For example, I(10.3) = 10, and F(10.3) = 0.3. For a negative-number example, I(-10.3) = -11 and F(-10.3) = 0.7. (Sorry if that seems counter intuitive, but I needed to have the property F(r+1) = F(r) in order to easily define "least element" later.) Now we define a less-than operator

@bachkhoahuynh9110

3 жыл бұрын

​@@danmerget Your ordering is not a well-ordering since the subset (0, 1) of R has no minimal element under the ordering

@danmerget

3 жыл бұрын

@@bachkhoahuynh9110 That's why I specified that the interval had to be "non-empty". If we use your criterion (i.e. that any open interval (a,b) must be non-empty), then it's impossible for ANY set to have a well-ordering, which renders the phrase "well-ordered" completely meaningless. PROOF: Suppose set S is well-ordered. Then given an interval (a,b) with a

@bachkhoahuynh9110

3 жыл бұрын

​@@danmerget I mean (0, 1) is a open interval in the standard topology of R, i.e., (0, 1) = {x in R | 0 < x < 1}. Well ordering requires every nonempty subset has the smallest element. For every a, b in (0, 1), we have a

@danmerget

3 жыл бұрын

@@bachkhoahuynh9110 But a major point in this discussion is that we're NOT necessarily using the standard topology of R; we're free to choose other orderings. And I'm not sure where you're getting "a

I don't think it's fair to say that "The well ordering principle is obviously false." It's just "an ordering", I see no reason at all why it should coincide with the standard ordering of the real numbers in any way.

@zapazap

Жыл бұрын

Espoecially since it clearly does _not_ coincide with the standard ordering.

Where does the cover from the video comes from ???

Is the axiom of choice actually part of Zermelo-Fraenkel? To my knowledge it was actually an optional extra one which when included turns regular ZF into ZFC

@An-ht8so

3 жыл бұрын

That is correct

your youtube chanel is just incredible, perfectly balanced !

Just... wow! 💜

An ordering where every open set of reals has a smallest element can actually be constructed as follows without the axiom of choice: Given x and y: 1) If x and y are both irrational, x

Amazing. Do you have any books to recommend?

Not all mathematical objects can be regarded as sets. Case in point, categories which are distinctly quite a lot larger than sets and we can talk about the category of all categories, but not really the set of all sets.

@MetaMaths

Жыл бұрын

I know, I apologise for oversimplifying. The margin of this video is too narrow )

@thestemgamer3346

Жыл бұрын

@@MetaMaths thanks for the reply, it be like that sometimes

@zemm9003

Ай бұрын

Not true. Sets are the atoms of human intuition. Categories are intuitive sets but are not sets in ZFC. The distinction is very important. Both categories and ZFC sets are two different facets of intuitive sets that serve different purposes.

Excelent video! Could you tell me what progam do you use to edit videos?

@tissuepaper9962

3 жыл бұрын

The graphics appear to be done with manim, and the video editing itself is quite basic, it could be done in any editor.

@adrielalberto6591

3 жыл бұрын

@@tissuepaper9962 Thanks man.

Really loved it . I'd share my opinion as well in the mean time in the future.

What’s the background music?

interesting, will defo watch again.

@dinglerdangler

Ай бұрын

watched it again. its pretty good

Loved it !

the video is neat and short, like math videos should be, but take note of what the other commenters said about the notation for tuples, triviality of aoc for finite sets, your explanation of well-orders etc

question. there is an axiom claiming the existence of the empty set. If we discard this axiom, what does the axiom of choice look like? I.e., does the axiom of choice depend on the existence of the empty set?

@kylecow1930

19 күн бұрын

the existance of the empty set follows from the axiom ∃ x (x=x) and the comprehension schema. Essentially once anything exists, so does the emptyset

can you do a video on first order logic?

I am for Axiom of Determinancy. Also, why did you use curly bracket for ordered pair? While it curly bracket do work if the domain and codomain are different, if the mapping is onto itself, that notation just won't work

@geraldthaler8880

3 жыл бұрын

But AD has itself weird consequences for infinite combinatorics. However "Definable Determinacy" is "true" in the sense that it follows from large cardinals and it is consistent with choice. This is enough to get reasonable behavior for reasonably defined sets. It excludes projective Banach-Tarski decompositions for example.

Nice video, and cool channel.

Personally I think the axiom of choice, similar to the continuum hypothesis is not something is necessarily true or false, but rather, it is a tool that can be used in a variety of different mathematical contexts.

Is the category of small categories a set?

@seneca983

4 ай бұрын

I'm pretty sure it can't be just like the class of all sets isn't a set.

The problem of finding a basis for any vector space is indeed puzzling, even in the enumerable case. Start with the field of rationals Q. The polynomials Q[X] are also enumerable and, as a Q-vector space, have the powers of X as a basis. No need of AC for that. Now consider the algebraic numbers over Q. They also form an enumerable Q-vector space. Yet we don’t have an ACTUAL privileged basis for them. We know it exists, we even know how to compute one, by enumerating algebraic numbers and checking if they are linearly independant from the previous ones. But we cannot describe such a basis in its entirety.

No one is stoping anyone to create a whole math ignoring the axiom of choice. The trickiest part is convincing the community to work on the crooked version mathematics

For 4:00, there seems to be an easy definition to make it well-ordered, just let x ∈ (a,b), and the less-or-equal-than is defined by comparing abs(x1 - (a+b)/2)

I thought axioms were unprovable "premises" we just chose to assume is true (or not) and mathematics is what happens when we explore the consequences of the axioms we've chosen to use...

@ahoj7720

Жыл бұрын

Unfortunately whatever the set of axioms you choose, as long as it allows usual arithmetics, there are true statements that are unprovable. This is Gödel’s incompleteness theorem.

A lot of people have pointed out a variety of mistakes in this video, but in my opinion the biggest error - one that I see made very often - is the idea that choice is something that we should decide is valid or not. In reality, there is no problem with the current state of affairs where we study mathematics both with and without the axiom of choice. There is no conflict here, simply two different systems both being studied.

@MetaMaths

14 күн бұрын

Why do you see this as an error ? The idea of this video is to present this decision as a philosophical challenge.

If you actually dig into the proof of Banach-Tarski, then the argument I find most paradoxical is not the application of choice; it is the fact that if you have the set of all sequences of cardinal steps (left, right, up, down) that end with a step up, you can step them down to get absolutely all sequences of cardinal steps.

@gernottiefenbrunner172

3 жыл бұрын

You actually get all sequences of cardinal steps, or all finite sequences?

@benjaminpedersen9548

3 жыл бұрын

@@gernottiefenbrunner172 Yeah sorry. I am only talking about finite sequences.

@austingubbels

Жыл бұрын

What does it mean to "step them down"?

@benjaminpedersen9548

Жыл бұрын

@@austingubbels I simply mean to add a down step to the end of all the sequences, but the down step cancels out with the up step at the end of every sequence. The context is stepping on the surface of a sphere. Watch the Vsauce on the paradox if you want more details.

The Well Ordering Principle puzzled me since I began my math degree in 1978 and still does :) Because it says something exists without saying how the hell you can get it!

@pecfexfextus4437

3 жыл бұрын

maybe u should become a constructivist :)

@joshuascholar3220

3 жыл бұрын

most numbers can not be constructed.

@JonathanMandrake

3 жыл бұрын

The least productive type of proofs are those that simply prove something without giving any other information, for example proving that something exists but without information regarding how to get it. The best type of proofs not only prove something, they also give additional information for using the proven statement to do calculations or additional proofs.

@raffaelevalente7811

3 жыл бұрын

@@JonathanMandrake Indeed. The fixed point theorem is great, because its proof gives also a way to find the FP!

@ExplosiveBrohoof

3 жыл бұрын

That's AoC for you. It assumes that you can choose things, even when there's no way to do so.

For the open set (a,b), can't we make a bijection f: (a, b) → R₊U{0}, and then take the usual ordering of R₊U{0}, and define a well-order 𝒓 on (a,b) by taking x 𝒓 y in (a,b) if and only if f(x) ≤ f(y), then having f⁻¹(0) as the least element of (a,b) ? Like, yeah, the usual ordering for R implies that there is no least element in the set (a, b), but that doesn't mean that we can't 𝗱𝗲𝗳𝗶𝗻𝗲 a well-order on this set, as the ordering relation that we defined above is a well-order. What the well-ordering theorem states is that 𝘦𝘷𝘦𝘳𝘺 𝘴𝘦𝘵 𝘤𝘢𝘯 𝘣𝘦 𝘸𝘦𝘭𝘭-𝘰𝘳𝘥𝘦𝘳𝘦𝘥, and that's what I constructed above for the set (a,b).

@imengaginginclown-to-clown9363

2 жыл бұрын

What you defined isn't a well order. It is an order that has a least element. Well order means that _every subset_ has a least element. Consider for example (0, ∞) ⊆ [0, ∞). This does not have a least element. Thus ([0, ∞), ≤) is not well-ordered. Moreover it is impossible to define a well order on any continuum-size set. We can only prove its existence. This follows from the independence of the existence of a well order on ℝ from ZF (see Jech). Otherwise, if it was definable, we could prove its existence in ZF alone and that would be a contradiction.

@JadeVanadiumResearch

Жыл бұрын

@@imengaginginclown-to-clown9363 Minor correction, the independence result implies that ZF can't prove any particular formula is a well order, but it doesn't imply that every formula *isn't* a well order. Under the axiom of constructability (V=L), there is a specific formula which you can write down and then a specific proof which shows that this formula well orders the entire universe of sets. This formula still exists in ZF, it's just that the associated proof is invalid (since it assumes V=L, which isn't an axiom of ZF). Since Choice is independent of ZF, we can't prove the formula is a well order within ZF, but since V=L is independent of ZF, we also can't prove that it's *not* a well order. Otherwise I agree with everything you said.

Love your video🤓🤓 Please do a video on topology

The Choice Theorem is born of the inconsistency (not incompleteness) of Arithmetic. The theorem poses the undefined "plus one" of Addition against the definable prospects of Multiplication. To redeem Choice, the Theorem must include an Axiom of Hierarchal Iteration, so as to secure rigor among the applications of Choice.

Your definition of a function as a set is wrong. A function is not a set of sets as you put it, but a set of ordered pairs, and an ordered pair is a very specific set of sets, namely: (a,b) = {{a}, {a,b}}.

@brandonwalker5011

3 жыл бұрын

Thats still a set of sets but I agree with you in principle.

@zapazap

Жыл бұрын

A function in lambda calculus is well defined, but is not a set of ordered pairs. And what you claim that an ordered pair _is_ is merely one way of encoding an ordered pair. Cheers!

It sounds like a dispute over the metaphysics of what a "set" really is. Given a set S, if you can "see" S can you also "see" its constituent members? Or when you "look" at any set does it appear atomic, with no inner structure, regardless of how complex its constituent membership graph looks? What does it really mean when we say a set "contains" another set?

@GabriTell

29 күн бұрын

I'm not specialized in logic or theory set, but I think a reasonable way to think of sets is as a kind of properties class. In second order logic, you can quantify properties, and the information of any set "A" can be encoded in one single property "Ã" such that "Ã[x]" ⇔ "x∈A" (so it wouldn't be that crazy to formalize sets just as "A=Ã"). I'm a Math Degree student and I have a good foundation in logic, but it's a very extensive branch and I would recommend that you ask specialized people in logic and set theory. 👍

The Banach Tarski paradoxon sounds like something that would depend on what types of metric or topological transformations you're allowed to do. Given enough allowed transformations, we can do something close to that IRL with closed balloons (by changing their temperature). The well-ordering theorem sounds like something that would be near-impossible to disprove since it's really hard to check _all possible_ orderings of the reals. Interesting that it can be generalized to the Axiom of Choice. Zorn's Lemma is obviously true.

@lukasrollier1004

3 жыл бұрын

The Banach Tarski paradox specifically talks about isometric transformations, i.e. taking the pieces and moving them about or turning them without deforming, shrinking or expanding them in the intuitive sense. That is: only acting on them by orthogonal matrices and translations. Also: "Zorn's lemma is obviously true" LOL

@iwersonsch5131

3 жыл бұрын

@@lukasrollier1004 Does it happen in a space we can't imagine then, similar to that hypersphere that's boxed in by touching hyperspheres in the corners of a hypercube, and is actually larger than the corner hyperspheres with enough dimensions? Cause what you describe sounds like doing it in 3d with finitely many parts should be blocked by some invariants unless i'm misunderstanding something

@lukasrollier1004

3 жыл бұрын

In a dimension higher than three, we can take isometrics that only turn in the dimensions we're used to and leave everything else intact, so higher dimensions follow from the regular Banach-Tarski paradox. Also "Zorn's Lemma is obviously true" LOL

@iwersonsch5131

3 жыл бұрын

@@lukasrollier1004 That doesn't mean that the product of two of those orthogonal matrices have to be a 3D turn, or that the same invariants exist

@lukasrollier1004

3 жыл бұрын

No, the (tensor)product or two such transformations would be 9-dimensional. I only suppose that we turn in three dimensions and leave the rest untouched. Btw: what was that about Zorn's Lemma?

My favourite thing equivalent to the Axiom of Choice is that every fully-connected graph has a spanning tree.

@egor.okhterov

3 жыл бұрын

Why so complicated example? Every non-empty graph has a vertex.

What is a vector space?

Now you got me really confused. How can an open interval have a maximum? But then again, how can the axiom of choice be false? It seems so intuitive.

@ga35am

3 жыл бұрын

An open interval doesn't have a maximum, definitely not considering the usual order of the real numbers. What happens is that AC guarantees some other order for which the open interval has a maximum.

@ga35am

3 жыл бұрын

@@PefectPiePlace2 I think that the things you said are equivalent. I'm playing cards now. Hope to remember to explain. And to be right. Hahaha.

@joaquingutierrez3072

3 жыл бұрын

Of course that is false. The usual intuitive order is not the only way to order the real numbers. Look at the natural numbers, for example, you can define an ordering by divisibility: a

@ga35am

3 жыл бұрын

@@PefectPiePlace2 wow. I really thought a lot, until I remembered the order guaranteed by the AC makes the set well-ordered, meaning every subset has a maximum or a minimum, you can adjust your order to be one or the other. But I still think we could argue without that information, kind following the idea of the proof of AC ---> well-ordering theorem. If I remember it correctly, it uses some idea of colimit in the category of ordered sets or something like that.

@ga35am

3 жыл бұрын

@@PefectPiePlace2 about "they are certainly not equivalent, the latter is much stronger", the fact that the latter is much stronger definitely do NOT implies "certainly they are not equivalent". There are several results in Mathematics that shows two statements, one clearly stronger than the other, as equivalents. Module theory has a bunch, especially because the properties about modules over a ring depend a lot of the properties of the ring as module over itself. But in all areas of Mathematics there is some results like that. It is one of the wonders of Math. Good study!

I've met with Jerry Bona and have had some professional interaction with him, as we work in roughly the same area. He's an incredibly nice guy.

I despise the Axiom of Choice. Although Gödel proved that it is consistent with the other ZF axioms of standard set theory, his proof relies on a significant loophole in the axiom of Powerset. The axiom of Powerset asserts that every set has a powerset, but it does not clearly define the powerset of an infinite set, such as N. Consequently, the ZF axioms permit the construction of a set theory model where the "powerset" of an infinite set like N includes only the subsets with finite descriptions. This model is known as the Constructible Universe. The Constructible Universe has an extremely sparse notion of the powerset, where all sets are essentially countable. (Although the powerset of N isn't technically countable within the model, as the diagonal argument still applies.) Unsurprisingly, the Axiom of Choice holds true in the Constructible Universe, making it consistent with ZF. However, this only indicates that the ZF axioms allow for pathological models of set theory. Many people mistakenly believe that the Axiom of Choice has been proven consistent with the intuitive notion of sets and powersets, but this is not the case.

Really Insightful ❤✌...but i just have one question or maybe more than one😁: how do we know there is nothing more fundemental than a set in math ? How come set theory explains everything but it doesn't treat mathematical logic and statements ? Aren't relations between sets are mathematical ? So there have to be a more fundemental basis for set existnce and their properties and all math in general ? ..i would be glad to discuss such questions with u ❤🍀👍

@An-ht8so

3 жыл бұрын

Set theory is precisely what we call a theory in logic. There can be other theories than set theory. There are theories that are stronger than set theory, in which we can prove more things. For you last question, one important theorem of logic is one that states that if a theory is consistant, i.e. has no contradiction, then this theory has a model, i.e. there exists a mathematical universe whose objects obey the axioms of the theory.

@imengaginginclown-to-clown9363

2 жыл бұрын

Set theory can and does treat mathematical logic. For example the compactness theorem mentioned in this video* is a theorem of mathematical logic. It depends on what you want to choose as your foundation if you want some "more fundamental" structure than a set. There is for example type theory as well. Classically foundation is viewed and structured like this: 1. Establish the language of first order logic 2. Define inference rules (for example sequent calculus or natural deduction) 3. Formulate ZFC in this language 4. You can do all the maths in ZFC from there on *the equivalence shown in the video is an open question, why it is presented as a fact is beyond me

I think this is a weird case where intuition varies wildly between people. To me choice requires many other assumptions that define what choice that by the time you get to choice you have either constrained it from existing or make a long tedious specification of all the things it isn’t just to arrive at the axioms either are or arent a choice. If they are a choice, the system is predicated on choice implicitly and not in need of its own statement as it inherits it from a parent logical system. If it isnt possible, you would have ruled out all cases explicitly otherwise and the statement would thus be a corollary. The only axiom I could imagine any value in would be to define choice, specifically that it does not influence probabilities, it is constrained by them. I don’t think that’s necessary as computability constrains consciousness and that would be a higher order of logic than any statement about choice. The recursive expansion of a system of computing its own system means that it would at the very least add far more certainty than the time available to act on any non-deterministic elements. Strictly philosophically, I think the contemplation of free will is reducible to the necessity believing it exists. If it doesn’t, nothing changes because you always would come to that conclusion and if it does, you have maximized your agency. That does rely on believing maximizing agency is something that should be done, but the logic of free will inherits that judgment. In summation, I propose the only virtuous path is to type long winded rants in KZread comments about freewill’s irrelevance, pat yourself on the back and leave it at that.

You're doing a good job, keep going, you're succeeding to convey the beauty of math

@imengaginginclown-to-clown9363

2 жыл бұрын

Would be nice if more effort went into the spelling, actual fact checking and most importantly: actually getting a point across. The only point that is giving contra to the AC is the well order argument which doesn't work because that also applies to rational open intervals which are well-orderable.

I've just came up with idea how to order e.g. (0.5, 1.5) real numbers interval. Let us compare numbers by absolute difference with the mean number of the interval ends, in this case with 1. Additionally, if these absolute differences of two numbers match, assume the number which is less than the mean one (in normal sence), "less" than the one greater than the mean (so e.g. 0.8 > 1.1, 0.8 < 1.2). That way the minimum element is the mean itself.

@Achill101

3 жыл бұрын

As far as I can see, this should work, for open and closed intervals. But I'm not a logician. The problem with well-ordering is that ANY set of real numbers should have a well-ordering.

@geraldthaler8880

3 жыл бұрын

You need a minimum for _every_ nonempty subset, so that there exists no strictly decreasing infinte sequence of elements at all. So you need to start numbering all the elements of the set, the first one, the second one, ..., the omega-th one etc. We need to count far beyound the natural numbers (transfinite recursion) and in a sense choose arbitrarily which element we want to assign the next ordinal number. There is no definable wellordering of the reals.

@imengaginginclown-to-clown9363

2 жыл бұрын

If you can well-order a set X of size continuum (for example open or closed non-empty intervals) then you can well-order ℝ. There is a bijection f: X → ℝ. If we call the well-order on X by the letter R then we can define a well order S on ℝ by f(x)Sf(y) iff xRy for all x,y ∈ X. It is well known that ZF (meaning no axiom of choice) cannot prove that there is a well order of the reals (see for example Jech); so you cannot define one, because otherwise it would provably exist in ZF.

@gauravdoley1527

2 жыл бұрын

@@imengaginginclown-to-clown9363 I came to this video cuz I had an argument with my friend over ZFC icarus set and absolute infinity, he said since ZFC icarus set resolves cantor's paradox along with many, ZFC icarus set is actually bigger than absolute infinity itself, but from what i read absolute infinite should be the epitome of all the numbers, it breaks the notion of size itself, but I can't find anything on the net for confirmation, so I decided to study into ZFC and came here, but I'm really new to infinite series and cardinality, so would you pls clear my doubt on which is bigger, ZFC icarus or absolute infinite? Also if you can give reasonings on how, that would be really helpful, so do you know something on this topic? Would really help me out :)

@imengaginginclown-to-clown9363

2 жыл бұрын

@@gauravdoley1527 I'm not sure what sort of fringe pseudo-math your friend has been reading but he is certainly on the one side either wrong or not even wrong (in the sense that what he says does not make sense). First of all the not even wrong part: Absolute infinity is a concept from philosophy and it is not defined _at all_ in ZFC. If something like this _would hypothetically_ exist then it would be some sort of super Ord and an Icarus set would not be larger than that. But it is not even defined so the question is entirely ill-posed. I'm not sure what Cantor's paradox refers to here; it usually refers to the fact that some classes of sets are not sets, like the class of all sets (there is no set X such that for all sets Y, Y ∈ X). It isn't really a paradox in the sense that there is a contradiction here. If the existence of an Icarus set would imply that this is not the case then Icarus sets are inconsistent with set theory. This however is currently unknown since consistency of Icarus set is equivalent to consistency of I0. If you want to learn some set theory you should find a proper book about it. Consider for example Jech's set theory.

In my mind, there is an easy way to understand axiom of choice, and when you don't need to use axiom of choice. Let X and Y be sets. Statement 1 can be derived from ZF, whilst statement 2 requires Axiom of Choice. 1. Suppose P(x,y) is a predicate such that for all x in X, P(x,y) is true for *exactly one* y in Y. Then we can construct a function f: X -> Y s.t. f(x) = y iff P(x,y). 2. (Axiom of Choice). Suppose P(x,y) is a predicate such that for all x in X, P(x,y) is true for *at least one* y in Y. Then we can construct a function f: X -> Y s.t. f(x) = y iff P(x,y). See the subtle difference? As a bonus, to also understand why we need axiom of choice,, try to work out a choice function for the real numbers. If you are given any subset in R, can you find a rule to pick an element from that subset? On the contrary, you can do this with the natural numbers (just select the minimum). Accepting the axiom necessarily means there exists things you can't construct, which some may consider problematic.

@miloszforman6270

2 ай бұрын

_"Accepting the axiom necessarily means there exists things you can't construct, which some may consider problematic."_ Yes, that's kind of problematic. Consider the powerset of the natural numbers. What is a subset of the latter? To define one, we need a rule, that is, a finite string of characters. Now the set of finite strings made of a finite set of characters is countable. The powerset of ℕ is said to be uncountable (Cantor's diagonal argument). That doesn't mean that the powerset is larger (it isn't) but that there is no rule which can tell us which finite string is a valid rule. So the powerset of ℕ can't be constructed - it has a kind of diffuse "existence".

Depends on the circumstance I'd say. There certainly are cases where it may be more useful to work in ZF instead of ZFC. I very much value that we can have a well-ordering of larger cardinals in ZFC. I did not quite get your point about the well-ordering of R contradicting our intuition of how there always is a smaller number below, how is the idea of smaller and smaller real numbers contradicted?

@MikeRosoftJH

9 ай бұрын

Obviously, the usual order on real numbers is not a well-ordering relation - for example, the open interval from 0 to 1 has no minimum under this relation. In fact, an exact opposite is true: the usual order on real numbers is a dense order - given any two different real numbers x and y there exists a real number strictly between them, such as (x+y)/2 (and by extension there exist infinitely many such real numbers).

I attained my level of incompetence with undergraduate Maths (I did graduate, with a degree in Maths, ahem). I'll leave these questions to the Big Brains. Just realized that Mathologer has discussed this axiom. Review time.

I am a little confused by the part about the well ordering theorem... We can prove that there is no way to order the reals, right? If the axiom of choice implies that we can order the reals, does that not mean that we have a contradiction and have disproved it? But you had also said that it was not possible to disprove the axiom of choice from the other axioms. What am I missing here?

@robertgamer3112

3 жыл бұрын

Our typical way of ordering the reals is not well ordered, but that is not to say that there is an ordering that works completely differently that is in fact well ordered. The same is true for the rationals. By taking our ordering to be the standard then the rationals are not well ordered. But we can put a different ordering on the rationals so that they are well ordered because they are bijective with the natural numbers.

@Jesse_Carl

3 жыл бұрын

@@robertgamer3112 Interesting. I am currently taking a proofs class, and we just did the diagonalization proof that the reals are uncountable. How does that proof depend upon the size of the numbers? I understand what you are saying about the rationals, but I am still a bit confused

@An-ht8so

3 жыл бұрын

@@Jesse_Carl What he is saying, is that for a countable set such as the rational, you do not need a powerful theorem to find a well ordering, as you can transpose the well ordering from N to your set. For the real numbers you cannot do that, and the axiom of choice is needed. However the proof does not depdend on the size of the set in the sense that you could use the axiom of choice to find a well ordering of Q, although it would be needlessly complicated.

And we got our version of Euclid's 5th Axiom.

@zapazap

Жыл бұрын

It's not the only candidate. Consider the claim that there is no cardinality between aleph_0 and aleph_1.

@MikeRosoftJH

Жыл бұрын

@@zapazap There by definition cannot be any cardinality between aleph_0 and aleph_1. aleph_1 is the smallest well-ordered cardinality after aleph_0; there by definition can't be any cardinality between the two. Cardinality of the continuum is not necessarily aleph_1; that this is the case is the continuum hypothesis. (Cardinality of the continuum can be aleph_1, or aleph_2, or aleph_37, or almost any other cardinality. For a somewhat technical reason, it can't be aleph_ω; it can be lesser or greater than that, though. And in absence of axiom of choice it's consistent that continuum can't be well-ordered, in which case its cardinality is not equal to any aleph number.)

Proving continuum hypothesis , proving inconsistency in ZFC , constructing ZFC from naive set specification , resolving Russell's paradox , constructing infinite number system , construct and ensure overall consistent mathematical universe and developing arithmetic system - edition 8 May 2024 DOI: 10.13140/RG.2.2.21713.75361 LicenseCC BY-NC-ND 4.0

This axiom is not of "indisputable importance", because the cases where it is used in practical applications, it isn't the set-theoretic axiom of choice, but the second-order arithmetic axiom of choice, which is not controversial.

0:32 Just nitpicking here, but I think there should be a difference between ordered pair notation and set notation.

@MetaMaths

2 жыл бұрын

You are right, my bad. What do you think about the video anyway ?

@ethannguyen2754

2 жыл бұрын

@@MetaMaths I think it was a good video overall. I hadn’t understood what was so controversial about the axiom of choice until I watched it. It really doesn’t feel like the real numbers should have a well order to me, but I guess the axiom of choice essentially demands it.

@gauravdoley1527

2 жыл бұрын

@@MetaMaths I have a question which is not really related to this specific video. I came to this video cuz I had an argument with my friend over ZFC icarus set and absolute infinity, he said since ZFC icarus set resolves cantor's paradox along with many, ZFC icarus set is actually bigger than absolute infinity itself, but from what i read absolute infinite should be the epitome of all the numbers, it breaks the notion of size itself, but I can't find anything on the net for confirmation, so I decided to study into ZFC and came here, but I'm really new to infinite series and cardinality, so would you pls clear my doubt on which is bigger, ZFC icarus or absolute infinite? Also if you can give reasonings on how, that would be really helpful, also we have a debate and i can't loose cuz I'll have to give him a party if I do :')

Set theory is not able to express all of mathematics. Firstly, you need logic before you can even talk about sets, so mathematical logic is not reducible to set theory. Secondly, there are mathematical objects that can not be expressed in terms of sets, namely proper classes. Famous examples are the class of all sets and the surreal numbers. There are other approaches to foundational mathematics like type theory and category theory which have fewer problems. Also, there are more axioms which are independent from ZF, like the continuum hypothesis. In modern math it depends on the area of research if you employ the axiom of choice or not. In analysis it is generally assumed implicitly, in finite algebra it is not (since you don’t need it). The nice thing is that all mathematical statements are of the form: If property A holds then also property B holds. So you can safely assume the axiom of choice when proving something and only if someone wants to use your result, they have to decide if they are willing to assume it themselves.

@An-ht8so

3 жыл бұрын

I have a genuine interrogation about category theory. From what I know (which is little), there does not exist of formalisation of category theory that is not based on set theory. Therefore I don't really see how there could be fewer problems, and I would be interested if you had a resource with for instance a list of axioms for category theory.

@imengaginginclown-to-clown9363

2 жыл бұрын

@@An-ht8so See Proceedings of the Conference on Categorical Algebra, La Jolla 1965. Professor Lawvere has published a paper there called "The Category of Categories as a Foundation for Mathematics" which seems to be what you are asking for.

@imengaginginclown-to-clown9363

2 жыл бұрын

No, you are wrong, set theory is able to express all of the current mathematics. You can indeed formalise first order logic inside of set theory and this is indeed done in practise when working in logic. The question of "what came first" is irrelevant to the question of "can it be done in set theory". The thing about classes is also a bit redundant, we usually only care about definable classes and can handle them similarly to sets, because we can say x ∈ C iff φ(x) with a slight abuse of notation and work as usual. If you really need a set theory with classes you can also use NGB. "Fewer problems" is also a bit vague. Type theory sure is great for formalising with a proof assistant. Otherwise, we can interpret most type theories in ZFC and ZFC within most type theories, so that's not really "better" or "has less problems". I also don't see how you can claim that category theory has less problems as a foundation when this is not really a developed theory.

Could someone please explain why/how there is any doubt over the Axiom of Choice being true? Surely it's not difficult to simply choose any element from each set in a collection? How is there a possible problem with that?

@MetaMaths

8 ай бұрын

The doubt comes from our intuition, if you like. Certain consequences of AoC that are mentioned in this video appear counter- intuitive

@miloszforman6270

2 ай бұрын

_"Could someone please explain why/how there is any doubt over the Axiom of Choice being true?"_ It gets kind of diffuse if infinitely many sets are involved. Let's say that we partition the interval [0, 1] in several sets of points. We can do that by defining that any two points should belong to the same set if their distance is a rational number. Quite easy a partition, isn't it? But now we create another set which consists of exactly one point of each of these previous sets. How do we choose this one point? We need an algorithm to do that. Unfortunately, _there is no such algorithm._ Now that's where things get ugly. Some math people say: we don't need no fckg algorithm, because the axiom of choice tells us that this set does exist: we _postulate_ that the set exists. - But other math people say: no algorithm, no set. Peculiarly, they are the minority.

أنا أعتقد أن بديهية الاختيار تشبه المسلمة الموازية لإقليدس، هناك أنظمة متسقة بافتراض أنها صحيحة وأنظمة متسقة أخرى بافتراض أنها خاطئة ولكن لم نوجد تلك الأنظمة بعد

I'm team axiom of choice all the way ! just cuz somethings weird to us doesn't mean it's wrong

@ArielLorusso

3 жыл бұрын

More like obvious but not redundant as it may seem

Wait, what? Is the well-ordering theorem equivalent to the axiom of choice? In that case both must be false?

@MetaMaths

Жыл бұрын

Well ordering implies axiom of choice

@kevalan1042

Жыл бұрын

@@MetaMaths ah ok. Well a false proposition implies anything, doesn't it? Ex falso quodlibet.

@MikeRosoftJH

Жыл бұрын

@@kevalan1042 OK, but I am not sure what you mean by axiom of choice or the well-ordering theorem being false. Axiom of choice is equivalent to the proposition that all sets can be well-ordered. And set theory without axiom of choice can't prove axiom of choice true or false (that is, assuming set theory itself is consistent). There's a model of set theory (such as the constructible universe) where axiom of choice is true; and there's a model where axiom of choice is false.

@semicolumnn

Ай бұрын

Well-order theorem isn't trivially false. In fact, AC => WO, and WO => AC, which means it is independent from ZF.

My understanding was that you only need the Axiom of Choice for INFINITE sets. For finite objects, things like every (finite dimensional) vector space has a basis are provable without it. The real problem I have with the Axiom of Choice is really a larger problem with inherently ill-defined and non-computable objects. Because such things generally void the warranty on pure logic itself, given that applying logic (deduction, induction, etc.) IS a form of computation. Too much of pure math has turned into questions akin to "If I put an infinitely hot burrito into an infinitely cold cooler, what temperature does it get"? As in, questions whose entire premise is logically flawed and therefore any claimed answer will be too. The Axiom of Choice or a good competitor is necessary to retain some sanity when you start playing heavily with all these infinities, and the fact you still get paradoxes is really a reflection of the fact that true reality doesn't allow infinite things.

@altrag

3 жыл бұрын

> Too much of pure math has turned into questions akin to "If I put an infinitely hot burrito into an infinitely cold cooler, what temperature does it get"? Sure, but we do that because every once in a while you'll get a perfectly crafted cheesecake out of it, which makes absolutely no sense and yet is somehow perfectly consistent. Its those completely unexpected results are what moves the highest ends of math forward.

@MuffinsAPlenty

3 жыл бұрын

"Too much of pure math has turned into questions akin to "If I put an infinitely hot burrito into an infinitely cold cooler, what temperature does it get"?" Nonsense. This is only "too much of pure math" if your opinion of pure math comes from popsci sources talking about the foundational problems people were grappling with in the early 1900s. "and the fact you still get paradoxes is really a reflection of the fact that true reality doesn't allow infinite things." This is good insight. All "paradoxes" resulting from the axiom of choice depend on the axiom of infinity as well. And as some like to point out, sometimes rejecting the axiom of choice has results that are equally "paradoxical" to accepting it. For example, it is consistent with the negation of the axiom of choice that the set of real numbers can be partitioned into strictly more nonempty subsets than there are real numbers.

To me, who csn do much arithmetic abd geometry, this iscs load of silliness

It boils down to being a constructivist/structuralist or not. Without defining a structure of the mathematical object (in this case the set), you cannot have an algorithm for choosing an element.

I don't really think that the well-ordering theorem is 'obviously false', neither I feel that it's 'obviously true', however, as far as the open sets of reals are concerned idk what the fuss is over since it's not about ordering them necessarily by the relation of '

Anything self consistent is legitimate in their appropriate context.

This is my favourite comment section on the entire website

My abstract algebra professor once said: "remember that, in this class, we're believers. We believe in the axiom of choice". I agree with her

@Achill101

3 жыл бұрын

You don't need to be a believer, you can be a formalist: Assuming the Axiom of Choice for this class, how far can we get? For another class: Assuming the Axiom of Dependent Choice, how far can we get now?

@Achill101

3 жыл бұрын

@Me Too - what's wrong with a game, especially such a beautiful and useful one as mathematics? Or do you think seeing math that way would disrespect its truth? . . . In history, formalism has probably been the leading choice of mathematicians since 1900, following Hilbert and Bourbaki.

@mjtsquared

3 жыл бұрын

@Me Too Math is just a chess game. Axioms are treated no more than the fact that chess pawns can only move one square at a time-a fact that you don't believe nor refuse, but accept simply because the rules said so! :)

@zapazap

Жыл бұрын

Sounds officious to me. I would stand up and say "I do not. Does that preclude me from getting an A?"

After seeing that every things we define is based on assumptions called axioms, I am having a mathematics existential crisis.

@michaeldamolsen

3 жыл бұрын

You are in good company. Check how Hilbert felt about Gödel's incompleteness theorems.

@An-ht8so

3 жыл бұрын

Axioms of ZF are merely definitions of what a set is. Every definition is formaly an axiom. If you're not puzzled by definitions, you should not be puzzled by axioms.

@imengaginginclown-to-clown9363

2 жыл бұрын

@@michaeldamolsen Hilbert knew about the thing that Ranjit mentions here. The incompleteness theorems go a lot further than this.

@imengaginginclown-to-clown9363

2 жыл бұрын

@@An-ht8so That's not true. Definitions are usually shorthands for longer formulas while axioms limit the class of models which satisfy your theory.

@An-ht8so

2 жыл бұрын

@@imengaginginclown-to-clown9363 What you're describing is the difference between conservative extension and proper extension of a theory, but they're still langage/theory extension, new symbols, new axioms. They can be shorthands, but they can also have real theoretical value. For instance, Henkin witnesses are absolutely core to proving the completeness theorem, The truth is "definitions" are not a logical concept

I am of the side that is true, combined with the side that works. If something is obviously true, and it works, thus has practical use, then it is fine with me. You should try it in daily life too.

"...which side are you on?" Do I have a choice ?

At 0:40 you wrote a function as the set of tuples but not ordered pairs as it is, one has {a,b}={b,a}

@MetaMaths

3 жыл бұрын

I agree, the burden of this error will pursuit me forever... Thanks for being attentive ! I love careful viewers.

@zapazap

Жыл бұрын

I always assumed the term 'tuple' to imply ordering.

The interval (a,b) doesn't have a smallest element in the usual order of R. Contrary to the statement in the video, the well ordering principle doesn't contradict this fact, it only asserts that there exists on ordering of R for which there will exists a smallest element for (a,b), ... obviouslly not the same ordering scheme then the usual for R

@MetaMaths

2 ай бұрын

I meant that it contradicts our normal intuition

Of course open intervals of reals don't have a least element when we use our normal definition of ordering. The WOP says that there is a well ordering, it doesn't necessarily have anything to do with the order we use normally. That's like saying because the negative integers clearly don't have a least element, the WOP shouldn't apply; but that's nonsense, we can just define "less than" to be what we normally call greater than, and this gives a well ordering of the negative integers. Of course no such easily definable well ordering can exist for the reals; this is exactly the point of the axiom of choice; but it shouldn't be unintuitive that one could exist.

Perhaps I am missing some background context, but pertaining to the well-ordering principle, how does the ability to make comparisons guarantee a smallest element? If there are an infinite amount of elements, we may not have a minimum. Additionally the reals are not a countable set, and I thought this was conclusive and provable via cantor diagonalization, there is no way to number them.

@benjaminpedersen9548

3 жыл бұрын

Cantor diagonalization proves that the set of reals is uncountable, i.e. not in bijection to the natural numbers. Look up ordinals to find a way to number past infinity. The video is misleading in its explanation of the well-ordering principle; it should be more clear about the fact, that the principle orders the reals in a very different way than the usual ordering; simply pick a number after the other ad so much infinitum that you cannot imagine and then some more. Again look up ordinals to get a sense of this.

Nice

One day, I'll wake up in the world where pure mathematicians realized the fallacies of transfinite induction, and all unbounded sets are necessarily the same cardinal infinitude, and Leibniz has a brand of coffee... I'm going back to sleep, now.

@dkzVirtual

3 жыл бұрын

You only need look where the answer are. We first need to critique with the historical hammer.

2:08 'Independence' is misspelled.

Perhaps

I don't understand what is the problem of reordering (a;b). Let's just redefine < operator for point (a+b)/2 and say that (a+b)/2 is less than any number in (a;b) except itself. Viola, we have a smallest element in open set. Problems?

@YAWTon

3 жыл бұрын

Problems? Yes. The order that you defined is not a well-order, because not every subset has a least element.

@luck3949

3 жыл бұрын

@@YAWTon Aah, Now I see where is the tricky part, thanks :)

So what, so not every vector space has a basis? Maybe like a zero space, or an infinite space doesn't have one? What's the closest thing to "every vector space has a basis" that you can say without the Axiom of Choice?

@Leidl.Michael

3 жыл бұрын

I think without the axiom of choice, you are forced to work with boring vector spaces in maths like R^n where you are able to imagine a basis. Just imagine the vector space of all bounded functions from fixed set M to R, so {f: M->R | there exists C>0 with f(M)

1:34 Joke's on you. I paused the video to understand the notation, and it made sense to me.

@MetaMaths

Жыл бұрын

Good to know we have mathematics speakers among us

0:30 - {(a, b), (c, d), ..., (e, f)} would be correct, or in only of sets expression { {a, {a, b}}, {c, {c, d}}, ..., {e, {e, f}} }

If you pick a side, that’s a….choice!

- The background piano music was annoying as all hell. But most information was also audio so you couldn't mute the racket... - However this was an introduction to a new field of study for me an may merit further exploration. - I am intrigued by the two equal balls constructef from one that seems to be a flaw in logicical reasoning if it is true.

We could use Axiom of determinism to solve all problems axioms of choice claim to be solving and not fall into paradoxes.

@MikeRosoftJH

2 жыл бұрын

Sorry, I am too busy splitting the continuum into more than continuum-many disjoint, non-empty subsets! Or, as I like to visualize this: imagine Cantor's Hotel - competitor of Hilbert's Hotel, where rooms are indexed by real numbers. The hotel has infinitely many floors, and infinitely many rooms on each floors: on each floor there are numbers which differ from each other by a rational number. In absence of axiom of choice there can be more floors than rooms! ("Say what! Of course there are no more floors than rooms." Okay, what does this mean? "It means that floors can be mapped one-to-one with a subset of rooms." And can you do that? "Sure, just pick a single room from each floor..." Oops. That's precisely what axiom of choice says. And, in fact, axiom of determinacy makes this impossible!)

I'm siding with 1-naturals!

My question is if someone learn set theory could they learn all the other branches like analysis,algebra, I read this is not true because analysis came before set tehory

@zapazap

Жыл бұрын

I suggest that all mathematics presupposes some pretheoretical notions -- a collection of things being one such notion. Ie sets, but in a naive rather than 'set-theoretical' sense. So sets are involved, but not set theory.

@semicolumnn

Ай бұрын

Set theory will probably not help nor hurt you in getting wherever you need to be. Algebraists, topologists, number theorists, what-have-you-ists, definitely don't deal with these foundational issues (They usually just pick whatever choice axiom that makes their job the easiest. For example if you're working in Linear Algebra you would accept AC since it gives that any vector space has a basis.) but it's not the worst thing to do with your time.

Saying that everything in mathematics is a set is like saying that everything in software is just code. Nah. There are computations. Programs are written, run and they perform something. Sets are code. It's about what you do with the sets. There are no mathematical things (objects). You need to imagine those static abstract eternal things only to have mental handles during the process. And the process isn't just your individual thinking. It's also communicating it, teaching it, testing, applying it and more.

#3:35 @3:35 I think you meant infinite there

I think that the axiom of choice is even too loose. It should be: "Given a set, it is always possible to choose ANY of ALL its elements". But is it possible to choose any e.g. real number? I have doubts, is it possible to choose a non-computable irrational number?

@normm5025

3 жыл бұрын

Yes. If the input to the choice function (implied to exist by AC) is the set of all non-computable irrational numbers, then the output (i.e. the choice) will be a member of that set, i.e. a non-computable irrational number.

@headlibrarian1996

3 жыл бұрын

Doesn’t non-computability imply transcendence?

@normm5025

3 жыл бұрын

@@headlibrarian1996 Yes. I was just quoting from the original post. "Irrational" is (very) redundant in the phrase "non-computable irrational number" ; )

@andsalomoni

3 жыл бұрын

@@normm5025 How do you build the set of all non-computable irrational numbers?

@normm5025

3 жыл бұрын

@@andsalomoni I don't know what you mean by "build". You can arrive at them by removing the set of computable numbers from the set of all reals. Would you say you could build the set of all computable numbers? If not, then you also can't build the set of all non-computible numbers.

Pretending I understood after 0:55

3:16 if two theorems are equivalent, how can one be weaker than the other?

I think the Axiom of Choice needs some nerfing so that Well ordering appears only where it makes sense. Well, nothing new to Set theory. ZFC did the same for Set membership rule in Naive Set theory.

The AoC is obviously true for finite collections of sets and perhaps also for countably infinite collections of sets. But it's not obviously true (to me) for uncountably infinite collections of sets and I believe this is where all the shenanigans come in. It can be dangerous to claim something we can't prove is obviously true. IMO the Banach-Tarski paradox proves the AoC is false in a world where you cannot magically double the volume of a sphere. So if we are basing things on what we experience in the "real physical world" then the AoC is obviously false. The most I can say is you are free to assume AoC is true and develop math that way and you are also free to assume AoC is false and develop a different math this other way. One of the main things we have learned from mathematics is we cannot trust our intuition for telling us what is True or not True. That is why we have math. Math is wonderful and beautiful because it's filled with non-obvious and non-intuitive results.

@freddyfozzyfilms2688

3 жыл бұрын

You can prove that a statement is unprovable (goldiel did this with his incompleteness theorum i think). I think that's what metamaths meant