How the Axiom of Choice Gives Sizeless Sets | Infinite Series
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Does every set - or collection of numbers - have a size: a length or a width? In other words, is it possible for a set to be sizeless? This in an updated version of our September 8th video. We found an error in our previous video and corrected it within this version.
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In this episode, we look at creating sizeless sets which we call size the Lebesgue measure - it formalizes the notion of length in one dimension, area in two dimensions and volume in three dimensions.
Written and Hosted by Kelsey Houston-Edwards
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Vsauce
âą The Banach-Tarski Paradox
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boy I miss "Infinite Series..." I wish PBS would bring it back
Why did they end the series?
@@thatchessguy7072 they said something about using their available funds most effectively.
This host made the show work. It fell off when she left.
@@thatchessguy7072 The host had other things to do (namely finishing PhD) and this show was much less popular in general (it was too abstract and hard to understand compared to the other shows, namely Spacetime to be sure) so it just didn't make the cut. There also are other really good maths chanels like Numberphile so competetion from the English didn't help either, I guess.
Me too, mate.
"Pretty much any set that you can think of has a size." (It's the sets that you can't think of that get you.)
Or the set of sets you can think of
@@leandro8897 Can you think of the set of all things you can't think of?
@@Pablo360able actually, we do it all the time. Whenever you think about the things that you don't know, your thinking about the complementary set of the things you know.
@@Pablo360able the set of things I have to do right now but am procrastinating doing
What about the set of all sets you can't think of? Is it countable? If it is uncountable is it a set that you cannot think of? If that's the case then the set of all unthinkable sets is not complete since it does not contain itself by definition.
âLet us know in the comments your favorite consequence of the Axiom of Choiceâ
The single greatest call to action in the history of CZcams đ
Like Optimus Prime's famous dialogue: Autobots Rollout
My favorite is how proofs now avoid using it.
@@mikemondano3624 AC has been embedded enough into modern mathematics that I don't think people have such big issues with it. There is a laundry list of nice statements that require it, such as Tychonoff's theorem (equivalent); that every vector space has a basis (equivalent); that every ring has a maximal ideal; that every field has an algebraic closure; that the algebraic closure of *Q* is unique; cardinal trichotomy (equivalent); that every infinite set has cardinality greater than the natural numbers (i.e. Dedekind infinite); that *R* / *Q* is the same size as *R* ; and so on. The Axiom of Dependent Choice (DC), a weaker axiom that is still independent of ZF, can be written as the statement that every tree of height omega (essentially countably infinite height), that has no terminal nodes, has an infinite branch. DC makes for a nicer proof of Hilbert's basis theorem. The list could go on. These facts are very commonly used or at least accepted in day to day work by algebraists, topologists, and the like.
To be contrary, my favorite consequence of the axiom of choice is that it allows us to compare the sizes of sets. Given two sets α and ÎČ, either |α| †|ÎČ| or |ÎČ| †|α|. This is the notion of đ€đąđłđ„đȘđŻđąđ đ€đ°đźđ±đąđłđąđŁđȘđđȘđ”đș, and it's equivalent to the axiom of choice.
How so?
@@quantumgaming9180Axiom of choice is equivalent to the proposition that every set can be well-ordered. Now assuming axiom of choice, cardinality of any two sets is comparable (because any well-ordered set can be mapped one-to-one with an ordinal number, and by well-ordering of ordinal numbers the cardinality of ordinal numbers is comparable). So suppose that some set can't be well-ordered. Consider the set of all order types of well-orderings of a subset of our set. (Equivalently, the set of all ordinal numbers which can be injected into the set.) By axiom of powerset and separation, this is indeed a set; and it can't include all ordinal numbers because the class of ordinal numbers is a proper class rather than a set. It follows (by well-ordering of ordinal numbers) that there exists the smallest ordinal number which can't be injected into the set, i.e. which has strictly greater cardinality than any its well-ordered subset. On the other hand, by assumption our set can't be well-ordered, and so it can't be injected into any ordinal number (because that would define a well-ordering).
It follows: if every set can be well-ordered, then cardinality of any two sets is comparable. And if some set can't be well-ordered, then there exists a well-ordered set whose cardinality is incomparable with it.
@@quantumgaming9180 Hartogs' lemma (in set theory) is a theorem of ZF that says for every set A there is a well-ordered set B such that |B| is not less than or equal to |A|. The way we do this is to define B to be the set of all ways to well-order subsets of A. It turns out well-orderings can themselves be well-ordered; when you look at two orderings R and S, we say R
â@@quantumgaming9180 I was about to comment "probably due to the well ordering theorem" but it seems like "the set of all cardinals" actually isn't a real set, rather it's a proper class...
@@minimo3631 That's basically it; if all sets can be well-ordered, then cardinality of any two well-ordered sets is comparable. And if there exists a set which can't be well-ordered, then there exists a well-ordered set whose cardinality is incomparable to it (by the construction of a 'Hartogs number' of the set).
Wanna see an anagram of Banach Tarski?
Banach Tarski Banach Tarski
Can't believe it took me 30 seconds to get that :-|
Anthony Khodanian I don't get it.
ophello the Banarch Tarski paradox is that you can take one item, take out a bunch of pieces, then put them back together to create multiple of the original. It's like if you could cut a pizza into 8ths, make two piles of 4 pieces each, then have two pizzas exactly the same size as the original.
An anagram is a word or phrase with the same letters as some other word or phrase.
Does this help? Have I sufficiently ruined the joke by explaining it?
Yup. For some dumb reason I thought it was supposed to be a palindrome. I am an idiot.
Daniel Kunigan I thought that you weren't able to actually measure anything of both results
"The Axiom of Choice is obviously true, the well-ordering principle obviously false, and who can tell about Zorn's lemma?"
đđđ
âI like it, it makes algebra more interestingâ
(Iâm pretty sure this isnât the exact quote and donât remember who said this but I know I read it a while ago)
I read that somewhere but can't remember where
My favorite consequence of the Axiom of Choice is mathematicians flipping their shit
But the set in the video is proved to be non-measurable. It's not a case of "we don't know". It means that the definition of size called the lebesgue measure just cannot be applied to that set. Other definitions of size may apply. For instance, the set still has a cardinality, which is often used to measure the size of a set in a different sense.
Not necessarily. You don't seem to know much mathematics.
Mathematics concludes nothing of the sort. "Time" is not a mathematical concept, it's a physical concept. You seem to be confusing mathematics with physics.
gespilk I suggest you learn a bit more about mathematics before saying stuff like that.
If you say stuff like "it's just a sophisticated way to say we don't know" then evidently you didn't understand the mathematics underlying it.
Perhaps what your commenting on is the free usage of words like "size" and "time" which don't have the same meaning in mathematics as in real life. What's important to understand is that when the video here says "size", what they mean is "measurable". When we say "time" in mathematics we don't mean "time" in real life. Some physicist (maybe Feynman?) once said: "All models are wrong, but some are useful".
This is true in a sense in mathematics as well, but a bit different: Many models are true, they just don't model what you think. Obviously when you measure the size of a rock in real life then it has a definite size, but this video isn't talking about rocks. What this video says is that assuming the axiom of choice, any function (call it size if you want) that has a set of real numbers as an input and a single real number as an output (plus countable additivity and some other stuff) can't be defined for all sets of real numbers.
tl;dr: Size in math doesn't mean size in real life. Rocks aren't sets of numbers.
I miss this channel so much.
On the whole, this is quite good. However, I don't believe any credit is ascribed to the mathematician that discovered this set, G. Vitali. This measureless set constructed by the application of the Axiom of Choice to the equivalent classes is known as the Vitali Construction or Vitali Set.
Yes, Guiseppe Vitali should have been given credit, and the video is quite good.
What's also good is putting the Axiom of Choice in the TITLE of the video, to show how statements connect. I've seen other videos here that show the Banach Tarski paradox and don't give the Axiom of Choice a prominent place or even don't seem to mention it at all.
O
While it seems reasonable to discard the Axiom of Choice, the consequences of Axiom of No Choice are similarly weird: without choice, there exist vector spaces without a basis; there are two different non-equivalent ways of defining continuous functions; and the cardinality of two sets may not be comparable.
Also, without axiom of choice it is consistent that a set can be split into a collection of disjoint non-empty subsets in such a way that the collection has more elements than the original set!
Those problems seem less bad to me than the problems AoC creates.
3:35 it took me 5 minutes to realize you must mean "each" rather than "all". "all numbers show up in 1 bin" can easily be interpreted as "all numbers show up in the same bin".
Hedning1390 Oh, I was totally confused and just continued, assuming she misspoke. Thanks for the clarification.
Same here
Man, do I miss this series. By its very own name, it was a series which should never have ended. :'(
Well, anyway, I am sharing this video with one of my students because it is one of the best explanations that I have ever seen.
Nothing is infinite.
@@mikemondano3624: Not with that attitude.
@@curtiswfranks Not an attitude. A fact. You can look up "fact" in a dictionary.
@@mikemondano3624: * eye-roll *
â@@curtiswfranksThere will always be people like that.
Just subscribed and got a new math video in the notification bar.Thanks for making such an amazing video
Some of the best video content ever uploaded to the internet, thank you for creating this. Thank you.
I'm really interested in the Curry-Howard transform/correspondence/equivalence ... and would love to see some discussion about it. I know your probably not taking requests but that's my 2cents. Thanks.
My favorite consequence of the axiom choice is that you chose to make these videos for us. thanks.
Keep the deep learning coming! Thank you for the hard work!
Size doesn't matter
Wondered when this one was going to turn up!
:-)
Math joke: "What's an anagram of Banach-Tarski? Banach-Tarski Banach-Tarski."
Thanks, so much, I was expecting this for such a long time from anyone willing, honestly thanks!
Amazing video! Thank you and so sorry to see this channel is no longer updated :(
This video got me thinking about measures and infinite sets and I think it would be interesting if Infinite Series did a piece on an extended number system such as the surreals or hyperreals that include infinitesimals and what that implies for those systems in terms of if or whether they can support something similar to a Lebesgue measure. (Heck, the surreals are kind of cool in general.)
Beautiful math presented beautifully. New post on an old video, I hope this channel gets reborn.
me too
My favorite consequence of AC, or rather equivalence, is Zornâs lemma. I want my rings to have maximal ideals.
I would mention also Hamel bases and Hahn-Banach
As a person working on a scholarly paper, sometimes I feel like I know something. Then I see this, and just have to go back to bed. I should just publish before watching any more science videos.
Until you got to the word "measure" I thought you meant cardinality, and was confused and intrigued.
That's an opposite result. From axiom of choice it follows that every set can be well-ordered. But assuming its negation, there exist sets which can't be well-ordered, and therefore their cardinality is not equal to any cardinal number (as long as we define cardinal numbers as initial ordinals and therefore only consider well-ordered cardinal numbers). Another statement equivalent to axiom of choice is that the cardinality of every two sets is comparable: given sets A and B, either A can be mapped one-to-one with a subset of B, or B can be mapped one-to-one with a subset of A. (If both is true, then the sets have the same cardinality. I am using a non-strict definition of a subset; a set is a subset of itself.)
For example, it is consistent with set theory without axiom of choice that real numbers can't be well-ordered.
from another source: I found the answer in the paper "Measure and cardinality" by Briggs and Schaffter. In short: not if I interpret positive measure to mean positive outer measure. A proof is given that every measurable subset with cardinality less than that of â has Lebesgue measure zero. However, they then survey results of Solovay that show that there are models of ZFC in which CH fails and every subset of cardinality less than that of â is measurable, and that there are models of ZFC in which CH fails and there are subsets of cardinality â”1 that are nonmeasurable. So it is undecidable in ZFC.
If it was intended that our sets are assumed to be measurable, then the answer would be yes by the first part above.
In other words, cardinality can be a measure but it's not meaningful to discuss of it like that and thus we render them unrelated.
Same. When thinking of set size, cardinality comes to mind first.
Same, I was disappointed to find out that wasn't the case.
Yeah thank u for that. I am also confused about her definition of size of the set
My favourite consequence of that axiom is that if we accept it, we can prove that each set admits a well-ordering relation. This means that, each subset of R has a minimum (of course, the relation might not necessarily be the usual less-than-or-equal-to, and in fact, with R it is not!)
Yeah, with it, all sets, regardless of how they're constructed have a well order.
However, as it involves the axiom of choise, we just simply can't know what that order is
Kinda hard to work with, as it gives very little new information
Solid reason to reject AoC.
Mine, too! An oldie but a goodie. If I recall correctly, first proved by Zermelo, who did not realize that he had implicitly assumed a new axiom.
What a great video, wonderfully explained and illustrated. If this was TV....
This just came up in my recommendations. For one sparkling moment I thought you were back. Ahhhhhhh!
Guys!, please read the description, they found anerror in the previous video and fix it in this one.
What really alleviates my sense of anxiety over the paradox is your soothing voice and your calm and relaxing way to explain mathematics. Thank you Kelsey, as long as you guide, I have no fear of diving into the math universe.
What a well produced and great series. Such a shame that they stopped.
Oh my god, a series on math! I'm in heaven :D
The Axiom of Choice is one of those wonderful mathematical statements that seems very intuitive, but leads to very counter-intuitive constructions; weird things happen when infinity comes out to play.
(hopefully) answers to the challenge questions:
1) Assume there exists an S_i and S_j that are not disjoint. Then there is a number of the form q_a+r_k, and also of the form q_b+r_m, with q_a and q_b in S. Since this means q_a+r_k=q_b+r_m, rearranging this equality yields q_a-q_b=r_k-r_m. Since the right hand side is rational, q_a and q_b are from the same partition of [0,1], so only one of them can be in S, which arrives at the contradiction.
2) Consider x in [0,1]. x is of the form s+r_i, with r_i rational between -1 and 1, and s in S, by the construction of S. Then x is in U(S_k)
For anyone looking into this, this particular set is called the Vitali set.
Thanks.
Thanks for this video!!! I was really struggling understanding the Vitali set
I picked a bad video to watch late at night...
Have to revisit this..fascinating...
[Q] What's an anagram of "Banach-Tarski"?
[A] Banach-Tarski Banach-Tarski
Favorite consequence of Axiom of Choice ...Tychonoff Theorem being equivalent to it.
Reminder to PBS that there are folks who still miss Infinite Series. _Bring it back._ đ
Thank you for this amazing video!
My favourite (well, probably still after Banach-Tarski paradox) consequence is Zermelo's theorem about well-ordering of any set - I believe it's the least trivial to prove equivalent definition for axiom of choice
To well order a set S, choose one element \phi(U) from every nonempty subset U of S. Then define a sequence by induction--- the first element phi(U) (where the subset is the whole space). Then the second element is phi( U / {phi(U)}). You continue, the next element is always the choice from the set minus the sequence constructed so far.
This sequence can be continued by transfinite induction to all ordinals. So either you get an embedding of all the ordinals into S, or S is well ordered by this operation. The ordinals are a proper class, so it must be that S is well ordered. That's the complete proof of Zermelo's theorem.
That's interesting, thank you, but we need it the other way round as well - any set is well ordered implies choice.
Annnnnd writing this it just occured to me that if any set is well ordered then we can just pick its smallest member. Does that work? Do we need to pick a well ordering? Which may require Choice in the first place and so be circular? Help!!!
@@steviebudden3397 your intuition about the circularity is correct - that's why, however you want to present it, the axiom of choice has to be an axiom and not just a theorem (the only things you could derive it from as a theorem amount to equivalent axioms anyway, like "any set is well-orderable" or "the cartesian product of infinite non-empty sets is non-empty")
and yes, it works because the ordering gives you a specific way of picking elements (might as well be the order-least, but it could be whatever since the ordering allows you to specify) - without assuming choice or something that implies it, we can loosely say "pick an element at random," but that does not have a rigorous meaning for an infinite set without any precisely known/defined structure
you don't need to pick a well ordering and in fact even with the choice axiom it's entirely unclear how a well-ordering of most sets would work (again, why it needs to be an axiom) - try imagining for yourself how you'd make a well-ordering on an open interval of the reals, for example (by definition has no least or greatest element in the usual sense, but with choice / well-ordering principle you assume there is some way)
stuff like this is why there were some opponents of the axiom a hundred years ago, but denying choice has even weirder and less intuitive (and much less mathematically useful) consequences - without choice there are cardinalities which are impossible to compare and decide which one is bigger than the other or if they are the same size at all, for example - so choice won the day
For the first part of the challenge problem, suppose there is an element, say x, in the intersection of S_i and S_j for some positive integers i and j. Then x = s_n + r_i and x = s_m + r_j for some positive integers n and m. But then s_n + r_i = s_m + r_j, which implies s_n - s_m = r_i - r_j, which is rational. Now assume for contradiction that i and j are not equal. Then n and m cannot be equal either, for if they were, then s_n + r_i = s_m + r_j would yield r_i = r_j, which would contradict that i and j are not equal. Since n and m are not equal, s_n and s_m must belong to different equivalence classes. But we know that s_n - s_m is rational, and so they belong to the same equivalence class. This is a contradiction. Hence i must equal j. Therefore, if the intersection of S_i and S_j is nonempty, then i=j. Contrapositively, if i does not equal j, then the intersection of S_i and S_j is empty, and thus S_i and S_j are disjoint.
For the second part of the challenge problem, let x belong to the interval [0,1]. Then x belongs to the equivalence class, which we will call E_x, of all elements of [0,1] which differ from x by a rational number. Since S contains one representative from each equivalence class, S contains some element s_i from E_x. Then s_i differs from x by a rational number, say r_j. Since s_i and x both belong to [0,1], r_j belongs to [-1,1]. Therefore, x belongs to S_j by definition. x was chosen arbitrarily, and hence for every number in [0,1], there is some positive integer j such that x belongs to S_j.
Minor point expanding on your answer to the second part: r_j, with value equal to x - s_i, must be between -1 and 1 as both x and s_i are in (0,1) so can't be more than 1 apart. Hence S_j exists. This is the key point that was wrong in the previous version of the video where a negative r_j had no corresponding S_j...
Ah. That's a major point rather than a minor point. I believe Folland defines things differently in his textbook so as to deal with this issue. Thank you for pointing that out.
EDIT: No wait, that is a minor point. It's completely fine for r_j to be between -1 and 1. But it's good to point that out.
Hey For Your Math,
Congratulations, you are the winner for this week's challenge. Please send your address with shirt size @ pbsinfiniteseries@gmail.com, so we can ship a Infinite series shirt to you!
This video help me a lot ! Thank you
In the midst of my first year graduate study in Mathematics, one of the professors stated that the purpose was to educate our intuitions. He then produced the Cantor ternary set. Yup. My intuition felt educated...
Calling the Lebesgue measure âsizeâ is a little misleading because thereâs a bunch of different ways to define size of a set. The three that come to mind most readily are cardinality, measure, and ultrafilters, but Iâm sure there are others. This was an awesome video though.
Yes. It clickbaited me even thought my field is analysis. Iâve known about the Banach Tarski Paradox for about 38 years.
This is wonderful.
Actually, I am a mathematician and I had to figure out all these things by myself before I got my degree. And I envy those that will after the release of this video đ
I mean, you probably didn't have to figure this out quite by yourself, you used the knowledge of previous mathematicians and your professors :) At least for me I saw this construction in a measure theory course in engineering school.
my favorite consequence is the construction of such real function which maps each non-trivial interval onto the real line.
Obviously this video was in response to my question regarding how to characterize the topology of countably infinite sets :P So thank you very much this was very helpful!
obviously
Please, finish your thesis finally and come back. Please!
Im sorry to say but its over, its done. Im sad myself. This channel was so amazing, putting out math content that's actually interesting and advanced and not just "how calculus works for 5 yearolds". The only channel that used to be on par with 3blue1brown :(.
@@berserker8884 czcams.com/video/3cI7sFr707s/video.html
@@berserker8884 The only channel? Mathologer is excellent, and Epic Math Time can be interesting as well.
Hum... there's a problem with the link to Vsauce's video: it's not long enough...
Here it is in full length: czcams.com/video/s86-Z-CbaHA/video.html
Thanks for flagging this Amaras. We just corrected it in the description.
You're welcome. Always glad to help! :)
I tend to give links in their short form: czcams.com/video/s86-Z-CbaHA/video.html czcams.com/video/ZUHWNfqzPJ8/video.html (Mathologer) is also worth watching on this topic.
I really miss this series!
Outstanding video lecture.
Is the problem with the axiom of choice that you can't do it operationally ?
Rather like summing an infinite series, where you have to specify a procedure ( e.g. the limit of partial sums ) before you can say what an infinite sum actually means ?
That depends. Because if you assume the axiom of constructibility (which is a stronger axiom than axiom of choice, because it implies axiom of choice, generalized continuum hypothesis, and many other statements), then it is possible to write an explicit formula which can be used to well-order the class of all sets (i.e. so that every set of sets has a minimum), and consequently to well-order the reals. This gives a method to select an element from any set, and so to select one element from each of infinitely many sets. (Recall that in the set theory everything is a set; for example, natural numbers can be defined like this: 0 is an empty set, 1 is the set {0}, 2 is the set {0,1}, and so on - every number is a set of numbers smaller than itself.) [Without the axiom of constructibility, the formula is still well-defined, but it can't be proven that it well-orders the class of all sets.]
Shouldnât sqrt(2)/2 be bigger than sqrt(2)/3?
I also noticed that. And yes, it is 3/2 times bigger, if anyone is wondering see 4:00 to 4:10
Don't know who edits these videos, the content is great but they sure do a shitty job.
I imagine the animators aren't mathematicians
Shhhh. They'll remake the video again.
>_>
Again, I miss this show.
My favourite consequence of the axiom of choice is that every ring has a maximal ideal.
need to make sure that the ring is unital, but ya (no ugly rngs).
So, since there are uncountably infinite bins that S picks numbers from, S contains an uncountably infinite number of points. Thus, when we add up all the sizes of the points, we are adding up an uncountably infinite number of zeros, which presumably is indeterminate and why we consider it sizeless.
Honestly, to me the surprise isnât that we consider S to be sizeless, but that we consider [0,1] to have a size. Like, whatâs preventing me from taking all the points in [0,1] and stretching them all so that theyâre twice as far apart and now span [0,2]? Itâs not like stretching them creates any additional points, nor does it create holes since the real number line is infinitely dense.
We can only work out the size of a set by adding together the sizes of the sets in a countable partition.
We consider [0,1] to have a size because Lebesgue showed us how to construct a "measure" on the real numbers which assigns to each "measurable" set a size in a way that respects properties that we would typically expect of such a measure (such as translation invariance and additivity for countably many disjoint unions), and because the set [0,1] is "measurabe" in the Lebesgue sense.
The set S is not "measurable" in the Lebesgue sense, nor in any other sense which would allow us to construct a measure with the properties that we would expect of a measure, for the reasons given in the video (i.e. you can form countably many disjoint sets by translation which combine into a set that should have a "size" between 1 and 3).
I saw this video a long time ago and understood very little. I came back today and after learning a hell of a lot and I see where the issue lies. The issues lies in the level of infinity. S1 contains rational numbers and thus has 2 levels of infinity, one for the numerator and one for the denominator and thus this is considered a secondary infinity. But when we get to S2, we are adding transcendental number square root of 2 which is a number that cannot be represented because it would require infinitely many digits to do so, thus this is a primary infinity. Thus we have an unending number that did not belong to the rationals that we can now use as a starting point for this new S2 and so on for S3 etc. If anyone is interested, I will take some time to go on. Of not I will leave it at this.
Please do go on. It would be great to have an eli5 (Explain Like I'm 5 years old) breakdown of how they built this up
I posted the answer to your challenge question on your previous video that you took down.
Even though the question was wrong in that video, I both formulated and proved the correct version of the question as it was stated in this video.
I also proved that the question was wrong.
We noted. And were impressed by your answer, as always!
Thanks!
superb presentation..
Does anyone remember what the error was?
I have the original video downloaded if its killing you =)
They claimed every number from 0 to 1 was contained within the union of the Si sets but it wasn't true, it was just a minor error though and changing the argument a little bit fixes everything. They fixed it here by changing the range of rational numbers indexing the Si sets from 0
Thanks morkovija and Hwd405 for watching both videos and explaining the problem we had with the original. We don't take down videos lightly but thought it was important to do so in this case.
PBS Infinite Series maybe you should have written an explanation of the error in the description? Would have been nice ;)
Pete Magnuson Lesbegue instead of Lebesgue
My favorite consequence of the axiom of choice is that I can and will have a nap now, and no one is going to stop me. Yay!
Awesome introduction to measure theory
The BT paradox arises from assuming every set is measurable. The axiom of choice in a way is the saviour here. It helps show that said assumption must be false.
My favourite consequence is that every nonzero vector space has a basis (equivalent to choice, actually)
Or, we might propose (in absence of axiom of choice) that all sets of reals have measure, and there's no Vitali set. A side effect of this proposition is that continuum can be split into more than continuum-many disjoint, non-empty sets.
@@MikeRosoftJH uh, I have seen that argument I think, can't recall the details.
4:00 - 4:10 The visualization is wrong: sqrt(2)/2 > sqrt(2)/3
Can't you people just shut up and enjoy the sexy girl saying mathy things?
@@trollop_7 no
@@noxaeterna8761
Your'e right. I may have overstated her physical attractiveness.
0th
Uncountableth
1th
Wow, great explanation. What do you use for editing the visuals, I mean those mathematical visuals?
Best explanationđ
As I asked the last time you uploaded this: What about infinitesimals?
AFastidiousCuber You mean in the construction of S?
Infinitesimals don't exist in the real numbers, thus they are not taken into account. You should study an extension of reals called superreal numbers if you're interested in them.
Without any specifics of what you're actually asking I can't really say anything else.
I understand that infinitesimals use different axioms, but, yes, I'm curious how those axioms would effect the construction and definition of S. Naively, it seems like that might make it more intuitive.
They would change everything. S is a subset of the reals with no Lebesgue measure. Lebesgue measure is a real valued function whise domain consists of sets of real numbers. If you introduced infinitesimals you would no longer be working with the reals or with Lebesgue measure.
It is somewhat problematic to use the real numbers - a fixed concept whose construction is heavily intertwined with set theory anyway - and expect something sensible when applying this to all sets. (ok this is only meant for Euclidean space, but the video failed to mention that - perhaps deliberately) I somewhat doubt one really needs the axiom of choice for creating iffy scenarios. You can goedelize set theory and then use that to create self-referential sets of points whose construction depends on the Lebesgue measure they'll end up with.
You also hit similar limits when looking at metrizability. So, it makes perfect sense to request a refinement of the reals when "measuring" complicated/large sets. Infinitesimals as in nonstandard analysis are such a refinement - but not the be all and end all, e.g. instead of using infinite sequences indexed by integers one can use larger ordinals as index sets, etc. when sets get really large.
I have read only the abstract of the paper at www.ams.org/journals/proc/1975-053-02/S0002-9939-1975-0382578-5/, but it says the author has found a measure defined on all subsets of R. However, the measure is not translation invariant. It is "nearly" translation invariant on bounded sets and "nearly" agrees with Lebesgue measure.
_Challenge problems:_
*1.)* Assume there are different i and j such that S_i and S_j=S_i+r are not disjoint. Then they share some element x. Because x is in S_j, x-r must (by definition of S_j) be in S_i. But x and x-r differ by a rational number (r), so x-r cannot be in S_i (because x is already in S_i). This is a contradiction, hence S_i and S_j are disjoint for all different i and j.
*2.)* Let x be an arbitrary element of (0,1) and c its representitive (according to the bins we created in the video). Then x=c+(x-c) is an element of the set S+(x-c), because c is by definition an element of S. We know further that x-c is rational (definition of the bins), so there is an index i such that x is an element of S_i.
exceptional video
Ohhhhhhhhhhhh, this is so interesting!
Wasnt this uploaded a few days ago lol
Yes it was TWPO. There was an error with the original version that we thought was necessary to fix so we corrected that video and reuploaded.
yeah ... I thought I'm having a de ja vu..
Now to test if i remember enough to tell what's been changed.... I'm guessing i will fail.
Kind of pointless (and, honestly, disingenuous) if you don't tell us what the error(s) was.
And speaking of fixing errors, when in the name of all that is holy will you ever fix the SUBCRIBE at the end???
My favorite consequence of the axiom of choice is that some papers suggested that the Banach-Tarski Paradox might be linked to the way particles can collide and form more particles than we began with. If that's the key to perpetual energy that would be super great!
You can get more particles but NOT more energy! So no new energy is produced in the collisions. YOu just re-partition the energy, but its total is unchanged. Sorry - still no free lunch!
Wonderful video! I have never seen non-measurable sets explanined so well, so kudos!
A little correction: I'd say that it is misleading to write S = {s_1, s_2, ... }, since it is not countable (9.24).
That's definitely a good point. It seemed like the most straightforward way to write it, without introducing any new notation, but I agree that's it's misleading. Thanks for pointing that out
@@pbsinfiniteseries Hi, PBS, doubt that you'll respond 3 years later but how is S uncountably infinite? If each of s_1, s_2, ... vary over the rationals (between -1 and 1), 4:59, which is countably infinite, then surely S is countably infinite. Each set s_i is uncountably infinite but I would have thought that S is countably infinite due to the way it has been constructed.
Thanks
@@vivanvasudeva3888 the s_i mentioned here is the small s i.e. the elements of S. while u are talking about the capital s S_i .you r right that each S_i corresponds to a rational so there number is countably infinite. But S itself has uncountably many elements which was the point noted above
@@vivanvasudeva3888 Split the real numbers (or an interval of reals) into equivalence classes using the relation: x~y, if x-y is a rational number. Every equivalence class is a countably infinite set (it's a shifted copy of rational numbers); it follows that there are uncountably many equivalence classes. The non-measurable Vitali set is a set which contains a single element from each equivalence class (assuming axiom of choice, such a set exists).
Well my favourite consequence of the Banach-Tarksi paradox is the Vsauce video talking about it
with paradoxes like these, why dont we drop the axiom of choice???
afterall, i thought mathematicians did not like contradictions
It's not a contradiction, it's a paradox, i.e. the result describes something beyond our intuition, which could still be true for some real objects out there (think about the properties of space-time or sub-atomic particles).
It is only a contradiction if you _insist_ that the Lebesgue measure of every possible set must exist. If you do that, then you reach the problem outlined in the video.
So, you have to throw out at least one of your assumptions to fix the contradiction. You could drop the Axiom of Choice or you could drop the assumption that every set must have a Lebesgue measure.
That choice of what to drop, however, is not as easy as you might think and the Axiom of Choice is controversial for that very reason. If you drop the Axiom of Choice, then every set in n-dimensional Euclidean space (so, R, R^2, R^3 and so on) is measurable.
However, if you drop the assumption that every set must be measurable, then you have to limit the Lebesgue measure to some subsets rather than all of them. That collection of sets is closed under complement, countable union and countable intersection and includes everything that can be constructed from intervals in countably many steps, so you don't actually lose that much.
Theophagous: It is a contradiction. One fact is that the size of set Union(Si) is between 1 and 3 and it contradicts the other fact saying that the size of each Si is the same as the size of S. The solution is that the question is wrong, we cannot ask the what the size of S is.
Well, sort of. It's a proof by contradiction which hinges on the assumption that S has a size to begin with. If you don't make that assumption, then there is no contradiction. What this proof shows is that the assumption that every set has a size is false.
But isnt that the definition of contradiction? That one of you assumptions is false. *edit* Definition was a bad choice of word, is lemma better? one of the lemmas of contradiction?
Dividing the interval [0,1] into equivalence classes the way it was done is explicitly assuming that irrationals are countable. If I am not mistaken, you seem to have started from a false hipothesis.
There are uncountably many equivalence classes, and you form S by making uncountably many choices. But that Axiom of Choice is okay with all that. But, I agree, that the "bins" metaphor was deceptive and made it seem like there were countably many.
But if there are uncountably many Si sets, how can you apply Countable Additivity to calculate the size of their union? See 1:09 and 6:26
Indded, you can use the axiom of choice to generate these bins. But the metaphor is bad as you say, and you are caught within it when you enumerate the bins in the video (first bin with 1/4, second bin with sqrt(2)/2, ...), but that does not compromise the argument we are discussing now.
I agree with observations 1 and 2, and since S_i are countable, although using a countable index to enumerate them is a mistake, but we already set that straight in the first round :)
Now take a look at what you said in 6:30, something like this: "Let's take the union of all the S_i. Because they are disjoint, by the property of _countable additivity_, the size of the union is equal to the sum of the sizes..."
I think that there is no way you can add an uncountable number of terms. In fact, measure theory is a way to work around that, using countable aditivity in a smart way (sigma-algebras).
The way I see it, the argument that the size of each S_i is zero is correct, since each possess a countable infinity of members. But you cannot invoke countable aditivity to argue that the size of the union is zero.
What do you think? Maybe I got something wrong.
We need to be careful. Each of the equivalence classes ("bins") is countable. Therefore, there are uncountably many equivalence classes. (A union of a countable set of countable sets is itself countable.) Now pick one representative of each equivalence class, and put them all in set S. (As mentioned, to prove that this set S exists requires the axiom of choice.) The set S is uncountable. Now shift the set S by all rational numbers between -1 and 1. We get countably many copies of the set S. The union of these copies must contain all real numbers between 0 and 1, but must itself fit in the interval from -1 to 2. Now we can use the property of countable additivity of the Lebesgue measure, and reach the paradox: the set S can't have a zero measure, and can't have a non-zero measure. (Therefore, the set S is not measurable.)
I agree. I failed to understand that the S_i were enumerated by the rationals. The argument is correct, my mistake.
I love this!
Wow, this is gold. I have a masterâs degree in math and never heard of that S thing, or more probably I did but forgot.
It just dawned on me that the Axiom of Choice can be "proven". If you use the AC then you choose to use it, if you don't then you also made a choice. This is a contradiction.
If anyone takes this seriously, I also sell bridges that you can put a toll booth on.
This is the Axiom of Choice of the Axiom of Choice
is that the Brooklen bridge ?
To *choose* is a slippery verb. A person can consciously choose. For example, chess players choose their moves. Can a machine consciously choose? Did Alpha Zero (artificial intelligence) choose its moves when defeating the best computerized chess engine (not losing in 100 games)? What is a mathematical *choice*?
Axiom of choice is a very specific mathematical statement: "Given any set of disjoint non-empty sets, there exists a set containing a single element of each of these sets." (An equivalent formulation is: given any set S of non-empty sets, there exists a function f which maps sets in S to their elements; that is: the domain of the function is S, and for every X being an element of S is f(X) an element of X. [The function must exist as a set - namely, a set of ordered pairs.])
Axiom of choice can indeed be "proven". For example, it can be proven in ZFC (rather trivially, because it one of the axioms of the theory). But there is a known result: in ZF (set theory without axiom of choice), neither axiom of choice nor its negation can be proven, unless set theory itself is inconsistent. (In other words, either axiom of choice or its negation can be added to the theory, and neither will cause a contradiction.)
Sure it is the Brooklyn Bride, and here's the proof: It is evident that the bridge on offer currently has no tollbooths. The Brooklyn Bridge, by observation, currently has no tollbooths. Therefore, the bridge on offer is the Brooklyn Bridge.
rad(2)/2>rad(2)/3, but this video indicates otherwise...
Where do they say this, I don't see it?
Daniel Grass Never mind I found it. 4:05
I wish this series back đą
nice video đđ» could you tell what tool do you use to make such videos ?
11:55 You can't literally see 3 dimensions either. Your retinas only see a 2 dimensional projection and your brain constructs a 3D image from it. But the 3D image only exists in your imagination. So if your imagination is rich enough, you might also see a 4D image in your imagination.
The cells in your retina "see" a sampling of photons across a large number of small but finite regions approximately arranged in a 2D surface. That's not the same as looking at a 2D projection...
rmsgrey you can see the direction that a photon came from, but you can't see the distance it traveled before it hit your retina. What you perceive with your eye is inherently 2D.
Can your retinal cells see the direction a photon came from? We can infer that they probably came through the pupil, but it's not certain.
And correlations between photons arriving at nearby retinal cells do provide distance information through blur-detection.
Photons also have phase and polarisation in addition to frequency - which is why holograms work - the eye responds differently to different polarisations; I've no idea whether the phase information gets picked up at all.
> blur-detection
This happens entirely in your brain, which was exactly my point.
I have no idea if your eye can see polarization or phase (I would guess not) but it has nothing to do with whether you see 3D objects as 3D or as 2D. What your retina detects is literally a pinhole (well, pupil-hole) projection.
I expect that if you keep analyzing this you will come to the conclusion that we really do not literally "see" anything. The eyes seem to collect data and the brain seems to arrange it into something coherent. The same is the case for many people with certain kinds of blindness who dream, of course, with the caveat that the brain of such a person may be making use of data collected from other sensors than the eyes, which also seems to occur when someone hallucinates, and of course if you and I were the only two people in the universe and each of us told the other he was hallucinating, then we would both be partly right and both be partly wrong, or, at least, there would be no way to determine for sure which of us is hallucinating and which is not, or else, which is telling the truth and which is not.
After learning about the -1/12 thing, I no longer believe that add a small number infinite many times results in infinity. Why not result in 2 :P
Similar things can be done. You'd expect 1^inf to equal 1, but a limit that approaches 1^inf can be evaluated to any number (0, inf), for example, the original definition of e.
Oh God, and I'm still trying to warp my mind around Riemann series theorem. I belong to the stone age.
Oh boi do I miss this series....
The assumption that the set has to have size bigger than one, is unjustified. Even if it contains every point between 0 and 1 those points have been defined to have size 0. You cannot conflate the ideas of points and sizes, then make a conclusion and expect it to be taken seriously
What set are you talking about? The set S has no size, so she did not assume it has size bigger than 1. The union of the sets Si contains the interval (0, 1) which has size 1. If S has a size. the size of the union of Si's must be greater than or equal to 1. That is not an assumption. It is a consequence of the definition of Lebesgue measure. A set cannot have a measure smaller than one of its subsets. In any event, the fact that the union of the Si's must have measure greater than 1 if S has any measure at all is completely irrelevant to the proof, which relies on the fact that the measure of the union, if S has a measure, must be less than or equal to 3, because the union is a subset of the interval (-1, 2), which has measure 3.
The size of an interval would be the sum of the sizes of its points only if an interval contained only countably many points.
In fact, you can show that, if the union of the S_i has a size, that size must be 2 - if you take a copy of the set and add 2 to every value, then every point in (0,1) is in the original set, every point in (2,3) is in the copy set, and every point in (1,2) is in precisely one of the original or the copy, so the union of the two is contained in (-1,4) and contains (0,3) so has measure between 3 and 5. Repeat the process to get n copies of the set (including the original), shifted by 0, 2, 4, ..., 2(n-1) respectively, and their union will be contained in (-1, 2n) and contain (0, 2n-1) so the size of that union will be between 2n-1 and 2n+1
As that union is made of n sets with the same size, that size must be between 2-1/n and 2+1/n. Since n can be any positive integer, the size of the original set can't be anything other than 2.
+msgrey There are uncountably many points in (1, 2), one for each number in S, that are not in U(Si) or U(Si) + 2.. Specifically, if s is in S, s +1 is in (1, 2), but not in U(Si) or U(Si) + 2.
Proof: Suppose s + 1 is in U(Si). Then s + 1 is in Sj for some rational j with -1 < j < 1. That means for some x in S, s +1 = x + j. So x = s + 1 - j. 1 - j is rational, so x and s are in the same bin. There is only one number from that bin in S. So x = s and j = 1. But j is strictly less than 1. Contradiction. Similarly, if s + 1 is assumed to be in U(Si) + 2, j = -1; contradiction.
You're regurgitating essential concepts-uncountability measurable as length, irrationality as uncountability, asymptotic-non-closure.... But a countable-infinite number of irrationals are, countable (e.g. âżâm/k including, all rationals),-AND,-nonterminating definitions (e.g. infinitely-repeating-decimal 'rationals') are infinitesimal-open-intervals NOT points...
(Mathematicians compare-closure of Lebesgue m/n to closure of Zeno 1/n or 1/10âż as the distance from Open to Closed is nonzero; and which-if-any-irrationals are uncountable; P.S. neophytes, please read extra-carefully: taking a Limit is one-step-beyond Open, one-step-beyond nonterminating, by definition-as for Zeno....)
Hard to even tell what you are trying to say since it sounds like you are confusing terms all over the place here but I'll pick one part. Non-terminating decimal expansions are numbers, not intervals, in the real number system. The unique number associated with a decimal expansion is defined to be the limit of the sum of the individual decimal units and such limits always exist and converge to a unique real number for all decimal expansions. There are no non-zero infinitesimals in the real number system and decimal expansions do not define a range of numbers, they define a single number.
Now there are other well defined number systems that have non-zero infinitesimals such as the superreals. But this video is talking about the Reals, not those other number systems.
Nice. Can you make a video on branch tariski paradox. One interesting thing is if we have a finitely additive measure defined on all open intervals of R. We can extend this measure to a finitely additive measure on powerset of R. We cannot do this in case of R3.
Proof every number n in [0, 1] is in one of the Si.
Every n is in one of the equivalency classes of âoffsets by a rationalâ in the interval.
Let the equivalency class that contains n be called C.
Call the element in S that is chosen from C, m.
By definition n - m = r is a rational number in the interval [-1, 1].
For each Si, m is mapped to m + r_i by some r_i in the interval [-1, 1].
The value r is in the interval [-1, 1] and the r_i consist of ALL the elements of [-1, 1].
Therefore, some specific r_i = r.
Therefore, n = m + r_i.
And Si contains m + r_i.
Therefore, some specific Si contains n.
I always hated that kid in math class that would interrupt the teacher and ask, "When am I ever going to use this?"
After watching this, I kind of feel like that kid now. :(
Yep. Been there!
Lucas Brown The whole point of It is to construct a robust notion of measure of sets, for any set. For example when you say that Pr('Heads')=1/2 in a coin toss experiment the Pr(âą) is a function that takes a set as input (Events) and outputs their measure (Probability measure) with respect to the sample space (Set of ALL Events). Measure theory is so useful that even Einstein had his share in the Probability Field with Brownian Motion, a model very fundamental to modern society.
To give you examples, there is a famous European Derivative pricing model that is based on the Brownian Motion previously said and every Hedge Fund firm use or has used the Black and Scholes Model. In physics Quantum Mechanics use Probability to describe particle movement and in computer science we have Quantum computers and fuzzy logic. All based on the robust notion of Measures on a explicit or implicit way. The question is not how you use It, is how It affects you!
The Lebesgue measure is used to define a more powerful form of integration than the Riemann integral typically taught in introductory calculus and analysis. So, studying whether there are sets where the Lebesgue measure fails does have a purpose!
You probably won't; people do, but you won't. But the great part is that's fine! You should still learn it if you think it's interesting. At this level math is less about utility and more about joy.
Please come back, you were such a great host
pretty good explanation
At 4:02 I will recommend that add some dots before and after the three bins...
I read an article about two mathematicians who seem to have unwound Cantor's diagonalization argument. In keeping with sets and measures, could you cover that soon?
Thanks for re-uploading, I was really excited when I got the notification for this video and was gravely bummed to learn that you'd taken it down.
morgengabe1 What do you mean "unwound"? Cantor's diagnolization argument that the reals are uncountable is well established, there's nothing wrong with it.
â@@Bodyknocki kinda wish i knew myself tbh. But cantor's argument only shows that he doesn't know how to count the real numbers with an algorithm that counts ints/nats/rationals. Not that it is actually impossible to count irrationals. If you transform a set, the number of elements in the new set will be the number of elements in the old set plus elements added minus elements removed. This can always be countable.
I think the best way to read the continuum hypothesis is as a suggestion to use the axiom of choice.
â@@morgengabe1 No, Cantor's Argument proves that there is literally no countable list of the Real numbers at all. It doesn't matter what kind of "algorithm" you use to create such a list, if the list is countable it will necessarily not include all the Reals and the diagnolization gives you a step by step way of constructing a counter example digit by digit.
@@Bodyknock no it doesn't. That's what people say it does but people also think that newtonian calculus rrsolves zeno's paradox. That's just sticking your head in sand.
All he actually establishes is that diagonalization is not an effective procedure to demonstrate that A0 is in one to one correspondence with A1. If you only try to diagonalize, it will always look like they have different sizes.
Mathematical objects aren't semantic invariants. That sort of thinking is precisely why it took so long to come to terms with the evitability of euclid's 5th postulate.
@@morgengabe1 No you are 100% wrong about what the diagonalization proof does. But Iâm not going to keep telling you why youâre wrong since you obviously refuse to listen to anybody who tries to explain it.
Missed this channel đą
It would have been helpful to tell us what the error was that you discovered. If it's something small, I don't need to watch this again... but if it's something major, then I would feel I should. My decision should be based on the seriousness of the error... but you've left it as a mystery!!
macronencer The error was basically they added only positive rationals between (0,1) in the original video. They corrected it here to be positive and negative rationals in (-1,1).
My favorite consequence of the axiom is that anything I chose is legit choice function. Just pick your favorite number in the set.