A nonmeasurable set
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- Опубліковано 21 жов 2018
- In this video, I show that there exists a non-measurable subset of the real numbers. In other words, that set is so weird that one can not assign the notion of a size to it. Its construction relies on equivalence classes and the axiom of choice. Enjoy!
Note: The thumbnail is taken from commons.m.wikimedia.org/wiki/...
At first I was confused about the size of the set not being equal to 0, a constant or infinity but then I realized I confused the size, the measure, with the cardinality of a set. Thanks for showing this interesting proof :)
Great video!!!! You're ideal for filling the gaps we don't have time to cover in class
Thank you Dr Peyam for this very instructive video, even it was hard to grasp for me (I will watch it again). But one sentence I will remember always:
"...and infinity is not between 1 and 3."
Thank you so much for spending time on this videos!
Dr. Preyam, this video of yours filled me with joy. I was struggling to understand this proof and then watched your video, you have cracked down the proof in a very simple way. It feels great to learn from you. I will be thankful to you if you upload lecture series on Measure theory.
Such a beautiful example of the transcendent nature of the axiom of choice... and of quantifying reality itself. It's funny, the axiom of choice is itself a choice.
Hi Dr Peyam, Just wanted to thank you for posting such a clear explanation of the proof!
Glad it was helpful!
It turns out that you can prove unmeasurable sets using Ultra Filter Theorem, which is strictly weaker than the Axiom of Choice! (you can get Ultra Filters from Choice but not the other way around).
I actually didn't know this, but we could actually live in a world where unmeasurable sets exist, but not every Vector Space has a basis 😁
Wow that’s so crazy!!!
The set of sets which do not contain themselves as an element! baha
If you want weird sets, then in ZF it is consists that there exists sets without linear order!
If A is infinite set such that there are no 2 disjoint infinite sets B,C such that B union C=A then A does not have linear order
That just blew my mind *set*.
Lots of Respect from India Professor
Thank you Peyam! Do you have a measure theory playlist? I'd like to watch more of your explanations on this topic.
Check out my Real Analysis playlist
@@drpeyam I will. Thanks, and I wish you a very productive 2021!
I am currently taking an analysis course in my theoretical physics master's degree and I find this topics really interesting
Great Video!
bravo ! thanks sir
Please, is the union of classes ( you put a greek letter) that divide the set [0,1] equal the set [0,1] itself?
This video is amazing 😃! Just one question: does this sort of mean that the size is infinitesimal but not zero?
In real numbers there are no non-zero infinitesimals. Intuitively speaking the set has an "infinitely small" measure, but that doesn't help much. The Lebesgue measure (under the assumption of axiom of choice) is countably additive. We have shown that a union of countably many copies of the Vitali set fits in an interval from -1 to 2, but covers an interval from 0 to 1. So let's enumerate the copies, and assume that the measure of the Vitali set is infinitesimal (less than any positive real number). Now observe: measure of the first copy is less than 1/4. Measure of the second copy is less than 1/8. Measure of the third copy is less than 1/16. Measure of the fourth copy is less than 1/32. And so on; the infinite sum is 1/2. But that doesn't make sense - how can the measure of the union be simultaneously less than 1/4 (and we could have picked an arbitrarily small starting value ε, yielding the upper limit 2*ε) and between 1 and 3?
Instead, we could propose some new measure which extends the Lebesgue measure but is only finitely additive, not countably additive; under this scheme the Vitali set would have measure 0.
Nice demonstration. It seems like there should be one one equivalence set with all rational numbers; the remainder contain all irrationals of the form r + q where r is irrational and q is rational. Therefore, just pick r as the representative of each equivalence set. Next you should demonstrate the Cauchy distribution, which contains an undefined mean and variance.
Which r? For each equivalence class there are infinitely many numbers to pick from. (Take the equivalence class of pi/4 - how can you decide whether to pick pi/4 or pi/4-0.2 or pi/4+0.1 or... And you have to pick a number for each of uncountably many equivalence classes. This requires axiom of choice; in absence of choice, it's consistent that there's no set which picks a single element from each equivalence class. It's also consistent that there are strictly more equivalence classes than real numbers themselves, in the sense that real numbers can be mapped one-to-one with a subset of the equivalence classes, but equivalence classes can't be mapped one-to-one with a subset of reals.)
Nice one
Seems interesting 🤔🤔🤔
Is an "infinitesimal size" acceptable? Just a size which makes the limit as N goes to infinity of N*C be between 1 and 3?
That doesn't help. Yes, intuitively the measure is "infinitely small", but let's see where it gets you: you have countably many sets. The first set has measure less than 1/4. The second set has measure less than 1/8. The third set has measure less than 1/16, and so on. So the union is no greater than the infinite sum 1/4+1/8+1/16+...; this sum is equal to 1/2. (You can make the upper limit on the total measure arbitrarily small by picking sufficiently small first limit.) Yet, the union is a superset of an interval of length 1, and a subset of an interval of length 3.
It would be better to say that the measure is 0, under some measure which extends the usual Lebesgue measure and which is not countably infinite. (Note that the countable additivity of the Lebesgue measure is a consequence of the axiom of choice. Without choice it's consistent that real numbers themselves - or any interval or other set with cardinality of the continuum - are a union of countably many countable sets. In this model Lebesgue measure is not countably additive: a countable set has measure 0 [this doesn't require axiom of choice], but the union of countably many these sets doesn't have measure 0.)
All geniuses are left-hander :D
Are there non-mesurable subsets of ℝ without the axiom of choice?
Surprisingly no!
Interesting how such simple axiom can lead to such weird results. It seems to natural for me not to be taken as an axiom + Zorn's lemma is equivalent to the axiom of choice right? Big one two
The entirety of axiom of choice is not necessary to prove that there exists a non-measurable set; existence of well-ordering of reals is enough. (More than enough: it's consistent that real numbers can't be well-ordered but there exists a non-measurable set of reals.)
On the other hand, you can't prove that there exists a non-measurable set in set theory (ZF) without axiom of choice.
This is the wildest proof I've seen in a while. I feel like I need to tell everyone about it
Trop dur pour moi actuellement mais merci quand même...
nice :)
The answer is yellow!
Is there a name to this non-measurable set? What is it called? is it the set of all cosets of Q?
Vitali set
@@drpeyam Thank you Dr. Peyam for the reply! I really love your videos!
"There exists a set IN NATURE" lol
Indeed strange that people feel the real numbers are real…. All math only “exists” as constructions.
what is the measure of empty set?
0
@@drpeyam does the cardinality also 0?
Yes
Any finite or countably infinite set has measure 0. (A set with - say - five elements has of course cardinality 5.) The Lebesgue measure is also finitely additive and, assuming axiom of choice, countably additive. So the union of countably many disjoint measurable sets has measure equal to the infinite sum of the measures of the individual sets. (In absence of axiom of choice this might not be true - it's even consistent that real numbers themselves are a union of countably many countable sets.)
Naked beauty. Thank you Dr Peyman.
I wish he taught Topology here at Rice and not Dr. Semmes :/
My mathematical opinions are quite odd. I don't think the Axiom is Choice is reasonable but on the other hand I'm perfectly fine with the Banach-Tarski paradox.
Interesting!!! Usually it’s the other way around!
"Axiom of choice is obviously true, well-ordering theorem is obviously false, and who can say about Zorn's lemma?" (Of course, the three propositions are equivalent in ZF.)
well if it is 'something between 1 and three' divided by 'countable infinity' it should be an infinitesimal surreal number :-)
Next let's measure the size of the set of all sets that do not contain themselves.
This set can be proven not to exist, so you can't ask what is its "size" (under whatever definition of size). In addition, Lebesgue measure is (by default) only defined on real numbers, or on n-dimensional real space. And the definition depends on the base set: the line segment (interval) from 0 to 1 has measure (length) 1 on real numbers, but measure (area) 0 on two-dimensional real space. If you want to ask what is its cardinality, then in a sense it has strictly greater cardinality than any set - precisely because it's not a set but a proper class.
Before I watch this video, I guess u are going to talking about Vitali set from the title.
After watching until 4:29, I decided not to believe in axiom of choice.
Then, do you believe that a set can be split into more subsets (disjoint and non-empty) than it has elements?
from Morocco..this set N is not measurable......and me i am miserable
Please a video about sureal number sets
en.wikipedia.org/wiki/Surreal_number