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- เผยแพร่เมื่อ 17 ก.ย. 2024
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from doing some straightforward point-set topology we can find
(1): If we have sequentially closed sets K_n where K_i < K_{i-1} and K_1 is sequentually compact and each K_i is nonempty, then we can construct a sequence a_i \in K_i, this has a convergent subsequence a_ji->a then this point is a limit of points in K_j for every j so is in every K_j since they are sequentally closed so is in the intersection so the intersection isnt empty
it's not too hard to see that closed intervals are sequentially compact so are sequentially closed by taking a sequence and constructing the sequence of floor terms {a_j: a_k >= a_j for all k>j} if this set is infinite we have a monotone, bounded sequence so it must converge to its supremum and if this set is finite then there is some final a_j, every other a_i must then have an element below it eventually so we can construct a monotone decreasing sequence which must converge to its infimum so closed intervals are sequentially compact so we can apply (1)
in fact we can probably do a more general fact from topology, given sets K_n, K_i
@@kylecow1930neither of which is her argument .
(1) a sequence
(2) contradiction
Are both rather indifferent to her more classical nested deduction of existence , without a sequential diagonalization .
It ‘ s interesting what youre doing :
‘ it ‘ s not too hard to see that closed intervals are sequentially compact ‘
Deducing something like that is one would use a diagonalization for ….
That looks like a diagonalization .
It ‘ s culturally desirable to be capable of that , even when it is flaunted around that one may simply deduce it … I would like to not say : ‘ by contradiction ‘
Thoroughly enjoyed this. It's been a few decades since I did this.
I did have to stop to think why Cantor's theorem implies the reals are uncountable though. I think it would have been good to briefly clarify that.
@@richardfarrer5616 thanks for the tip, Richard! You are right. We thought that the video was getting too long, so we decided to not explain why this theorem implies the uncountability of the real numbers, but I guess it would have made the final conclusion more clear 🤔
I'm not a math major, but your videos are fascinating and I can't stop watching
@@sphakamisozondi awesome!!! Let us know what kind of topics you’d be interested in 😎
Thanks for this one. Kronecker's attacks on Cantor, based on fear of transfinite numbers, is another despicable chapter in math history. This is kind of like what I emailed you guys about earlier, with the Pythagoreans murdering the guy who proved that the square root of 2 is irrational. Another one you can add is Saint Gauss, who obnoxiously dissed a teenage Janos Bolyai to his father, claiming that Gauss was the true inventor of all non-Euclidean geometry, to diminish the young Bolyai as a mathematician (It worked, unfortunately).
The math establishment does now oppose, and always has opposed, innovation that would shake up the powers that be. Look at the dogma that Kurt Godel had to put up with!
The YT algorithm still has me in its sights I'm glad to know! It's been rough the last few months.
I think I came up with a slightly different proof. If there is no such point c, then the complements of the closed nested sets form an open cover of the interval I1. Since I1 is compact, the cover must contain a finite subcover. But it's easily seen that a finite number of complements of closed nested intervals can't cover the interval I1, yielding a contradiction.
8:10 The minimum of B won’t necessarily be sup(A) unless we know these intervals get arbitrarily small.
@@jakobr_ yes, but we know that they are getting arbitrarily small… since we said that in the beginning
With luck and more power to you.
hoping for more videos.
@@Khashayarissi-ob4yj thanks!! What did you think about the video? 😎
7:30 You defined sup with inf and inf with sup, that's circular definition, you should have defined sup as the min of all upper bounds and inf as the max of all lower bounds.
Martin Paul Moore Timothy Lewis Jose
Also great video❤
7:17 Isn't the infimum of both (2,3) and [2,3) 2? So inf(I)=a whether the interval is open or closed at a.
@@Himanshu_Khichar yes, but you only know that because you know that the real numbers are uncountable and there for there is no lower bound of (2,3) greater than 2. But that’s the whole point of the theorem. At this point of the proof “we don’t know it”, we just know that there exists such infimum.
@@dibeos I don't see how the uncountability of real numbers has any bearing on whether there exists a lower bound greater than 2. Could you explain that? I suppose the theory covering definition and rules of infimum and supremum is well-established before proceeding to this theorem and stating that inf(I)
@@Himanshu_Khichar Thanks for the encouragement :)
The distinction between open and closed intervals doesn't affect the infimum in the specific case of intervals like (2, 3)or [2, 3], where the infimum is indeed 2 in both cases. Let me clarify what I meant by referencing the uncountability of real numbers: The point I was making isn't that uncountability directly affects whether an infimum exists. Rather, it’s related to the nature of real numbers as a complete ordered field, which ensures that any bounded subset of the reals has both an infimum and a supremum. This is key in establishing the existence of an infimum in cases like (2, 3).
Where uncountability comes in is more subtle: we rely on properties of the real number line when proving things like Cantor's Theorem, and the fact that the real numbers are uncountably infinite helps guarantee that no element smaller than 2 can be a lower bound for the interval (2, 3).
As for your second point: you're correct that the infimum and supremum are well-defined before applying the theorem. My intention in the video was to focus on the logical development and the intermediate steps of the proof, particularly that at certain stages of the nested interval theorem, we may not explicitly "know" the infimum yet but are assured of its existence by completeness. Let me know if my answer is more clear now and if it really helps you :)
@@dibeos Oh okay no problem. Thanks for elaborating. I guess one can always overlook such small details. Nevertheless, it was overall a wonderful explanation of the proof of the theorem by you, all the best once again.
You ‘ re certainly correct .
Clearly she is expressing the process of considering
(1) the space of all lower bounds
Before one defines the the infimum of the subspace as the
(2) supremum of all of these lower bounds : this is exactly equal to a
That is correct , and her text is technically incorrect .
If one cannot follow the reasoning , then there is no complaint , and all other considerations such as uncountability are quite irrelevant:
a posteriori .
a priori , you ‘ re free to concoct a butt - ugly complex of known results .
I think the little animations and explained notation was beautifully done. Very little left unexplained.
That being said, I’ve been watching math videos on and off for the past year so a lot of notation is familiar to me.
One suggestion that would really make it crystal clear is some examples for the conclusions made.
Maybe a single example during the walkthrough would provide a concrete thing to walk through as you prove the general idea.
@@jks234 thank you so much for the tip! Could you please point out (for example) where in the video we could have shown an example but we didn’t? I say that because I really tried to add as many examples as possible throughout the proof, since many people often ask us for examples in the comment section. Thanks anyway!!! 😎
@@dibeos What I was thinking about was the part where you… combine the infinum with the supremum and prove the theorem.
That part is all conceptual and I would personally probably want to sit down and put numbers in everything to see what is meant by it.
@@jks234 oh got it, so maybe we should have picked an example of sequence of intervals (with numbers) and calculate the sup and inf of A and B (numbers, again), and finally show that the interval [sup,inf] contains a real number that belongs also to the infinite intersection of the sequence of intervals, right? 🤔
@@dibeos Yeah. Actually, I can imagine that example doing an excellent job of clarifying the theorem.
The ultimate conclusion is that there is always a real number between two other real numbers right?
The concrete example would be trivial then.
@@jks234 yes, that’s exactly the conclusion! In other words, the real numbers are uncountable
You use inf in your definition of sup, and sup in your definition of inf, which is not good.
Also, I would have liked you to say something about the existence of sups and infs.
Finally, it is not clear to me why this shows that the real numbers are uncountable.
About the last one, my students showed me idea of such a proof, it's pretty cool. Suppose [0, 1] is countable, let's form a sequence x_i. Now say a_0 = 0, b_0 = 1. 0 and 1 were counted in a sequence at some point. So let's truncate the sequence x_i so that 0 and 1 are no longer in a sequence. Notice that we have thrown away only finite amount of points, therefore we can choose a_1 and b_1 in such a way that [a_1, b_1] doesn't contain thrown away values. Continue this process. By this theorem there is a number c contained by every interval [a_n, b_n]. Also c = x_k by assumption that sequence x_i contains all points from [0, 1]. But then c can't be in [a_{k+1}, b_{k+1}] by construction. We arrived at contradiction
Ah yes, that seems to work. Thanks
It's particularly interesting when you look at the detailed definition and then work out sup and inf of the empty set. It's a slightly surprising result.
I got lost at 8:16. Why does it follow? Also, I'd love the video to first establish why this conclusion is counter-intuitive.
@@danielchin1259 in this timestamp we say that sup(A)
@@dibeosI’ll keep saying this until it sinks in around the world, the mapping of reals to naturals is 1 to 1.
The only difference is the reals essentially have an infinite set of zeros before appending a natural number to the end 0.0…..01 for every integer.
We took the abstract concept of a unit and gave a single unit the mathematical value glyph of 1 in our current system.
at 7:22 ,Isn't Inf(I) a set ? so how can you compare it with a number?
@@charithreddy3327 inf(I) is a real number. It is the maximum lower bound of the set/interval I
Indeed when we know that any complete metric space (with at most finite isolated points) is uncountable AND the Cantor's intersection theorem can be implied for a general complete metric space, the main point of the theorem reduces to completeness of the Euclidean line! If the Euclidean line is assigned another metric which is not complete, then the Cantor's intersection theorem fails. So can't we put the blame for this theorem on completeness of one's preferred choice of metric over real line and not Infinity? 10:33
This is not really a metric fact. Ths result is true for nested compact subspace of any topological space, so this is independent of the metric
@@bastamtajik512 Good question! Seriously! This theorem holds more generally for any nested sequence of compact sets in a topological space, not just in metric spaces (as somebody else already said). Completeness of the metric on the real line plays a role, but it is not necessary, because the core result relies on compactness, which applies even in non-metric spaces. So, it’s independent of the specific choice of metric, and therefore it is a consequence of the weirdness of the “infinity”
I feel like this doesn't need a proof. "After shrinking a set infinitely many times there is something in the set. Yeah of course. That's a no brainer.
Specifically shrinking closed sets, but yeah
Still though, with bs like the banach tarski paradox and all, you gotta prove it yourself since you really can't be sure what you're gonna get when infinities are involved
It depends on what you want as your axiom of reals. You can pick this thrown as axiom (together with Archimedes principle stating that the set of positive integers is unbounded). If it's not an axiom, it should be proven
@@MetalMint Closed *compact* sets! The intersection of the closed sets [n,\infnty) is the empty set, even though it is a nested decreasing sequence of closed sets.
It is not obvious. There are many subtleties here. The theorem required the intervals to be
1. Nested
2. Closed
3. Bounded
Your intuitive explanation only uses the nested condition, but every condition is necessary. You can find sequences of intervals satisfying any 2 out of 3 with empty intersection.
Moreover, one must use a technical fact about the real numbers, because this is not true for the rational numbers. Witness
I1=[3, 4]
I2=[3.1, 3.2]
I3=[3.14, 3.15]
I4=[3.141, 3.142]
I5=[3.1415, 3.1416]
....
This has empty intersection when viewing these as intervals of rational numbers.
It's not entirely obvious. It doesn't hold for rational numbers, but it does hold for real numbers
Williams Daniel Lewis Joseph Anderson Jessica
@@CatharineAlbert-y6g beautiful names 😎👌🏻
@@CatharineAlbert-y6g Joseph Anderson? Wait... LEON!
how do you open by mispronouncing Georg?
bro... if you're that lazy you have nothing to say that's worth saying.
watch how simple it is to demonstrate that you have no idea wtf you're talking about:
you say that e is Real, right? since it's neither infinite nor infinitesimal, and has no Imaginary part, it must be Real by your understanding of how things work.
take the definition of e:
e = lim n -> pos. inf. (1 +1/n)^n
since 1/n is asymptotically infinitesimal here, let's replace it with an infinitesimal, c:
e = lim n -> pos. inf. (1 +c)^n
now, let's let n be some Transfinite Integer, so that we can get rid of the limit:
e > (1 +c)^n
this is infinitesimally less than e, so we can add c to correct this:
e = (1 +c)^n +c
now, since n is an Integer, we can apply the binomial theorem to expand this into a polynomial of base-c:
e = 1 +Ac +Bc^2 +Cc^3... +Bc^(n -2) +Ac^(n -1) +c^n
at this point note that all coefficients and exponents are Integers, which makes this a non-Real Levi-Civita series.
thus, your assumption that e is Real is incorrect. and as such, when you talk about the Reals, we must bear in mind that you fundamentally do not understand what you're talking about, and as such, you should be ignored.
all you've demonstrated is that the Reals are Archimedean. but you're also concluding from this that the Reals are continuous, which is a contradiction.
this contradiction is made plain by the fact that the Levi-Civita field and the Surreals are accepted to exist as extensions to the Reals on the Real line. but if these non-Real extensions to the Reals lie on the Real line, and all values on the Real line are Real, you've definitely fucked up.
not to mention, you count zero among the Reals, and this inclusion on its own renders the Reals non-Archimedean, but everyone just ignores that and calls it rigor.
it's also notable that the constructor for the Surreals cannot possibly work:
{ | } = 0
{ 0 | } = +1
{ | 0 } = -1
etc.
notice that what's supposed to be happening here cannot possibly be happening. { | } is taking the empty set as input, twice, and producing zero. this part is totally fine. the problem comes in the next step, where { 0 | } produces something other than zero, because zero is the empty set. so { | } = { 0 | } = { | 0 }, meaning that by the Surreal constructor -1 = 0 = +1, and ultimately all Surreals as produced by the Surreal constructor are zero. note that this does not mean that the Surreals themselves do not exist, merely that Conway's construction of them doesn't do what everyone claims it does.
so at every level, you're just fucking everything up. and if you cannot see this for yourself, let alone address it, you are simply too incompetent to be speaking on this topic.