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in fact we can probably do a more general fact from topology, given sets K_n, K_i<K_{i+1} where K_1 is compact and K_i are non empty and closed, we can see that if the intersection is empty the union of the compliments is the whole space so K_1 = U K_1 n K_i^c but since K_1 is compact and K_i^c are open there is a finite subcover K_1 = K_1 n (K_i^c u K_j^c u...) so taking the compliments we see that K_1^c = K_1^c u (K_i n K_j...) since K_i, K_j... are disjoint from K_1^c their intersection must be empty, but since they are all contained in eachother and there is finitely many this must mean the last one is empty which is a contradiction

@@ella-yt2xw thanks Ella!!! It means a lot to us!!! 😎💪

@@jakobr_ yes, but we know that they are getting arbitrarily small… since we said that in the beginning

@@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 🤔

@@CatharineAlbert-y6g beautiful names 😎👌🏻

@@CatharineAlbert-y6g Joseph Anderson? Wait... LEON!

@@sphakamisozondi awesome!!! Let us know what kind of topics you’d be interested in 😎

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”

@@expchrist thank you so much for your support!!! We will definitely post much more often about these subjects 😎

@@charithreddy3327 inf(I) is a real number. It is the maximum lower bound of the set/interval I

@@Khashayarissi-ob4yj thanks!! What did you think about the video? 😎

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

There isn't

@@theoneandonly-lu5cf Is that your claim specifically at the infinite intersection?

​@@ValidatingUsernameyes. If you take intervals with sequence of lengths with limit zero, there will be exactly 1 point in intersection (if we're talking about reals of course). For example, take I_n = [0, 2^{-n}]. The intersection will only consist out of 0

@@notEphim You understand that your interval has an infinite set of reals in the interval every step of the way all the way out to infinity right? What’s your proof that the interval closes to a point AT infinity 🧐

@@ValidatingUsernamefor every real number x other than 0, there is some natural number n such that x is not in [0,(1/n)] . Therefore, x is not in the intersection over all n of [0,(1/n)] . As this reasoning applies for all x other than 0, this infinite intersection doesn’t contain anything that isn’t 0. Also, because each of the [0,(1/n)] contains 0, the intersection contains 0. Therefore, combining these two things, the intersection has exactly one element, namely 0. Of course, you can also argue more generally that if the lengths of all the intervals in the sequence of nested closed bounded intervals, goes to zero, then the intersection will also be a single point. (If there was more than one point in the intersection, all the points between those points would also be in the intersection, and so the intersection would contain an interval of some length strictly greater than zero, but it has to fit inside each of the intervals in the sequence, which we assumed tend to zero, and so eventually the intervals in the sequence get shorter than the interval we obtained, yet must contain the interval, so we have a contradiction, and so the assumptions that the lengths of the intervals go to zero, and that there is more than one point in the intersection of all of them, are incompatible.)

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.

@@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

When we refer to simpler groups as 'building blocks', we touch on the idea that more complex groups can often be constructed or understood through combinations of simpler ones. As an example, every finite group can theoretically be broken down into a series of simple groups through a process called composition series. The specific groups you mentioned can be combined in a few ways. The most straightforward method is through the direct product, where you pair each element of A_5 with each element of {Z}_17, resulting in a new group where the operations are done separately within each component of the pair. There's another method called the semidirect product, which allows one group to dictate some of the structure of the other, possibly creating a non-trivial interaction between them. This can only happen if there's a suitable way (defined by group actions) for one group to influence the group structure of the other. So, constructing new groups from simpler 'building blocks' helps us understand possible group structures and their properties.

@@dibeos Would it be possible for you to make a video on the semi-direct product?

@@josephmellor7641 of course, we will include it on our list right now!! 😎

@@ravikantpatil3398 thanks!!! Let us know what kind of videos you’d like to watch in the channel 😎

@@fullfungo thanks!!!! Let us know what kind of content you are interested in! 😎

@@nycholasgr8112 yessss we will 😎

@@Prof_Michael yes, I’m actually working on a video that will be titled: All Types of Integrals in Analysis (or something like that). So as the title shows, we will cover a brief explanation of all integrals, including Lebesgue Integral and Lebesgue measure, of course 😎

@@WillJohnathan thanks for the advice! We will definitely slower the pace so that it is more accessible for everyone interested in learning math. Also, we will make a video about groups of Lie type, just as you asked 😎

@@dibeosLie groups are fantastic but definitely, for me at least, needed some extra time to fully grasp it the first time around - a slower pace would definitely be helpful!

@ValidatingUsername thanks for the tip!!!

I've found the same. But I wish they go deeper into the topics with more details and examples. Don't make it slower just for more audience (same type of surface level content is all over the place) !

@@joeeeee8738 yeah, it’s just that there is a looooot to talk about. So I guess we will try to pick an even more specific topic in group theory and give a bunch of examples

The classification theorem deals specifically with finite simple groups, not all finite groups. Emphasis on simple here. The simple groups have been completely classified, but finite groups in general (which can be built from these simple groups in more complex ways) are not fully classified. Hope that makes sense! :)

@@dibeos Not really, to "describe them all" we just construct all products of simple groups. We cannot easily see which one of those are isomorphic, be in that way we do describe them all! We just may get duplicates. So we could perhaps say that we are missing a unique standard way to describe each finite group as a product of simple groups in just one preferred way. (But I think the statement in the video suggests that it's worse than that... that's why I asked. 😇)

@@JosBergervoet > to "describe them all" we just construct all products of simple groups No, you can glue finite simple groups together in many different ways. A product is just one way of gluing groups together. It's very difficult to figure out what all the ways of gluing two groups together are.

@@logosecho8530 Then to avoid calling it a "product" of groups, let's say that every finite group has a composition series, en.wikipedia.org/wiki/Composition_series#For_groups . That's still gluing finite simple groups together, as you call it, so it still leaves unclear what we are missing: is there fear that we don't get all finite groups in that way? We could even go completely back to basics: every finite group can be "described" by its multiplication table. So some (tedious) procedure could just generate them all. As viewers of the video we are of course curious to know: what are we missing? What more would mathematicians desire after the classification? It sounds like something is finished (nicely explained in the video!) and something is still missing, but there we are left in the dark, which of course makes this second point unbearably intriguing... You will need to make a follow-up video!

Could one compare the sitation to atomic physics and chemistry? Being able to describe the behaviour of all types of single atoms does not mean that one can also describe the behaviour of all molecules.