The Bolzano-Weierstrass theorem, a proof from real analysis

[@zachstar]

Get the QED shirt over at STEMerch! stemerch.com/collections/qed I know this video format was different from usual, was just way easier to do this one on a whiteboard but will be back to normal for the next one! Also I know this proof wasn't exactly as complete as you'd see in a real analysis course but was trying to make it so students of all levels could understand the kind of thinking you need for a proof like this. P.S. For anyone wondering how this comment was posted so much earlier than the video release date, it's because I'm a wizard.

[@sumanporel9670]

I was wondering. But you are a wizard 😂❤️

[@k.o.3948]

very nice video :-) kind regards from Germany

[@pabloariza2295]

how was this posted 3 weeks ago

[@rickyjacobsen6922]

Loved the white board and very easy to understand, would love to see some more complex real analysis too, keep up the great videos!

[@leoG017]

Adami riccardo. Search for his channel and you will not be dissapointed

[@thomy2562]

There is a white board now? The budget must be thrugh the roof.

[@HorkSupreme]

Imagine the costs of having a black board.

[@zachstar]

Had to flex my luxurious lifestyle

[@GymCritical]

@@HorkSupreme imagine flexing with chalk?

[@alan2here]

Could really push the boat out with a Surround Prof cylindrical whiteboard.

[@devinotero1798]

@@zachstar i heard you where going to get a job in engineering or going back to engineering school soon. Whats going on with the channel?

[@EpicMathTime]

Man, I remember when a group of math majors I knew were taking real analysis and they turned this into a meme. At a restaurant "Because every bounded pizza contains a convergent sub-pizza." Watching Game of Thrones "Right because every bounded army contains a convergent sub-army" etc

[@pendragon7600]

this needs to be a thing

[@phoenixhorsley6658]

I took real analysis last semester and this instantly became a joke after we learned BW

[@NoorquackerInd]

Zach: "here's a proof from Real Analysis!" Me, nearly failing calc 3: _interesting_

[@twenty-fifth420]

Me who barely passed algebra and statistics trying to understand this: tinyurl.com/yyet4ne3

[@canyadigit6274]

Don’t give up bro

[@leandro8897]

And he gave the easy route hahahahaha if you read this, you'd be shocked

[@PapaFlammy69]

Watching these videos is better than Star, ngl

[@natesavage3576]

Oh My lord, you are in the comments, please bless me with your mathematical talent.

[@comaphygourav9324]

@@natesavage3576 1729

[@aristo7051]

“better than Star”? what’s that mean

[@clonebin0]

@@aristo7051 zach star Zech Star Sech Star SecStar ....Sex Star?

[@ryanheznts4540]

@@clonebin0pron directors going from star wars to sex wars as a plot

[@RC32Smiths01]

Things like this are what make math endlessly interesting, and anyone can get into it! Very wonderful proof.

[@MathPhysicsEngineering]

I also made a video about this theorem which is very visual and at the same time very rigorous, I believe you will like it. I made two proofs of the Bolzano Weierstrass theorem, one based on the monotone convergence theorem , which works only for the real line (without proper modifications) as it exploits the order properties on the real line. The video on my channel is called: "Easiest Proof of the Bolzano-Weierstrass Theorem" The other video exploits Cantor's lemma as is done in the current video: Visualized Proof of the Bolzano-Weierstrass Theorem using Cantor's lemma I'm currently recording a visual and rigorous course on calculus where I introduce ideas from topology and metric spaces early on look it up on my channel. I would also like to recommend my video called: Visual Proof of The Heine-Borel Theorem and Compactness of [a,b] It has some sound issues but I'm very happy with the visual quality and the rigor of the video.

[@Chris-b-2]

What is really nice about this method of proof for Balzano-Weierstrass is that it can then be used for the Heine-Borel theorem and Cantor's Nested Set theorem, so it is very useful pedagogically.

[@morgard211]

What's the name of this proof technique?

[@Chris-b-2]

@@morgard211 I’m not sure if there is a specific name for this type of argument. The closest I can come to would be from computer science: divide-and-conquer or recursive reduction.

[@quantumgaming9180]

I don't think you can use this to prove Cantor's nested set theorem since it's used in proving Boltzano-Weierstrass

[@Vex-sn7ne]

The fact that this man did this all in one take

[@SleepyHarryZzz]

I'm gonna blow your mind: lecturers do it in one take, too.

[@tomkermer33]

Holy shit that timing is perfect this is one of the things i just couldn't understand from the textbocks. I'm a First year physics student and this was the bane of my analysis 1 class.

[@skull4110]

Maybe start reading textbooks?

[@tomkermer33]

@@skull4110 yeah we went over it only briefly in class and i just didn't what the Textbook is saying with this weird intervalls but now it clicked for me.

[@windowsxseven]

@@tomkermer33 I'd like to interject for a moment. What you are referring to is, in fact, a textbook. But in your original comment you referred to it as a, quote, "textbock". Hence the first reply, which was a satirical comment hinging on the assumption that instead of reading a "textbock", you should've read a textbook, as if they were two different things. Don't worry, I won't blame you for letting high humor fly over your cranium.

[@NormanWasHere452]

Whattt, I’ve applied for physics and didn’t expect to be doing things like this.

[@tomkermer33]

@@NormanWasHere452 I'm not from the USA i'm from Germany so i don't know how similar the Curriculum is, so don't worry too much and see how it goes.

[@Entropize1]

The thing to keep in mind is you have to pick a term in your half-interval that comes after all the previous members of your subsequence in the original sequence. Make sure you understand why this is possible everyone!

[@user-ll1ow4xf8n]

because as you are constructing your subsequence, you have selected a finite number of elements, but your interval has an infinite number of elements, so you have an infinite number of choices for your next element

[@johnchessant3012]

That was a really good explanation! I definitely agree that these higher-level math classes are a totally different beast than the plug-and-chug problems in high school, and for me these are more enjoyable. It was the first time I realized math is an art.

[@RowanAckerman]

I am currently in Algebra II and I can confirm that I understood this.

[@xephyr417]

Interesting. It's almost frustratingly simple. It's just a rigorous method of proving you can find the infinite term and pick it no matter what. Intuitive once you see it, but would I have thought of it?? Idk!

[@dekippiesip]

It's also interesting how binary comes into this. If instead of I1 and I1' you would use I1 and I2, then I11 and I12, etc for the levels beneath, you can associate each such interval with a string of binaries that in the end will coincide with the binary representation of the number your subseries converges to if the starting interval was [0,1].

[@reppmag12345]

Not necessarily, an infinite sequence can have no repeating terms and this theorem and proof still apply.

[@RiemannThumbs]

My university is using Abbott's Understanding Analysis for their undergraduate Real Analysis courses and I really appreciate the simplified proofs of the text. In fact, the proof you covered is one of my favourites so far! Good shit, dude!

[@DarinBrownSJDCMath]

As Zach alluded to in his comment about Cauchy sequences, it should be noted that the proof depends crucially on the completeness of the real number system, which guarantees the existence of the limit of the subsequence so constructed. For example, in the space of rational numbers, the intersection of a nested sequence of closed intervals can be empty.

[@hebernelski8029]

Does he need it, though? Since each closed interval is being constructed with 2 rational numbers as the bounds, the intersection of nested sequences with rational bounds necessarily contains a rational number (since the bounds are rational).

[@DarinBrownSJDCMath]

@@hebernelski8029 Not true -- e.g. the nested family { [3, 4], [3.1, 3.2], [3.14, 3.15], [3.141, 3.142], ... } where the endpoints are decimal approximations to pi, has empty intersection in the space of rational numbers.

[@Javy_Chand]

I cant help, but notice that the Bolzano-Weistrass Theorem is just Bisection-search, but generalized for infinite sequences... Feels like Im learning.

[@tdmcnally33]

Every once in a while I’ll go through your videos and 3blue1brown’s videos and just watch it. I don’t understand anything you talk about, but I still watch it

[@admann24]

same

[@concorde747X]

I like how I am procrastinating my real analysis homework by watching KZbin videos on real analysis

[@lambdamax]

Is it possible for you to do more of these types of videos? The only reason I'm asking is because I graduated with a math degree 10 years ago and I had the hardest time with both of my real analysis classes. I still have ptsd. I never could gauge what was considered rigorous and what was not rigorous when doing proofs. When do you know that referencing a theorem is enough and when do you have to also prove the theorem you're using in the current exercise? Anyway, thanks for uploading these videos.

[@fetchstixRHD]

The "referencing theorems/needing to prove them" point was one of the most annoying parts of analysis I found. So often I would have my work with "your explanations are too long" at the same time as "these steps need to be justified", and it's like "well that's very helpful(!)" I still loved analysis and algebra the most of what I covered (and loved covering metric spaces with someone!) but I don't know if I could survive more maths...

[@positivegradient]

I am sure the profs quite often mix up the hierarchy of the theorems and just put the burden of remembering on the student. To teach a rigorous subject is a big responsibility. That is why these kinds of math topics are best learned from a good book, where all the checks have been made.

[@johncgibson4720]

I see you started out with engineering and digging yourself deep in math and no end in sight.

[@baticadavinci3984]

Man you should really pursue pure math, like for real bro. You're a true inspiration!

[@alfredholmes9899]

Other proof: pick the (possibly finite / empty) subsequence containing elements of the sequence that are greater than all the elements that follow. This sequence is decreasing so, if infinite, is a convergent subsequence. If this sequence was finite / empty then the original sequence has an increasing subsequence (each element is not an upper bound) and so this subsequence converges.

[@cr1216]

4:55 Actually you don't have uncountably infinite many numbers. A sequence only consists of countably infinite many numbers.

[@dougdimmedome5552]

Also notice how this idea of constantly creating nested intervals, all enumerated by natural numbers out to infinity at some point doesn’t terminate at least one value, and what that says about how the natural numbers map onto the real numbers, and also even more how this behavior of real numbers is necessary for calculus itself to work.

[@alan2here]

Both a bounded random sequence and a sequence of repeatedly adding an irrational and mod constant, should allow for every choice of where to converge.

[@omerd602]

Reminds me of competition level math, especially at higher levels like AIME, where sometimes you can be stuck on a problem forever but a single thought can give you a beautiful insight that helps you solve the problem

[@sumanporel9670]

Zach Star always makes videos on some interesting facts. I really like this channel..........

[@nirbhay.8400k]

Thank God someone is trimming the rigour and making Real Analysis understandable!!

[@h2ogun26]

Thank you. I had hard time understanding explanation of this on wiki but now I get it

[@gi0rg0sk]

I was just studying the proof and during the break came to see if you uploaded any new video. Guess I won' t have a break then. 😂

[@jorgealexandre4616]

I'm not a mathematician, but here is what I think: Our sequence is bounded within the interval of, let's say, the number x to the number y, x being greater than y. Perhaps we can find a value A such that our sequence never dips below A (no number is smaller than A). If can choose this number, than our sequence happens all between A and x. If you can't find any A, then the subsequence that we construct will tend to y. If you can find an A, choose the biggest you can. The subsequence will be tending towards this number. If you can always find a greater A, that means that the set ranging from y to A isn't closed (A.K.A [y. A[ ). If that is the case, it means A BELONGS to our sequence, and then, every other number is always greater than A, so you can make a subsequence that tends towards A. The only way this wouldn't work is if the numbers will only come to a certain minimal finite distance of A. If that distance is, let's say, 2, just choose B = A - 2. If numbers never come closer to this new choice than a distance d2, make B - d2, and so on. Since every time you walk a finite positive distance towards x, you will either find a good choice or will go all the way up to the end of the sequence without finding it. If the latter is the case, it means that the smallest difference (like in subtraction) between two elements on the sequence is whatever your smallest jump was. So the maximum number of numbers that the sequence can have is the ratio between module(x-y) and this smallest distance. Since this distance is a finite number (if it was infinite the sequence wouldn't be bounded, and if it was infinitesimal you would have found your limit already) then there is a finite number of points that make up the sequence. If all of them appeared finite many times, then the sequence wouldn't be infinite. So at least one of them is being repeated. In that case, you can make a subsequence taking this number (let's call it R) only and the sequence would look like {R, R, R, R, R, R, R...} and would tend towards R

[@jerrythe_boy6066]

came up onto my recommended about a year ago, as a freshman in highschool now studying real analysis, Bolzano weirestrass theorem got me really excited for some reasons

[@luisbelgois2102]

Here is a sweet excercise to the viewer: Since que sequence {sin(n)} is bounded, by the Bolzano-Weierstrass has a convergent subsequence. Well then, find it explicitly, meaning give a formula for the general term of the subsequence.

[@EpicMathTime]

Something that converges pretty quickly to 0 is to choose a_n as the numerator of the nth term of pi's continued fraction (all of which has an explicit formula). That is, pi's continued fraction gives the terms 3, 22/7, 333/106, 355/113, 103993/33102, 104348/33215,... so we are picking the sequence sin(3), sin(22), sin(333), sin(355),... The logic being that since these rational numbers p/q are approximations of pi, the numerator p is an approximation of qpi, ie, some integer multiple of pi. IE, the sequence of numerators mod pi approaches 0.

[@yonatanbeer3475]

All the zeroes of sin(n)

[@JacobGaiter]

Infact, you there is a sequence that converge] to every real number in [0,1]

[@bjornfeuerbacher5514]

@@yonatanbeer3475 The sequence sin(n) contains no zeroes at all. n is a natural number, and the sine of a natural number is never zero.

[@farhanislam8463]

As a guy who has read that book and attempted it's problems, I really enjoyed this. (The first chapter, 20 pages took me 3 days. At the end, I had a nervous breakdown)

[@baljeetgurnasinghani6563]

Actually, instead of using the Nested Intervals method, u can first prove : 1. The Monotone Convergence theorem ( A monotone sequence is convergent if and only if it is bounded) 2. The Monotone Subsequence Theorem ( Every sequence has a Monotone subsequence) These two statements are comparatively easy to prove and you can prove the Bolzano-Weierstrass theorem just by combining them. A bounded sequence will have a monotone subsequence. Since, the sequence is bounded, it’s subsequences will also be bounded. Since the Monotone subsequence is bounded, it is convergent. This is the way I actually studied the proof for the first time. The book also had the Nested Interval proof, but it was labelled as the alternate method PS : The book I was talking about was ‘Intro to Real Analysis’ by Bartle and Sherbert

[@epicmarschmallow5049]

Yeah this is how I was first taken through proving Bolzano-Weierstrass. After that we used this method, but I didn't like it as much

[@bjornfeuerbacher5514]

"These two statements are comparatively easy to prove " I don't agree. For me, the proof of Bolzano-Weierstrass was easy, but I don't see how one would prove these two statements!

[@jaimeduncan6167]

6:33 The fact hat the series is superior and inferior does not imply that the interval is infinite. A donned series like. (1, -1,1,-1, -1 ...) is an easy counter example. What is true is that if none of the intervals is infinite the series will repeat infinite many times, and once that happen is trivial to choose a sub sequence that converges.

[@markkennedy9767]

Really nicely done. This would have helped understand Bolzano-Weierstrass theorem so much better when I did it. Great stuff. Wish maths was taught first like this before the formality

[@alessandromestri9004]

*me, who did real analysis last year* Ah shit, here we go again

[@sjn7220]

I’ll stick to my fake analysis 😀.

[@MrKyltpzyxm]

As you can see from this graph, Santa Claus is clearly just a complex inverse derivative of the Tooth Fairy. In this essay I will...

[@peterkim359]

@@MrKyltpzyxm the space is said compact when you pass it through a trash compactor

[@Tubemaster32]

thank you zach, most useful! i appriciate the general information about what matters and what doesnt when it comes to the sequence

[@Tubemaster32]

i paused 8 min into the video, it becomes obvious and i got the concept ~ will enjoy the rest anyways

[@AndrewTyberg]

7:56 OH OKAY I see where you’re going with this!

[@richardbenes9]

What a great proof of the fact, that a video does not have to contain any efects, - not even cuts! - and yet it can be capable of holding viewer's attention for the whole time! Thanks for that!

[@sanjursan]

Yes, and without music also!

[@leirumf5476]

Fun fact, that ain't working if your set is not on a compact metric space. Example: let's define A = {(a_n) : a is bounded} and d:A×A →|R, d(x,y) = max{|x_n - y_n| : n in |N} Let (b_n) be such that b_n_j = 1 if n = j zero otherwise. Now, {b_n : n in |N} is bounded (it's distance from the sequence that's all zeroes is 1 for every element on it) but there's no subsequence that converges, since every element appears just once and the distance to any other element on the set is always 1. (Could be more rigorous, but this ain't no latex anyway)

[@DG-rl5ty]

can u make a discord server for us to discuss science and math

[@aperture0]

that would be dope

[@noellundstrom7447]

Interesting, I never knew how to prove that the interval [0,1] is compact even though I literally use it every single day. Cool!

[@mattgsm]

sigmoid function is bounded and yet it has no repeated numbers

[@marcuswright4874]

Pronounce it "Bowl -Zahn -Know-Veye-Err-Strazzz". The infinite/finite pigeon-hole principle: If an infinite no. of pigeons go into a finite no. of holes, then at least one hole will contain an ∞ no. of pigeons. I like this proof because if used the "Bisection Method", that your graphing calculator uses to find zeros, and is the "divide and conquer" strategy. This scheme appears many times in the subject.

[@Luihtz]

I also learned this proof. It is so constructive which is why i like it. Formally proving that the sequence is a Cauchy Sequence is kind of tricky. But of u know how, it is so good!

[@boldblazervids]

You explained this so much better than my first year prof did!

[@dakota5569]

I actually had a lot of fun in my real analysis class, even if I didn’t understand what I was doing until the end.

[@lamagicx]

wish this came out slightly early, just had my test and this was on it.

[@dominikraith1555]

Great video! But I'm curious wheather you can denote a sequence using {} brackets. It seems like something confusable with a set, which I think wouldn't work as a sequence? Sorry id that's a trivial question.

[@dominikraith1555]

my english really was bad. I guess it still kinda is

[@EyesOfFrozenMeat]

Just for the record, giving a proof in this way and actually writing up a formal proof are universes apart.... UNIVERSES!!!!!!!

[@EyesOfFrozenMeat]

@ゴゴ Joji Joestar ゴゴ Oh yeah, I mean in general though, not for this particular problem.

[@mathjitsuteacher]

Very nice video! There is another proof for this theorem which I like a lot. The proof consists in showing that every sequence has a monotonic subsequence which I think it's a very cool result. Once you have showed that, you only need to show that every monotonic and bounded sequence is convergent, which is easy because the candidates for the limits are either the inf if the sequence is decreasing of the sup if the sequence is increasing, of the set of terms of the sequence.

[@clayton2167]

Since you liked real analysis so much, You have to do functional analysis and complex analysis! I think since you like applied math, you'll like complex analysis most

[@theultimatereductionist7592]

This theorem makes use of the unstated assumption that we don't refer to a finite sequence as converging to anything. Otherwise, the theorem would be trivially true. We seek an infinite sequence that converges.

[@bjornfeuerbacher5514]

That's not an assumption, that's implicit in the definition of convergence.

[@HoloBoss]

we just did this in analysis 1

[@amyholmes8373]

Thank you so much! I have an exam tomorrow and your simple explanation helped so much!

[@DylanCVlogTV]

If sequence A converges, every sub sequence converges. If sequence A does not converge, the extremes of the sequence are bounded by at least two elements repeating infinitely and no elements ever repeating. In every case between those two extremes, there is an arbitrary value of n between two elements of the sequence that will always locate the next element in the subset that approaches the value of interest.

[@Jukka70]

I love seeing your videos, and I get to learn all different types of pieces in Math. I teach math to Adult Ed learners so the math level tends to stay from Algebra and below, though I do tutor Algebra II and sometimes a bit higher. I originally was going to be a math major, and had to take real analysis. It's funny how you mentioned the Real Analysis course, because I hated it, and left it twice. LOL, I think it's the only time I yelled at a professor. The nutcase talked to the birds outside, I swear. Anytime he got confused, he'd walk over to the window start to talk and then come back to the board, and then do it again. I sat there just wondering. Was happy that I no longer had to take that course. But to be honest, by watching how you explain things. If you taught the course and I was in it, not only would it be enjoyable, but it most likely would have started to make sense.

[@maxpercer7119]

great video, clear and insightful - relaxed voice and presentation.

[@wayneliang4524]

After watching this video, I realized the calc 1 I was doing and 2 that I am currently doing on open course in NTU is more similar to real analysis rather than calc1/2 in US

[@amaarquadri]

This is a shockingly simple and elegant proof!

[@dominiquefortin5345]

If all the integers from the interval are equally likely in a sequence positon then in an infinite sequence you will find an infinite count of the integer i (i being inside the interval). Because you can make a subsequence of only the elements equal to i and that a sequence with only the same integers converges to that integer, there always exists a subsequence that converges in such a sequence. QED. It also work for any real number because there are less infinite positons in a sequence than there are real numbers in the interval, so any real number will be present an infinite number of times in the sequence. (Correction: this doesn’t work for real numbers because it is not possible to pick enough real numbers to exhaust the R set while creating the sequence)

[@EpicMathTime]

What do you mean "if all the integers from the interval are equally likely in a sequence position"?

[@dominiquefortin5345]

@@EpicMathTime Pick an interval and make a set of all integers in that interval. This set is finite. If you make the sequence by picking an integer from that set at random, than for any position in the sequence you have an equal probability of finding any integer from the set.

[@EpicMathTime]

@@dominiquefortin5345 I understand the assumption you're making, but I don't understand why you're making that assumption.

[@dominiquefortin5345]

@@EpicMathTime I don’t quite know what you’re asking because writing is low bandwidth, but I’ll try an explanation on my idea and maybe you’ll be able to extract the answer to your question from that. My intuition was that by using a finite set (an interval contains a finite set of integer) to build an infinite sequence, I would be able to prove the theorem another way. If, while I’m building the sequence from the first position to the infinite position, I use a random integer from the set, I get the property that once I found the integer i at a position, I will find an infinite quantity of i in the following elements of the sequence. Combine this with the fact that the simpliste infinite converging sequence is a sequence where all it’s elements are the same and you prove that it is always possible to find a subsequence that converges. I was hoping that I could do the same for an interval of real numbers but I don’t get the property with this construction for reals.

[@EpicMathTime]

@@dominiquefortin5345 Alright, from what I understand, you are giving thoughts for a proof for the special case of sequences which take integer values. So, the first thing to point out is that your suspicion is correct; this argument cannot be extended to the reals. It relies on the range being finite. Second, you should abandon this strange restriction that the sequence has an "equal distribution" of all of the numbers in the range as entries. This is, of course, not generally true for sequences (consider the sequence 1, 2, 3, 3, 3...; 1 and 2 only appear once); and it doesn't really help with anything. So, here's my modification to your argument for bounded sequences of integers. Suppose we have a bounded sequence of integers. Then the range of the sequence must be finite, because there are only finitely many integers between any two real numbers. Since the range is finite, _at least one_ element x of the range appears infinitely many times in the sequence. So, the constant sequence of x is a convergent subsequence of our bounded sequence of integers. Hence every bounded sequence of integers has a convergent subsequence. *This all applies to any sequence with a finite range. As an aside, your thoughts about the sequences of integers in which more than two elements of the range occur infinitely many times shows that these sequences never converge. If you have a sequence in which x and y appear infinitely many times, then the sequence has a subsequence that converges to x, and a subsequence that converges to y. Hence, the sequence does not converge, because every subsequence of a convergent sequence converges to the limit of the sequence.

[@aylerayler]

I don't understand why it would converge on any particular number? Like I get how it can be endlessly split... bit why would it converge to any particular number? And why for instance were all the numbers he picked positive numbers?

[@bjornfeuerbacher5514]

"I don't understand why it would converge on any particular number?" Because the intervals get smaller and smaller, their length approaches zero. And an interval of length zero obviously can contain only one single number. (That's how you can visualize it, it's also possible to rigorously prove this - but you first have to understand stuff like Cauchy sequences.) "And why for instance were all the numbers he picked positive numbers?" Because in this example, the intervals which contained an infinite number of terms from the sequence all contained only positive numbers.

[@ayushdeep7900]

Very very creative!!! Btw, I really enjoyed you and the white board

[@Senya7893]

that is the best explanation I found so far. thank you !

[@rodrigocastillo9825]

Does it really need to be bounded by [- m,m]? Wouldn't work for any interval? [a,b]? a

[@aman-qj5sx]

Yes, it works for any interval that is clear by the same proof

[@EpicMathTime]

If it's bounded by any interval, then it's bounded by an interval of the form [-m,m].

[@sanjj_1]

@@EpicMathTime how

[@RowanAckerman]

@@sanjj_1 just add or subtract a constant.

[@EpicMathTime]

@@sanjj_1 This is obvious. For example, [1,3] lies inside [-3,3]. [-14, 4] lies inside [-14, 14], and so on. There isn't anything profound here. Any interval can fit inside an interval of the form [-m, m]. Explicitly, take your interval [a, b]. If |b| > |a|, then [a,b] lies inside [-b, b]. If |a| > |b|, then [a,b] lies inside [a, -a].

[@chewzhengwei5722]

Make a video about how to enter the space industry, like whether should we get a degree in aerospace or get a degree in mechanical engineering. Ohh and whether is it possible to get a master in aerospace by having a mechanical degree?

[@Jack_Callcott_AU]

I remember being blown out by this theorem when I was studying real analysis many moons ago. I wonder if it requires the Axion of Choice.??

[@frede1905]

But doesn't this proof require the principle that the real numbers is complete (that every Cauchy sequence converges), or alternatively that the intersection of an infinite sequence of nested closed intervals is nonempty (so there is an element of the real numbers that is common to all these nested intervals). The last property relies on the fact that closed intervals are compact. Proving either of these (that the real numbers is complete, or that closed intervals are compact) is not easy, and may even require the Bolzano Weierstrass theorem in the first place (although there are probably proofs of these facts that doesn't require that theorem). In any case, this proof seemed quite incomplete.

[@kaleabhip]

@11:28 "[For applied/school math] You know what's coming...**Deep sigh**....." Real analysis summed up in one line😆😆 Edit: this is why I love math Edit 2: As always amazing videos Zach

[@Myrslokstok]

I had like 12 different profs that I should know (how to do) this was one of them, but like writen in analazys and very hard to get. If I got all of them explained like this it would be supergreat.

[@hazevthewolf178]

Zach, if I may, allow me to point out some serious problems with your demonstration. The Bolanzo-Weierstrass theorem is about subsets of R_p (R X R....X R p times). From my old textbook Elements of Real Analysis, 2nd Edition, by Robert Bartle... Every bounded subset of R_p has a cluster point. On the real line, a cluster point of some subset, S, of R is simply a point y where one can always find a point x in an interval around y, no matter how small, that is in S. When one attempts to prove that a sequence converges, one looks at the range of the sequence, the * set * of all values that it actually manages to map to. Your proposed sequence is finite, since some element of a set cannot belong to a set more than once. Finite sets do not have cluster points and there are infinite sets which do not have cluster points. The set of integers is one such set as there are no integers between 1.5 and 2.5 other than 2 itself. The BW theorem doesn't apply to your set. Showing that a sequence converges amounts to showing that its range, is associated with a cluster point. Now, here's a sequence whose set is both infinite and bounded and its range set. {1, 3/2, 7/4...} and beginning with 1 = a_1, it's easy enough to see that 2 - a_n = 1 / 2^n, so this sequence is very definitely bounded. If you do the bit with the intervals, splitting the set into two parts at each stage, obviously, the interval on the right will contain infinitely many points and we can keep that up forever and always find a point in the nth subinterval that is a member of the range and less than 2. In other words, 2 is a cluster point of this set which happens to be the sequence of partial sums of the series... 1 + 1/2 + 1/4 + 1/8 + ... + 1 / 2^n + ... For fun, you might play with a circle in the real plane and draw the subintervals. As you do, you'll see an increasingly smaller square appear about some interior point of the circle, ever smaller, but never empty because your mapping into R_2 and the nested cells theorem. All in all, I enjoy many of your presentations and I'm a belled subscriber.

[@tetraedri_1834]

There are different versions of BW theorems, some more general than others. The one Zach proved is what most people encounter first, but it can be generalized to what you stated. For the purpose of giving a hunch of what proofs look like, I think it's a good idea to use examples that need minimal background, so doing the proof only in the special case of bounded sequences in R is a good choice.

[@EpicMathTime]

One often refers to the result you're talking about as "Bolzano Weierstrass for sets" and the one Zach proved as "Bolzano Weierstrass for sequences." The two are closely related. The issue is that you are applying Bolzano-Weierstrass for sets to the range of this sequence. This is a bit of a mismatch. The range of a sequence does not fully capture a sequence's information, so we should not expect to make conclusions about a sequence just by talking about its range alone. The proposed sequence has a finite range, but the sequence itself is not a finite sequence. When we select an interval, we pick the interval that contains "infinitely many terms of the sequence". I think you may be misinterpreting this; this does not mean that we pick the interval that contains infinitely many elements in the range of the sequence. It means we select the interval such that a_n is in that interval for infinitely many natural numbers n. That is, if we come upon an interval that contains 8, that interval contains infinitely many terms of the sequence, because a_n = 8 for infinitely many natural numbers n (namely, all even natural numbers greater than or equal to 14). I understand that this may be nonstandard language, because we aren't actually looking for an interval that contains infinitely many of anything in a strict set-theoretic sense. What we are really looking for is the interval whose pre-image under the original sequence is an infinite set of natural numbers.

[@djpatpat4763]

Why did I hear all this stuff in my first semester, while you guy's seem to be in higher semesters? (I'm a physics student in Vienna, Austria at the Technical University of Vienna and these are exactly the things we do in Analysis I. Btw I started this year)

[@mattiabruni5463]

I'm at my first semester of electrical engineering and we too (in Italy) start out with this kind of analysis. One of my prof.s said that in many places (US for example) the more abstract math is skipped in favour of more applied teaching, maybe that is indeed true.

[@TheRealInky]

This was really intriguing. Ty

[@paketbaand1523]

This proof only works in the real numbers, right? Since the subsequence you get is just a cauchy sequence and therefore not neccessarily convergent, except in a complete metric space.

[@EpicMathTime]

The theorem itself is only true in a complete metric space, so any proof of it will only work in a complete metric space.

[@returnexitsuccess]

Yes, easy counterexample in Q is to just take rational approximations of sqrt(2). Bounded sequence but doesn't converge in Q and no subsequence will converge in Q (because if a subsequence did converge in Q it would converge to the same thing in R, but the sequence converges to sqrt(2) in R)

[@EssedikAbdelRahimStudentBachou]

thats so cool for sure, but i always see people using sets to represent sequence, but the thing is i know from the definition that numbers cant be repeated so why dont you use ordered tuples that have the importance of both repitition and order

[@alan2here]

Presumably a sub-sequence of a sequence is a sub-list of a list.

[@brunomcleod]

1:18 Isn't -25 less than -10? Interesting video though, I watched this after the first video expecting it to be extremely hard to understand but it's quite interesting and not too complicated.

[@AssemblyWizard]

I'm confused as to what is the difference between calculus and real analysis is. I learned this in multiple calculus classes, and also what you're describing as "thinking outside the box" is what calculus class is. So is real analysis a subfield of calculus?

[@danielking8475]

How would you prove that the subsequences converge to the same limit using this technique?

[@BreezyTwice]

That was exquisitely explained. Amazing👌

[@Salmanul_]

I guess the title scared some people lol

[@EastBurningRed]

Is the proof still possible without the axiom of choice?

[@VaradMahashabde]

Man, I cane dangerously close to having the proof, did something similar in my calc 1 recently...

[@ramnair4402]

3:22 "nae niba nae nae niba"

[@Kudravets-Diana]

How to prove the first Wistras theorem? If a function is continuous on a closed segment then it is also blocked in the segment.. I see that i need to use with your proof in order to prove what I wrote down..

[@fatihsenguel]

why can you just pick the 6 as your first element? Dont you have to follow the indexes of your sequences due to the monotone property? Or am i wrong with that? In that case the first element of the subsequence would be a1 = 1

[@sanjursan]

How about displaying a statement of the theorem.

[@anshumanagrawal346]

What guarantees that you can keep doing this? What if both the intervals contain infinitely many terms? Wouldn't require like AOC?

[@sherkhanthelegend7169]

Plzz do math videos like this they are very interesting

[@okoyoso]

That's pretty neat! I just started learning real analysis.

[@KaliFissure]

Makes one want to do this for each of the trancendentals. Using modulo 10

[@juancorral7216]

I don't quite follow something. The statement that needs to be proven is "Prove that any bounded sequence has a subsequence that converges". Is {8, -8, 8, -8, ... } not a sub-sequence of the sequence you defined as (a_n)? And isn't that sub-sequence non-converging, since you said the sequence (a_n) itself also didn't converge because of that same repetition? One of these two questions has to be false, but I intuitively see both as being true, which would contradict the proof.

[@zachstar]

You have to prove that there exists SOME subsequence that converges, not that every subsequence converges.

[@juancorral7216]

@@zachstar Oh, I see. Thanks for clarifying!

[@tylern6060]

We literally did this in class yesterday.

[@alexander_adnan]

There’s many ways to prove the Bolzano weistrass theorem in R... but only one sexy way to prove it for C ( using topology results. And the Cauchy sequence behavior

[@lokegustafsson247]

Could one not generalize this exact proof with an oct-tree?

[@lokegustafsson247]

I mean quadtree, complex is 2d of course... But in general, this should generalize to R^n with bounding boxes and space partitioning?

[@HeavenlyPhilosophy]

Proof I thought of while the video was paused at 5:34. Edit: Interesting. The proof I thought of was not the same as the one in the video. Imagine that you run your finger from right to left across the interval. Eventually, you will pass an infinite number of terms in the sequence through that interval, since the interval contains an infinite number of terms of the sequence. Now, there has to be a specific point where one passes an infinite number of terms, such that if your finger is any to the right of that point, it has only passed a finite number of terms. That's because you can't have half of infinity or something like that. It's a sudden jump from having passed a finite number of terms in the sequence to an infinite amount. There might be multiple points where this happens. Now, there can be no two of these points right next to each other, because that is impossible on the real number line. For any two points on the real number line, there is always a third in between them. Now, we can create an interval that is inside the interval of the original sequence such that it contains one of these infinity-passing points and no others. Now, there will be an infinite number of terms of the sequence in this interval, because if your finger passes an infinite amount on that point, it's still going to within that interval. None of the points changes. So, we can create an interval of arbitrary smallness (as long as no other infinity-passing points are inside it) such that the interval contains an infinite number of terms. We can make the interval smaller, but it will still contain an infinite number of terms as long as that point is inside of it. This is getting to the definition of a limit. We take all the terms in this interval and order them by closeness to the infinity-passing point. If two points are of equal distance, one can have an arbitrary order. Now, it will be the case that for any distance epsilon from the infinity-passing point within the interval, one can create a smaller interval containing all the points within that distance epsilon. That new interval will still have an infinite number of points because it contains the infinity-passing point, and it will exclude the first terms in the new sequence one by one because you are shrinking the interval around the infinity-passing point. So, it is the case that for any arbitrary small distance epsilon, there is a term number delta such that all the terms after term delta will be within distance epsilon from the infinity-passing point, which means that the series we created converges on the infinity-passing point. And thus, The Bolzano-Weierstrass theorem is proven.

[@suryansh4715]

Can you make a comparison video between computer engineering and computer systems engineering

[@brockobama257]

I’m currently in real analysis and the rigorous proof is easier than comprehending the concept. So I don’t actually understand what I’m doing.