Real Analysis, Lecture 11: Compact Sets

Real Analysis, Spring 2010, Harvey Mudd College, Professor Francis Su. Playlist, FAQ, writing handout, notes available at: analysisyawp.bl...

[@frajirek90]

that feeling when you kinda understand compactness

[@jeremiahkeenan7572]

Great lecture! Thanks to proff Su and whoever posted this. Really important to students like myself who lack an understandable professor.

[@medhasinha8348]

I have literally never commented on a youtube video before but thank you so much for your lectures they have actually saved me in my real analysis class

[@QGF1]

Fantastic lectures. Much more helpful than my professor (at a research university).

[@Bignic2008]

Great videos. This guy is much more clear than my analysis teacher (though it might be too late to save me before the exam :( )

[@yash9shhreya]

Same

[@eduardoescoto8806]

Hope it went well bud, im probably in the same boat hahah

[@leadh99]

same : (

[@tianchenzheng7464]

omg same here. Exame tomorrow and I just discovered this guy.

[@YoYo-nt7yf]

Self studying because of C19 n my prof. Did didnt take any online classes for Analysis. My last sem. Exams gonna be nxt week. Im so lucky to be able to learn from this guy.

[@lijie2511]

Great lecture! Wish I had a teacher this good in college. Would have been nicer if the video quality is better.

[@bvanpelt8]

A subcover is a subcolletion of the open sets which cover the compact set. An easy idea is to consider a closed interval [a,b]. An open cover is any union of open sets (open intervals will work here) where [a,b] is a subset of that union. For example, you could have the interval [0,1] covered by the open intervals (-1/2, 1/2)U(1/4, 3/4)U(3/8, 3/4)U(2/3, 11/10). A cover could initially have an infinite number of sets. A finite subcover simply has a finite number of sets.

[@dekippiesip]

@45:00, wouldn't it be a lot easier to just exhaust the space X? Take any point x inside X and then cover X by considering the balls of radius n centered at X, and take the union over the naturals. Now You have a cover of X, hence of K. But since K is compact, you could extract a finite subcover, now take the biggest ball in that subcover, since it contains all other balls it contains K and hence K must be bounded. A much easier proof if you ask me!

[@tejasnatu90]

Think of Compactness by drawing parallels with vector spaces ... VECTOR SPACE ----------------- > METRIC SPACE VECTORS ----------------- > OPEN SETS SPANNING SET -------------------- > OPEN COVER SUB SPANNING SET(BASIS) --> SUBCOVER FINITE DIMENSIONAL VEC SPACE --> COMPACTNESS

[@yhuiuc]

@1:01:00 How many people say E is open in Y? I think E is neither open nor closed, since the points on the edges of Y are not interior points by definition.

[@Anthony-db7ou]

Yao Hu that didn’t make since to me either. The interior should be open, and the boundary closed. From what I know of open sets.

[@sergiohuaman6084]

@14:20 "argh, I actually have to do math" made my day!

[@YoYo-nt7yf]

Lol The IRONY 😂😂

[@LaureanoLuna]

I'd say an easier proof that every compact set K is bounded would be: pick a point p of K and consider an infinite set S of open balls Bi with center p and doubling radii; S is a cover of K and if K is unbounded S has no finite subcover, for if a finite subcover (containing e.g. Bk as its largest element) existed, K would be included in Bk. So, K is bounded if compact.

[@LaureanoLuna]

Seems to me this proof holds for any topological space with a metric-induced topology.

[@sgtcojonez]

Great lecture, but too bad that the camera is a potato.

[@studentmele]

Agree with Laureano, We can prove boundedness of compactness by proving that if a set is Not bounded it is not compact (hence, if is compact, it is bounded). The open cover, Vi(0) (index i by radius around reference point) where i goes to infinity can be assumed. This open cover, covers the entire metric space, hence is an open cover. However, since the assumed set is not bounded, non of the open covers can be eliminated from the set (no finite subcover). Hence desired....

[@bentupper4614]

I believe there may be a small error in the proof at 1:09:10. Prof. Su says, "E open in Y means every point x [in E] has a neighborhood that’s completely in Y and in E." He writes: N ⊂ Y ⋂ E. But it's possible that some x ∈ E will be a limit point of X \ Y, in which case no neighborhood of x will be completely in Y ⋂ E. Did he mean to write/say: N ⋂ Y ⊂ E ?

[@giovannacorsi8387]

Given E⊂Y⊂X, Y ⋂ E = E. The definition of Prof Su is correct. A set E is open relative to Y if, for every p in E, I can find a neighborhood of p which is completely contained in E (relatively to Y). I think that what confuses you is that the set E could include points which are on the boundary of Y, so that, apparently, for those points you cannot find a neighborhood which is completely contained in Y ⋂ E. The point is: Neighborhoods of p in E relatively to Y have a different shape than neighborhoods of p in E relatively to X. Think about this exmple: consider X to be R^2 and Y to be R. The interval (a, b) is open relative to R but it is not open relative to R^2. In R a neighborhood is an interval, while in R^2 a neighborhood is a disc. When you consider the Cartesian plane and you move along the x axis you are moving along the real line, which is you subset Y. If you are in Y, Y is you metric space and you can only move along the real line. If you are in X instead you can move in the x-y plane, therefore, as son as you move up or down from the x axis, you get points which are not in (a,b). This means that E could be open relative to Y even if it contained points which are apparently on the boundary of Y. If you want to prove that E is open relative to Y, when moving around p in E, you can only move in Y not outside of it. I do not know if I have been clear but English is not my first language. Bye!

[@Augustine_354]

Fantastic stuff. In particular as complement to Baby Rudin.

[@kristertrandal3287]

Remember that if Rudin is closed, the complement is open. :)

[@Lyagami31]

wonderful...teacher is enjoying teaching class and he is interactive while writing the proofs...:)

[@fahezunahmed3605]

The lecture is excellent, but the video is too blurry to watch and note down.

[@sophiamayne-deluca4623]

Definition at 26:53 but the lead-up is well worthwhile

[@janhenrikwiik8219]

Cuneytushka: I understand what you mean, but I am pretty certain his argument holds. Here is why: you cannot point to 0 and 1 being examples of numbers in the set (0,1) not being interior, because obviously neither 0 nor 1 is actually a part of that set. The idea is actually really simple; for any point in Y, since we can FIND that point in Y, it's either interior or it's a boundary point. It is a boundary point if every ball around it, according to the metric space we're working in, contains points both from Y and not from Y. But since we are working in a new, inherited metric space that only contains points in Y, no ball contains any point that is not in Y, so every point is interior. That is of course not the same as Y as a set being open or closed in X, though this is where the confusion arises, because the way he points at a point 'on the line' between X and Y, it seems like Y is assumed to be closed, or at least not open. If Y is open, you cannot talk to any point at the edge, because it's not part of our inherited metric space.

[@babyylovezmusiq]

You make so much more sense than my teacher

[@MrRk540]

These lectures really need an HD upgrade!!!

[@imapublicagent]

In R, (0,1) does not have any edge points, but [0,1] has two edge points, and both are interior.

[@bapaden]

These lectures are great! My real analysis professor sucks so I just watch these videos instead.

[@Caulfield789]

@Robert: By a subcover, we just mean that you pick some of the sets in your original cover and throw the other ones away. For example, I can cover [0,1] with the sets (-1/2,1/2), (1/4, 5/4), and (-1/2,5/4). So the collection V = {(-1/2,1/2), (1/4, 5/4), (-1/2,5/4)} is an open cover of [0,1]. But note that the collection V' = {(-1/2,1/2),(1/4,5/4)} is also a cover. And it's a *subcover* of V because I obtained it from V by just tossing out (-1/2,5/4). Does that help?

[@delayyu7958]

Nice lectures on analysis! But when can the lectures on Real Analysis 2 be posted?

[@brandomiranda6703]

What I find really confusing about his proof is that it seems that he constructs a open (sub?)cover to show compact sets are bounded, however, what doesn't quite seem right to me is that we need to show things are bounded for **every** open cover, not just his construction...not sure if anyone got how that part works...how does he make it such that his proof works for **any** open cover of K?

[@you_tube618]

We only use a specific open cover because we're not actually trying to prove a theorem about the open covers of K, we're trying to prove a theorem about K. Which open cover you happen to choose doesn't change anything about K itself.

[@fred321cba]

I don't think the proof outline @1:10:02 completes the argument. You then have to argue that for all the new points added to each neighborhood (i.e. points that are in X but not in Y), each one also has its own neighborhood that contains only points in G. You need to do this to show that all points in G are interior points (thus showing that G is open). Proof outline as follows: You can argue that every additional point will have its own neighborhood as follows. For each additional point g that is in the neighborhood of Nr(x) for a given point x with radius r, choose a neighborhood for point g with a radius of r - d(x,g). This neighborhood will have a positive radius since d(x,g) must be less than r (since g is in the neighborhood Nr(x)). All points in this neighborhood must be in G since the neighborhood is a subset of Nr(x) and Nr(x) is a subset of G. Thus all points g are interior points in X and thus G is open.

[@mambu6]

This means if im given any open cover of K(for example A _i where i is out of an index set I) i can chose finitely many A_i's out of that open cover which already suffice as cover of K

[@brandomiranda6703]

At first I thought that compact sets being bounded due to the finiteness but then I realized that K the set to be covered could expand to infinity...but that brings problems cuz then there exists covers that extend to infinity that couldnt be bounded...

[@Caulfield789]

@Robert: "...a subcover is simply a union of open sets covering another set." No, this is not quite what I mean. The subcovering covers the *same* set - namely, [0,1] - but the number of sets doing the covering is smaller than the number of sets you had in the original covering. The key idea is that you covered [0,1] with more sets than you needed to. In my example, using (-1/2,5/4) in the covering was unnecessary.

[@sahilguleria4979]

Thank you sir for such a wonderful lecture series, i really learnt a lot and enjoyed too while learning analysis from you 🌟🌟🌟🌟🌟🌟🌟, thank you so much

[@hamedgholami261]

54:00 Relative open sets

[@jintelcore2duoe8400]

at 1:02:38 why is the set not open in X?

[@kristertrandal3287]

+Jason Ngo The set E is not open in X because the points on the "straight edge" boundary are not interior points of E with respect to X. That is, any neighborhood (again with respect to X) around the points lying on these straight lines is not completely contained in E. Sorry for a late answer, you have probably figured it out by now :)

[@PRAKHALGOYAL]

I still don't understand. At 1:02:15 prof says that the the curvy set is open in Y. But the point on the 'straight edge' is not an interior point (I think he mentioned that that edge in included in the curvy set). If points on the 'straight edge' are not interior point then why is the set open in Y?

[@kristertrandal3287]

The clue is in how you define the neighborhoods. Notice in my previous comment that I speak of "neighborhood with respect to X", which means that the neighborhood itself must be CONTAINED in X when being a neighborhood with respect to X" (Don't confuse this with the definition of open sets). So, clearly E is not open in X (choosing neighborhoods contained in X), because you can pick points on the straight edge boundary and have your neighborhood go outside E (but still remain in X, which is crucial). Now, the set E is open in Y because now your neighborhoods are restricted to Y (that is, they have to be contained in Y in order to be neighborhoods with respect to Y), and any point in E will have a neighborhood contained in E (the definition of a set being open). For example, the points on the straight edge boundary are interior points of E when regarding neighborhoods with respect to Y. A neighborhood around a point on this boundary will be a half-disc, because the neighborhood itself must be restricted to Y. As can then be seen from the figure, this neighborhood will clearly be contained in E also, which is the criterion for being an interior point. When all points are interior points, the set is open.

[@PRAKHALGOYAL]

Thanks man. Got it :)

[@360MrCarter]

the example about open set being relative to metric seems wrong. R^n intersect R^m = Null Set if n!=m for all non negative integers, it makes no sense to talk about the open interval (0,1) in R to be in R^2.

[@jonathantraugott700]

He gives an example of an open set as "All triples of not co-linear points in the plane." What does this mean? Evidently we're not defining a metric on triples of points.

[@HarpreetSingh-ke2zk]

Hello Prof, its been a luck to have your lectures. I am looking for stochastic calculus, the prerequisite brings me here. Being in the same situation for many hours, your videos dimensions settling towards the expected vision. Thanks.

[@spydr4111]

@xatzin2004 If you take the limits of each endpoint as n goes to infinity,you will see that the unions of the open sets converge to (0,1). Thus you can always start counting at some arbitrarily high n because the limit will still converge.

[@vecter]

Is the following proof valid that all compact sets are bounded?: Pick any point x in the metric space and let our cover be the infinite collection of concentric open balls of radius 1, 2, ... around x. This covers K b/c d(x, y) < infinity for all y in K so we can at least one of the balls in our cover will eventually contain any given y in K. By the compactness of K, our cover has a finite subcover, which we pick to be the biggest ball around x that contains K. (I'm out of characters remaining)

[@jonaskoelker]

I'm persuaded, and I was about to post the same proof. I call it "supernova", as the point x "explodes" all over (the metric) space. I might add 1 to the largest radius in the subcover just to not mess up strict vs. weak inequality, but I'm not sure it's necessary. Actually, it's exactly because I'm not sure and I'm too lazy to check that I would do that :-)

[@navid7491]

one is open. the example is correct. from any n to infinity open 1 would be covered

[@brandomiranda6703]

What does it mean if u need an uncountable index set...?

[@dopplerdog6817]

My other car always has a finite subcover for any cover, so it's a compact car.

[@milksushi6640]

He says compactness doesn't depend on the space that you're in, but does it depend on the metric that you use? Like if I have a set in R^2 that is compact with the regular metric, does that mean it's also compact using the sup metric (or any other metric) instead?

[@depressivepumpkin7312]

hello from 2021 :) as a first thought, it should not matter, because he uses only generic properties of metric, without using any actual implementation. as a second thought, he said it didn't matter what __metric__ space you are in, and metric space consist of a space and a metric, so...

[@52bicycles]

I wish I had this guy for analysis.

[@michelleslee8040]

Such a charming Prof!

[@Carlosrv19]

Excellent explanation...! thanks.

[@한슨-f8j]

fantastic lecture !!Best ever

[@yku01993]

Thanks a lot. You are an amazing teacher!

[@panchajanya91]

Amazing teaching !!

[@mambu6]

No it isnt the definition in my book is perfectly clear def: A subset K of X is compact if every open cover of K has a finite subfamily which is also a cover of K

[@TerresEndormies]

"This is no different from the definition you had before." Except how it's different. The definition of "ball" has changed, and I can't read your mind. I'm fine with the adjustments, but a reasonable person should not be "with you" when you first appeal to the boundary of Y being interior, because it does not follow by deduction. It's pretendsies until you declare yourself. I am with you when you clearly acknowledge your alterations the the definitions, to which I am happy to agree.

[@scientificpages]

Big + plus for several first sentences.

[@yangliu2686]

this is fantastic!!!!! Math is fun, guys!

[@debjyotimaiti8146]

Ha ha ha.this is ridiculous

[@vishwanathlohar7326]

Good and helpful lecture

[@ahmedalghamdi3073]

so good lecture

[@Pradeep1989]

Nice lecture. Bad camerawork.

[@stephensaku827]

A helpful lecture.

[@DUNGSI27]

Incredible lecture 5*

[@nidakay6229]

Very very helpul!!

[@emanuelguzman-garcia3567]

Here I am. 30 minutes into a a math lesson I know NOTHING about. I don’t even know why I watched this. I don’t go to college nor do I take any math clases 😂. But now I want to know what these “sets” are.

[@zhenhuijiang5317]

Great professor

[@zwazwezwa]

Great teacher

[@HanzAlbertNguyen]

Professor Su might be not the best mathematician , but he has pedagogical talent!!! compared to many math professors i have known

[@elishadayag219]

Prof Su graduated from UT Austin when he was 19 and got his doctorate at Harvard. He's a pretty smart fella.

[@emlmm88]

Hanz Albert Nguyen Wow, that's a backhanded compliment if I've ever heard one.

[@inf0phreak]

While it's true that compact sets are bounded, the notion of boundedness in a metric space is near USELESS. Here are two examples why: a) Any infinite set with the discrete metric is bounded, but obviously not compact or in any sense small. b) Any metric space (X,d) has an associated bounded metric, d' defined by d'(x,y) = min(d(x,y),1). Under this metric *any* subset of X is bounded. In short: Boundedness only makes sense if you have a norm or a bornology (that's a real word. q.v. Wikipedia).

[@shawnwilliams77]

Yes; this sounds right.

[@dopplerdog6817]

Yo mama has no finite subcover.

[@gandalf85]

Are you with me?

[@peterultraman]

thank you very much!

[@kuldeepbansal2477]

Clearity bada

[@khonghengo]

great

[@asterixx6878]

My teacher skimmed through all of this in less than two minutes!

[@alainliu7601]

太伤眼睛了

[@Hosseinshafii]

the camera is awful! whats the point of uploading the video if no one can see anything ????

[@MathematicsFoundation]

video quality is bad