Real Analysis, Lecture 13: Compactness and the Heine-Borel Theorem

[@hamedgholami261]

Recap 1:03 Closed Intervals are compact 6:27 Questions about proof 22:08 Closed and bounded sets are compact in R^n (Heine-Borel) 25:40 Question about proof 32:31 Examples of closed and bounded sets which are not compact 36:30 A set is compact iff every infinite subset of it has a limit point 44:43 Corollary (Bolzano-Weierstrass) 1:00:20 Cantor's finite intersection property 1:03:03

[@milksushi6640]

To be clear, the forward proof at 50:00 works because the set E is closed. It does not have a limit point in K, and it cannot have a limit point out of K because then that limit point would also be a limit point of K, and therefore a part of K (since K is closed). Therefore, E has no limit points, and therefore contains all its limit points and is therefore a closed subset of K (which we have proven means its compact).

[@brandomiranda6703]

Sorry I didn't quite understand your explanation, but with his proof E does not have to be closed. The argument is by contradiction. Assume E does not have a limit point in K. Then each x in K is not a limit point so every ball around x misses E. Thus, the balls around point in E are the only open balls that cover points in E and only cover 1 point from E. This can belong to some open cover of K. This is problematic because only the balls around points in E cover E. If we remove any of them, then we don't cover E and thus not cover K, which means we constructed some open cover of K that does not have a finite subcover (FS).

[@thelastcipher9135]

Wow. Thank you for that. And we can also say, since E is closed then E^c is open thus E^c U V_q's is an open cover of K with no finite subcover - contradiction. The frequency of the necessity of going back to the definitions is really high with proofs regarding compactness, ugh. But, it's fun!

[@user-rn9ww2bh8q]

First of all, thank you Pf. Su for providing such a great lecture. I just don't quite understand the example explained at 42:50, where we prove the set of "spikes", A, is closed. I think when he says each spike has a ribbon (a neighbor) around it that does not contain other spikes, it only means that every element in A is not a limit point of A. However, does it necessarily mean that A is closed (every limit point of A is not in A)? I thought we should start with an arbitrary function, say f, and show that there is a ribbon around f that does not contain any spikes. Could someone explain what I am missing here? Thanks in advance.

[@chriss6356]

when the student asks "why wouldn't this proof work for open intervals" and the teacher responds with "because the intersection could be one of the endpoints", it should be "the intersection could be empty". the proof that nested intervals are non-empty relies heavily on the inclusion of the endpoints because the supremum which is used in the proof is no longer necessarily less than the "b's"

[@brandomiranda6703]

interesting, I didn't notice that he said that!

[@sergiohuaman6084]

@@brandomiranda6703 @22:50 answering to some student question

[@56jmoney]

@48:45 an easier way to say it is, each q in K can be isolated by a neighborhood N(q)

[@brandomiranda6703]

56:40 make sense to me but wasn't sure why the following was correct: by contradiction assume K is not bounded. Assume also that every infinite subset has a limit point. If this is true then we can construct an infinite set (say the one he suggest |x_n|>n). This doesn't have limit point because limit points stop at some point (i.e. are fixed) but we assumed it did which is a contradiction =>

[@henrytian8612]

At 1:08:07, why is it that the intersection of a finite collection of Ks must be empty? All he used to arrive at that conclusion was showing that there exists a finite subcover {Ua1,...,Uan} of any K.

[@elizabethstevenson9707]

If the union of sets gives everything then the intersection of their complements gives nothing, it seems to suggest. I think that’s only true for disjoint ye oh well

[@XerulaGraphics]

In the final proof, that of the finite intersection property for arbitrary metric spaces, it took me a minute to see why the argument holds. He says assume the union of the K alphas is empty for the sake of contradiction. But for contradiction we would negate the antecedent. Here we are negating the consequent, so our strategy is actually a proof of the contrapositive. Also, in places he mistakenly writes E for K and z for q in the second-to-last proof, which really confused me for a while.

[@matthewdo8104]

for the proof at 21:50, why do we need to construct N_r(x)? wouldn't G_\widehat{\alpha} play the same role as N_r(x)?

[@findclue]

Thanks Pf. Su. I think you made some typo for the theorem right before the Bolzano-Weirstrauss one, or I understood the argument wrong. When we assume K is not closed. We want to show that there exists p in K such that it is a limit point of E, but it looks like you are concluding by saying "then ..." instead of "WTS", and I think such an E = {x_n} as you defined, but instead of z, I think we should stick to the letter p. Tell me if my argument sounds wrong.

[@timothyfortune3922]

In the first part--i.e., forward direction or only if-- of the proof at around 45 minutes, we don't know that the infinite subset E is compact do we?

[@brandomiranda6703]

What the conceptual idea behind HIS proof that compact sets are closed? Is it simply that the outside of K is open? There seems to be something crucial for that to work...

[@robertfulton6397]

I believe that in earlier lectures, prof Su showed that the complement of an open set must be closed, and the complement of a closed set must be open. In this case, by showing that the "outside" of K (the complement of K) is open, he uses his previous work to conclude that K itself must be closed. hope this helps.

[@jonalderson5571]

4:00 his proof is a little bit hand-wavey

[@gregoryzhang743]

Hello, I think z appeals with ambiguity. and also do you know what the limit point under this theory mean,(if K is compact, every infinit subset E of K has a limit point in K) it seems to be there exists one Limit point in every infinit set E that is in K in the theory, but in the backward proof, it seems to be the professor is looking for a point which is the limit of infinit subsets E rather than each limit point in each E. any comment?

[@brandomiranda6703]

Whats a product of compact sets?

[@jamesh625]

I don't see why the final proof is a proof by contradiction. Rather, he seems to show the contrapositive. Anyone care to explain?

[@josegallego-posada2066]

We want to show that the intersection of the K_a is non-empty. Assuming that it is empty leads to the conclusion that a *finite* intersection of Ks is empty. But this contradicts our hypothesis about finite intersections of Ks being non-empty.

[@studentmele]

In reference to the theorem that "Set K is cmpt every infinite subset E in K has a limit point in K". It seems to me that backward is not true, how about K is entire R. K is infinite (hence not compact) and any infinite subset in R has a limit point in R . However if we replace "every subset infinite E in K has a limit point in K" by "every subset E in K has a limit point in K\E" then it makes perfect sense.

[@sylvainbolduc9392]

I totally agree ! Its a shame, i trusted this guy till there... The backward cant be true...

[@dcpidude]

It's not true that any infinite subset in R has a limit point in R. A simple example are the integers; Z is a infinite subset in R, but has no limit points (at least as defined in Rudin; I know that in some references such as Pugh, limit point is defined differently. For the purposes of this course, a limit point must be distinct from the point about which a neighborhood is being considered). See en.wikipedia.org/wiki/Adherent_point and en.wikipedia.org/wiki/Limit_point

[@gregoryzhang743]

Ye, its been too long. K is compact if every infinite subset of K has a limit point in K. that is the argument.

[@findclue]

Then I wouldn't comment anything more about it. I probably asked very nitty picky question which might not be very important for the validity of the proof.

[@sachinmehra3543]

why we proof lemma

[@duckymomo7935]

Sachin Mehra Lemma is really just a ‘weak theorem’ for all intents and purposes, there is no actual distinction We prove a lemma To show a weaker result and use that result. In a future theorem.

[@sahilguleria4979]

Thnx a lot

[@yku01993]

thanks a lot Sir! khan

[@dopplerdog6817]

Highest tuition fees in the US and they won't buy a decent camera

[@mj6456]

Ok... KAYYYYYY