Real Analysis, Lecture 12: Relationship of Compact Sets to Closed Sets

[@arjunbanerji]

Great teaching Prof. Su! I like that friendly atmosphere of the class!

[@giovannacorsi8387]

Beautiful lecture! Professor Su has explained everything with clarity and precision :)

[@shakesbeer00]

Thank you very much Dr. Su!

[@trukkstop1]

At 23:00, begins proof that Compact sets are closed. An alternate proof begins by assuming that set X is not closed. Then there exists a limit point p of X with p not in X. Every ball around p contains points of X, and I think you can use this to construct an infinite cover of X with no finite sub-cover. I have been trying to understand how these proofs work if the metric is the "Discrete Metric", where d(x,y)=1 when x~=y. It seems confusing. Last comment: at 1:12:30, his proof that R is uncountable works very much like Cantor's diagonal proof. The nested intervals constructed show why the Rationals do not have the "least upper bound" property, without resorting to any algebraic properties of the Rationals.

[@jonhillery7736]

You can construct such a cover using these points, but not the obvious one: if you just take the balls around p, then if you take any particular N_r (p), only finitely many of the points of X (of the ones that are found in these neighborhoods of p) will be outside of N_r (p), so you could easily have a finite subcover. However, let's instead create a sequence of points "converging" to p (convergence is defined in a later lecture, but it is not important for this argument). To do this, just take balls of radius 1/n around p for all n and find the point of X guaranteed to be in the ball. You can then remove the points that show up multiple times in a row (if the are any). Call this sequence {y_n}. Around each y_n in the sequence let E_{n}=N_{r_n} (y_n) where r_n = d(y_n, x) - d(y_{n+1}, x). Therefore, each y_n in this infinite sequence is in a different ball than any of the other y_n. We can extend this cover to cover all of X by just taking sufficiently small radii around each point that was not in this sequence, so we have constructed a cover of X that has no finite subcover. Note that the reason this argument fails if p is in X is that the cover we constructed could not be extended to also cover x without creating some overlap that allows us to take finitely many of the neighborhoods.

[@nabilakdim2767]

@Hugh Jones : We got rid of the dependence on the embedding space by defining compactness (vs open/closed subsets, both notions depend on the bigger set), but we had to keep the same metric in order for things to hold for compactness, I think we do the same and we get freed from the used metric when we define topological sets.

[@matthewdo8104]

Shouldn't the radius at 40:40 be min{d(p, q_i) / 2}?

[@brandomiranda6703]

What about his proof is special about closed sets? Why wouldn't it work if the sets were open? Is it because the "last set" I_infinity could be such that the lower interval equals the upper interval and thus its just empty?

[@sergiohuaman6084]

37K views for an Analysis course. I am starting to know KZbin for education and I'm lovin' it!

[@HotPepperLala]

0:53:28 when break is over

[@SaulBerardo]

I believe there is a mistake in the statement of the "Nested Interval Theorem" in the end of the lecture. Instead of stating that "Nested closed intervals in R are not empty", probably he meant that "The *intersection* of nested closed intervals in R are not empty". It is the theorem 2.38 in Baby Rudin (3rd edition).

[@sergiohuaman6084]

very famous property indeed in first classes of Analysis, I learnt. NI property is Theorem 1.4.1 of Understanding Analysis by Abbott. Also important is the Archimedean property.

[@dariodemattiesreyes3788]

Hi everybody, thank you very much for the lectures. They have been of great help to me. I have a problem with this lecture 12. It suddenly stop working at around 0:46:15 and I can't see the remaining of the video. I think the video could be damaged or something like that. Could you help me with this problem? These lectures are very important to me. Thanks.

[@Fematika]

It's a problem with your computer.

[@sovankumardas5200]

How to prove it by limit point

[@brandomiranda6703]

Why do we need to start with an open cover of B rather than with K?

[@BareClause]

You should make some lectures on compact lie groups

[@inf0phreak]

48:30 can any set be given a metric s.t. that it's compact? No. A compact metric space is seperable, and metric spaces are 1st-countable Hausdorff. It's a theorem that a seperable 1st-countable Hausdorff space has at most the cardinality of the continuum. Hence I^I, the set of functions from the unit interval to the unit interval, does not admit a metric under which it is compact --- though it is compact in the product topology by Tychonoff's theorem.

[@studentmele]

Nq(r/2) and Np(r/2) intersect in the point that is in the middle??

[@jamesh625]

Recall the definition of a ball: Nr(x) = {y : d(x,y) < r}. So neither Nq nor Np contain a point y such that d(y,p) = d(y,q) = r/2. Therefore, they do not share a point. Hence, they are disjoint (they do not meet in the middle or anywhere else).

[@sahilguleria4979]

Thank you 😊

[@dopplerdog6817]

Does a compact car stop being compact if the doors are open?

[@brandomiranda6703]

Is every closed subset Y open in Y? It seems yes cuz if we perturb points in Y and only move to points in Y then we are open in Y...right?

[@mingzeyu7525]

Every metric space is open and closed relative to it self.

[@mingzeyu7525]

Or Clopen

[@yku01993]

Thanks a lot!

[@sakhilefatyi7561]

thanks alot

[@zwazwezwa]

Damn it, I blinked

[@sergiohuaman6084]

jajajaja @1:13:15

[@guytrolol5303]

NICE MEME :D