Real Analysis, Lecture 13: Compactness and the Heine-Borel Theorem
Why Mathematics Stack Exchange recommends it
Mathematics Stack Exchange users link it when explaining or clarifying compactness, finite subcovers, limit points, and the finite intersection property, often pointing to specific parts of Su's proofs.
Recommended by 1 people across 1 different months, 2013 to 2018.
What it is
A university lecture from Francis Su's Spring 2010 Real Analysis course at Harvey Mudd College, covering compactness, open covers, and the Heine-Borel theorem in the style of Rudin's text. It is a full-length chalkboard lecture with a companion playlist and notes.
What people said
Recommends kahen · 2013-04 · Mathematics Stack Exchange · on “Does the union of open neighborhoods of all points in a metric space cover the metric space?”
In response to your latest edit: I see you can make open covers of $M$, but I see *nothing* in your question that hints at how you want to extract finite subcovers. Might I suggest you watch Professor Su's presentation of compactness on Youtube. Lectures 11 [LINK] , 12 [LINK] and 13 [this video] . If I recall correctly, he's teaching from "Baby Rudin".
Cites user241043 · 2015-05 · Mathematics Stack Exchange · on “$K$ is compact iff every infinite subset $E$ of $K$ has a limit point in $K$”
[this video] The professor did this proof at 59:00, but I don't quite understand the way he did this. Basically he's using Heine-Borel Theorem: If $K$ isn't bounded, choose $\{X_n\}$ such that $|X_n|>n$ (what is this??). $K$ has no limit point. if $K$ isn't closed, there exists a limit point not in $K$. Choose $\{X_n\}$ s.t. $d(x_n,z)<1/n$ (what is $z$??). $\{X_n\}$ has limit point at $p$. Cans someone help me to elaborate it?
Cites Charlie Parker · 2018-04 · Mathematics Stack Exchange · on “Why is it obvious that if the intersection of all sets is empty then its complements cover the whole space?”
I was watching Harvey Mudd's real analysis video [this video] talking about the finite intersection property. The first step of the proof relies using contradiction and thus starts by saying that the intersection of all the sets is empty: $$ \cap_{\alpha } K_{\alpha} = \emptyset $$ Then the professor says the following (perhaps paraphrased a little bit): If a point is in the intersection then it means that point is not in any of the complements. So if there is no point in any intersection then some point must be in some complement. The first part of the statement seems obvious to me, if one is in all the sets then thats the same as saying that point must be in every set simultaneously. If one is