Properties of Compactness

Рет қаралды 7,455

3 жыл бұрын

Compact sets enjoy some mysterious properties, which I'll discuss in this video. More precisely, compact sets are always bounded and closed. The beauty of this result lies in the proof, which is an elegant application of this subtle concept. Enjoy!
Compactness Definition: • Compactness
Topology Playlist: • Topology
Subscribe to my channel: / drpeyam
Check out my TikTok channel: / drpeyam
Follow me on Instagram: / peyamstagram
Follow me on Twitter: / drpeyam
Teespring merch: teespring.com/stores/dr-peyam

Пікірлер: 30

@carterwoodson8818 3 жыл бұрын

Coming here after tonight's stream, this is so good thank you Dr. Peyam!

@coreymonsta7505 Жыл бұрын

The proof for compact implies bounded I had in mind was to consider the open cover consisting of all open balls (any radius works) centered at some point in E. The lemma to use for this proof is that the finite union of bounded sets is bounded (and the super trivial result that open balls are bounded). Which is all fine but your way involves a cover which isn't just completely abstract lol

@AhmadAhmad-qx6fp 3 жыл бұрын

There's compactness.. And then.. There's para-compactness To which Mary Rudin (of Rudin fame), proved in late 60s, that Every Metric Spaces is para-compact Such a beautiful proof in its own right

@thedoublehelix5661 3 жыл бұрын

I wonder how they came up with the finite subcover idea. I would've thought it was impossible to define compactness purely as a topological notion.

@drpeyam 3 жыл бұрын

I wonder that too! Maybe from algebraic topology, or a mathematician shuffled a deck of cards?

@kamilziemian995 3 жыл бұрын

I'm not sure, but I belive that first step was Borel's proof that when you have (in first version only countable) cover of closed interval on real line, there is always a finite subcover, what is know now as Heine-Borel theorem. And these open the way to proving the propeties of closed interval just using notion of existence of finite subcover. Having that there is not hard to imagine that they discovered, that it is a defining property of such sets. en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem

@dgrandlapinblanc 2 жыл бұрын

Ok. Thank you very much.

@Demki 3 жыл бұрын

Small correction to the definition of boundedness: We do not require that x is in E, for example the empty set is bounded and closed (as it is compact) but there is no x in the empty set. The proof that a compact set is bounded doesn't require that x is in the set either, you can take any x in the space.

@aneeshsrinivas9088 2 жыл бұрын

Why can't you just add on that the empty set is considered bounded to the definition directly?

@juanaedo6870 3 жыл бұрын

Justo estoy empezando con mi curso de topología.

@drpeyam 3 жыл бұрын

Fenomenal, buena suerte!

@rishabhbhutani5835 3 жыл бұрын

13:13 ---- LOL btw beautiful proof

@arturcostasteiner9735 3 жыл бұрын

Another proof that E is bounded: Pick r > 0 and let U= { B(x r) | x is in E}. Then, U is an open cover of E and, since E is compact, U contains a finite subcover U'. Since U' is a finite collection of bounded sets, the union of its members is bounded. And since E is a subset of this union, E is bounded. Actually, this shows E is totally bounded: for every r > 0, E is covered by a finite collection of balls of radius r.

@xshortguy Жыл бұрын

is it worth looking into compactness in spaces without distances, or are those just parlor tricks?

@drpeyam Жыл бұрын

Of course, compactness also works in pure topological spaces, with coverings

@tomkerruish2982 3 жыл бұрын

It's possible I've missed it, but do you ever mention that "closed and bounded" is equivalent to "compact" only in the finite-dimensional case?

@drpeyam 3 жыл бұрын

Yes, in the heine Borel theorem

@tomkerruish2982 3 жыл бұрын

@@drpeyam Thank you. I recall being surprised that it didn't hold in the infinite-dimensional case, requiring the set to be totally bounded rather than simply bounded.

@bramlentjes 3 жыл бұрын

Good observation! In an infitine dimensional space the closed unit ball is for example closed and bounded but not compact (Riesz) :)

@citizencj3389 2 жыл бұрын

@@bramlentjes it is locally compact I believe.