Every Closed Subset of a Compact Space is Compact Proof

[@samir.mahto.925]

Thanks Sir From INDIA🇮🇳

[@jasspreet200]

some love for you from India ❤❤, thanks

[@krisgearhart7427]

Oh my god, I wish I had found you during my undergrad. HAIL!

[@TheMathSorcerer]

Hail!!

[@curiouskoala411]

really good video. thanks!

[@ericryu599]

its pretty funny how this is recommended to me, because i'm covering compactness of topological spaces right now lol

[@TheMathSorcerer]

lol ya it’s weird, happens sometimes

[@alijoueizadeh2896]

Thank you.

[@pez4]

All Hail the Math Sorcerer!!! You have saved me once again!!!!! Thank you so much!!!

[@TheMathSorcerer]

You are welcome!

[@aboyhya612]

I believe that you used the result that A is compact in the proof.

[@squarerootofpi]

Hey thanks! I don't know if I speak for everyone but as a beginner in proof writing one nagging doubt I have is "how much is enough?" When books write proofs, or when professors write them, the written word is almost always skeletal and accompanied with verbal commentaries, like you said, to save on writing so much. But as a student who submits a proof (say, as part of a test), how skeletal can the written portion be, considering whoever grades the work won't hear you speak, for it to be mathematically complete?

[@TheMathSorcerer]

as an undergrad, include every possible detail, even if it's overkill!

[@ntvonline9480]

Thanks for the compact set videos! Hearing about them is easier and more understandable than reading the text. I would bet that you have never featured our text on one of your best books videos. The author and his brother unfortunately both write unreadable texts.

[@TheMathSorcerer]

what text is it?

[@ntvonline9480]

The Math Sorcerer Analysis with Introduction to Proof by Steven R. Lay. His brother David wrote the Linear Algebra text we use at NOVA. Their father was also a famous mathematician.

[@TheMathSorcerer]

Interesting I don't have that book yet!

[@ntvonline9480]

The Math Sorcerer You can have my copy once I am done with it. 😄

[@TheMauror22]

But how does it follow that X/A is open?

[@mohammadamanalimyzada1462]

love you buddy thank you I wish i could join your channel but this service is not available in my country

[@TheMathSorcerer]

That's ok:) I will keep posting, thank you for the support my friend!!!!

[@antoniodeoliveiranginamaub2845]

thanks, it is well understood

[@J_Stockhausen]

Hi, thanks for the video. I still have a question, where do we make use of the hypothesis that A is closed. Or why is this not true when A is open. Thanks ! Edit: It's because A closed implies A^c open so it can be added to the open cover. :B

[@mikee-fl8ex]

Ur so good man😂

[@jasspreet200]

sir you are looking like my favourite Newton 😊❤

[@richardbloemenkamp8532]

I think this proof is insufficient or insufficiently clear. I would have preferred a proof by contradiction, something like: 1 Assume that O_I_A is a cover without a finite subcover that covers A: We need an infinite number of open sets from O_I_A to cover each element (point) of A. 2 Since X = A u (X\A) and A is closed we have that (X\A) is open we can add (X\A) as one extra element to this cover and arrive at a cover of X. let's call is O_I_X. 3 Since (X\A) is completely disjunct from A, O_I_X does also not contain a finite subcover that covers X = A u X\A. 4 However since X is compact O_I_X must have a finite subcover. 5 We have a contradiction therefore (1) is not possible and therefore A must be compact. My step 3 still seems a bit weak to me.

[@laishrammaikelsingh3860]

U_(alpha) s are open in A as it is open cover for A . For them to be in the open cover of X ,they have to be open in X first . Are they open in X too ??

[@motherlandlover147]

You look scientist Newton

[@nated8531]

I don’t like this proof primarily because of when you say A contained in unions union X\A seems more like a contradiction

[@nicholashayek5495]

First-year math undergrad here, so take what I say with a grain of salt: suppose A is a subset of the Reals. Then A is a subset of A U R\A, which is the mechanic at play that you see as contradictory... my point being, A can still be a member of a set which contains its compliment, i.e. A := {1, 2, 3} can be a subset of S U N\A, where S is {1, 2, 3, 4} and N\A is {4, 5, 6, ...}. In fact, since in the example X\A *cannot* be a superset of A, then all other elements of the set (O1, O2,..., On) must necessarily be the superset.