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?
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.
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.
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
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 ??
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.
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.
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its pretty funny how this is recommended to me, because i'm covering compactness of topological spaces right now lol
lol ya it’s weird, happens sometimes
really good video. thanks!
Thank you.
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You are welcome!
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?
as an undergrad, include every possible detail, even if it's overkill!
I believe that you used the result that A is compact in the proof.
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.
what text is it?
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.
Interesting I don't have that book yet!
The Math Sorcerer You can have my copy once I am done with it. 😄
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That's ok:) I will keep posting, thank you for the support my friend!!!!
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
But how does it follow that X/A is open?
thanks, it is well understood
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 ??
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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.
I don’t like this proof primarily because of when you say A contained in unions union X\A seems more like a contradiction
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.