For the proof of (c) @ 1:04:10 I would suggest adding a note to cover the trivial case where there are no points in the intersection (i.e. the intersection is the empty set) in order to be thorough. For this trivial case, the theorem also holds since the empty set is also "open".
Zaid, I'm not quite I know what you're arguing, but the goal at 54:30 is to only show there exists at least one infinite intersection of open sets which is not open. There are, naturally, infinite intersections which are open. For example, the infinite intersection of (-n,n) as n goes to infinity.
It is becoming more and more clear that I am wasting my tuition money. My professor is a highly decorated math genius, but he is totally autistic and speaks in monotone.
From 11:00 mark, talking about closure of set is closed. The "fictitious" point P is badly drawn. Existence of such isolated point does not contradict that such set is not closed. There are two cases, either closure contains points that contradict closeness (which is what the picture says) or don't contain enough points to ensure closeness. The picture drawn is talking about the former case, that is, closure of a set creates a fictitious point P as demonstrated in the picture
Given that a metric space, X, in its entirety, is both a closed and an open set, we must be careful in saying that the Rationals Q are not closed. As a subset of the Real metric space R, Q is not closed. As a metric space itself, Q is closed, and thus Q contains all its limit points. In fact, an arbitrary subset S of R is a metric space by itself, containing all its limit points, with the metric of R restricted to S. The fact that Q does not have the "least upper bound" property ["lub-property"] does not tell us that Q is missing any of its limit points. Thus, the lub-property is a condition requiring the existence of a limit point, when certain sets exist.
Sometimes the explanation is confusing. Too many "this" and "that". The video has a really bad quality and the professor keeps using the words this and that to refer to statements that the person on You Tube is not able to read.
His proof for closure being closed seems silly and over complicated, he defined A closure to be the union of A and A' so it has all the limit points by definition. Not sure what there is to prove...
It's not obvious that adding limit points of A won't introduce new limit points (limit points of limit points, if you will) If you read Baby Rudin(the textbook) Rudin actually give a nice & clear proof of this result using the open/close complement theorem, so it is not trivial
ah, I see. Thats seems as an odd intuition because closure is defined as having ALL limit points so those should be included recursively if that existed. I will check Rudin's proof.
Though the weird thing to me is that he proof takes like 10-20 minutes to explain. There must be a conceptually concise way to explain why that (weird) scenario wouldn't happen. Do you know it?
I saw baby Rudin's proof. It doesn't seem to me to capture the intuition that you mentioned (that no new limit points are not introduced). I see why ur complaint that E' unioning might include new limit points, but Rudin's proof depends on a proof that depended on the definition he gave for limit points so it doesn't feel that special, especially if he defined E' to be all the limit points of E **by definition** (then there are all there by definition). I guess I understand ur 1st complaint better, it doesn't seem that weird to suggest that new limit points might be created or something.
I struggled in the first minutes of the lecture. But that's what lectures are for! To expand your understanding. Thanks, professor Su.
sergiohuaman6084
34:35 The relationship between open and closed sets.
For the proof of (c) @ 1:04:10 I would suggest adding a note to cover the trivial case where there are no points in the intersection (i.e. the intersection is the empty set) in order to be thorough. For this trivial case, the theorem also holds since the empty set is also "open".
Zaid, I'm not quite I know what you're arguing, but the goal at 54:30 is to only show there exists at least one infinite intersection of open sets which is not open. There are, naturally, infinite intersections which are open. For example, the infinite intersection of (-n,n) as n goes to infinity.
It is becoming more and more clear that I am wasting my tuition money. My professor is a highly decorated math genius, but he is totally autistic and speaks in monotone.
35:02 if u want the proof closed iff open
From 11:00 mark, talking about closure of set is closed. The "fictitious" point P is badly drawn. Existence of such isolated point does not contradict that such set is not closed. There are two cases, either closure contains points that contradict closeness (which is what the picture says) or don't contain enough points to ensure closeness. The picture drawn is talking about the former case, that is, closure of a set creates a fictitious point P as demonstrated in the picture
Can you expound upon his proof? I don't get his logic.
Thank you, yor lectures are very useful
If phi is clopen then the argument at 54:30 is wrong .
Given that a metric space, X, in its entirety, is both a closed and an open set, we must be careful in saying that the Rationals Q are not closed. As a subset of the Real metric space R, Q is not closed. As a metric space itself, Q is closed, and thus Q contains all its limit points. In fact, an arbitrary subset S of R is a metric space by itself, containing all its limit points, with the metric of R restricted to S. The fact that Q does not have the "least upper bound" property ["lub-property"] does not tell us that Q is missing any of its limit points. Thus, the lub-property is a condition requiring the existence of a limit point, when certain sets exist.
Sometimes the explanation is confusing. Too many "this" and "that". The video has a really bad quality and the professor keeps using the words this and that to refer to statements that the person on You Tube is not able to read.
His proof for closure being closed seems silly and over complicated, he defined A closure to be the union of A and A' so it has all the limit points by definition. Not sure what there is to prove...
It's not obvious that adding limit points of A won't introduce new limit points (limit points of limit points, if you will)
If you read Baby Rudin(the textbook) Rudin actually give a nice & clear proof of this result using the open/close complement theorem, so it is not trivial
ah, I see. Thats seems as an odd intuition because closure is defined as having ALL limit points so those should be included recursively if that existed. I will check Rudin's proof.
Though the weird thing to me is that he proof takes like 10-20 minutes to explain. There must be a conceptually concise way to explain why that (weird) scenario wouldn't happen. Do you know it?
I saw baby Rudin's proof. It doesn't seem to me to capture the intuition that you mentioned (that no new limit points are not introduced). I see why ur complaint that E' unioning might include new limit points, but Rudin's proof depends on a proof that depended on the definition he gave for limit points so it doesn't feel that special, especially if he defined E' to be all the limit points of E **by definition** (then there are all there by definition).
I guess I understand ur 1st complaint better, it doesn't seem that weird to suggest that new limit points might be created or something.
The proof for E-bar being closed is a bit messy not that its not good its just he goes back and forth on different "topics" in an unstructured way
Not at all. Your failure to understand his very slow (my only complaint) lecture is most definitely your issue.
miloinindo ur so smart!, good for u