Dr. Peyam, you are a blessing to me. I now understand mathematical concepts fully. You have made me a pro in mathematics....i wish i can have a one to one live interaction with you. I need your motivation and mentorship please.
Thanks a lot, Actually I have enrolled in IIT(One of the prestigious institutes in INDIA) in dept. of Mathematics as PhD student but I am a major in statistics, and we have compulsory Maths Courses that every student from maths dept. have to take during coursework, but actually, I am only one in my class from stats background and during teaching metric spaces my professor thinks that everyone has a major level of knowledge in mathematics, Its very hard transition, and sometimes feels like I will be kicked out, but thanks to you for explaining it so easily
It's worth noting that connectedness is really a property of spaces. Otherwise [0,1] u [2,3] would be connected, since the components aren't open when considered as subsets of R.
@@hrs7305 But for E = [0,1] u [2,3] as a set in R, there aren't open sets A and B such that E = A u B. The union of a finite number of open sets is open, and E is not open when considered as a subset of R. The point is that when the definition refers to "open sets A and B", it means "open" as sets in the space E, where, for example, there aren't any points not in [0,1] within 1 of a point in [0,1]. What doesn't matter is what space with that metric you consider E to be a subspace of. It's kind of a dumb technicality, but if you don't handle it, people are going to wonder how non-open sets can be disconnected, when the answer is just that every set is open when you're considering whether it's connected or not.
@@iabervon note that when you are using open set definition of connected it has to be applied considering E as the space itself and definitely [0,1] and [2,3] are open in E Anyway only intervals are connected in R
25:30 I thought that γ([0,1]) was just the path in C and not all of C? It seems like γ should be a function mapping all of R or a connected interval of R (that includes [0,1]) to C. Anyway, thanks a lot for these videos. Very interesting.
He means open in the subspace topology on [a,b], not open in R. That basically means we ignore the rest of R and take [a,b] to be the whole space. So is 'a' a boundary point? That would mean every neighborhood of 'a' contains a point outside of [a,b]. But there are no points outside of [a,b]. Therefore 'a' is not a boundary point.
@dwaipayansharma2282 any finite set is compact. Indeed let X = {x_1, x_2, ... ,x_r} be a finite topological space. Let U be any open cover for X. As U covers X, there is an open set U_i such that x_i is in U_i, for all i =1,2,...,r. But then U_1, ... U_r is a finite open cover for X! As U was arbitrary, X is compact.
Dr. Peyam, you are a blessing to me. I now understand mathematical concepts fully. You have made me a pro in mathematics....i wish i can have a one to one live interaction with you. I need your motivation and mentorship please.
Keeping us connected is your gift to us, sir! 😁
Thanks a lot, Actually I have enrolled in IIT(One of the prestigious institutes in INDIA) in dept. of Mathematics as PhD student but I am a major in statistics, and we have compulsory Maths Courses that every student from maths dept. have to take during coursework, but actually, I am only one in my class from stats background and during teaching metric spaces my professor thinks that everyone has a major level of knowledge in mathematics, Its very hard transition, and sometimes feels like I will be kicked out, but thanks to you for explaining it so easily
I know your pain 👍
It's worth noting that connectedness is really a property of spaces. Otherwise [0,1] u [2,3] would be connected, since the components aren't open when considered as subsets of R.
Nah they are disconnected you can prove as long as your metric is same it doesn’t matter in which space you set E is embedded in
@@hrs7305 But for E = [0,1] u [2,3] as a set in R, there aren't open sets A and B such that E = A u B. The union of a finite number of open sets is open, and E is not open when considered as a subset of R. The point is that when the definition refers to "open sets A and B", it means "open" as sets in the space E, where, for example, there aren't any points not in [0,1] within 1 of a point in [0,1]. What doesn't matter is what space with that metric you consider E to be a subspace of.
It's kind of a dumb technicality, but if you don't handle it, people are going to wonder how non-open sets can be disconnected, when the answer is just that every set is open when you're considering whether it's connected or not.
@@iabervon note that when you are using open set definition of connected it has to be applied considering E as the space itself and definitely [0,1] and [2,3] are open in E
Anyway only intervals are connected in R
Connectedness seems important for continuity and homotopy
This is amazing and mind boggling, the proofs made me want to dance! hahaha
Thank you very much. Interesting.
"Connexité par arcs" : WOW. Vous parlez français, Dr Peyam? Super vidéo, comme d'habitude ;)
Oui en effet 😁
Thanks prof!!!
Thanks alot. This helped me so much :).
So a set is disconnected if you can break things up into two tracks that can be followed, but multi-track drifting is not allowed.
25:30 I thought that γ([0,1]) was just the path in C and not all of C? It seems like γ should be a function mapping all of R or a connected interval of R (that includes [0,1]) to C. Anyway, thanks a lot for these videos. Very interesting.
How was A open in the example for intermediate value theorem
Isn’t a in the boundary because of f(a)
I have the same question
He means open in the subspace topology on [a,b], not open in R. That basically means we ignore the rest of R and take [a,b] to be the whole space.
So is 'a' a boundary point? That would mean every neighborhood of 'a' contains a point outside of [a,b]. But there are no points outside of [a,b]. Therefore 'a' is not a boundary point.
@@martinepstein9826 okay I had no idea of that, thank you
{a,b} connected and compact?
Compact but not connected
@@drpeyam it is not connected because it is a union of two disjoint sets namely {a} and {b}..right?
But why it is compact?
@dwaipayansharma2282 any finite set is compact. Indeed let X = {x_1, x_2, ... ,x_r} be a finite topological space. Let U be any open cover for X. As U covers X, there is an open set U_i such that x_i is in U_i, for all i =1,2,...,r. But then U_1, ... U_r is a finite open cover for X! As U was arbitrary, X is compact.
The random intro rofl
모든 수학적 사실이 진짜라면 좀 무삽긴해요.
In your example, A U B = [a, b] - {c}, not [a, b]. OK?