Hi! I’m revisiting general Topology concepts and your videos are really helpful for getting all the ideas back. It’s really important for people all around the world to have nice material like this for learning (I would have loved this when I first studied Topology), and I wanted to say thank you for uploading it. Cheers from Argentina.
Thanks for the viedo. I have a question at 12:57. Given that Vy is open and Y is compact, why is the intersection of Vy and Y necessarily open? The same trick is used later on.
The intersection of Vy with Y is open in Y under the subspace topology by definition. While it may not be open in X, it doesn’t matter because Vy intersected with Y only needs to be open so we can have an open cover for Y.
@@ethannguyen2754 I am still confused about this. As an open cover, each component needs to be open. It is true that Vy \cap Y is open by referring to a topology defined on Y, but this would be a different topology from the one defined on the entire Hausdoff space X.
I find something here that seems to contradict (my) common sense - but I'm probably wrong: So if Y is a compact subset of a Hausdorff space X then it must be closed in X. On the other hand, if some other Y is a closed subset of a compact set X, then Y must be compact. What about an open subset of a compact Hausdorff space? Can it be none-compact? If so, how can a subset of a compact set not be compact?
Sorry, and thanks for spotting my mistake! Let me correct what I said, (1,2) and (1,3) are not covers for [1,2] since they don't cover 1. The way I defined things, (-1,3) is not a cover for [1,2], but some people define covers differently where it is.
Compactness being a generalization of finiteness is a very nice insight.
my thoughts exactlly
Hi! I’m revisiting general Topology concepts and your videos are really helpful for getting all the ideas back. It’s really important for people all around the world to have nice material like this for learning (I would have loved this when I first studied Topology), and I wanted to say thank you for uploading it. Cheers from Argentina.
Thanks a lot for a clear and detailed explanation
We would like more with examples if possible
Thanks for the viedo. I have a question at 12:57. Given that Vy is open and Y is compact, why is the intersection of Vy and Y necessarily open? The same trick is used later on.
I think a cover should be defined as it contains X, not equal.
The intersection of Vy with Y is open in Y under the subspace topology by definition. While it may not be open in X, it doesn’t matter because Vy intersected with Y only needs to be open so we can have an open cover for Y.
@@k.y.5845 Yup, that is another glitch throughout the video.
@@ethannguyen2754 I am still confused about this. As an open cover, each component needs to be open. It is true that Vy \cap Y is open by referring to a topology defined on Y, but this would be a different topology from the one defined on the entire Hausdoff space X.
Great video, spent a couple days trying to understand compactness
Excuse me but i think you have a little error in minute 2:36, a family is a cover for X if it contains X, not if it is equall to it
This. I also want to know if this is a mistake?
In this case X is the whole space, so the union of open sets cant be bigger than X itself.
Btw it is not necessary X to be in the covering sets.
I find something here that seems to contradict (my) common sense - but I'm probably wrong: So if Y is a compact subset of a Hausdorff space X then it must be closed in X. On the other hand, if some other Y is a closed subset of a compact set X, then Y must be compact. What about an open subset of a compact Hausdorff space? Can it be none-compact? If so, how can a subset of a compact set not be compact?
Think about (0,1) in [0,1] closed in R. [0,1] is compact, but its subset (0,1) is not
Is (-1, 3) a cover for [1, 2]?
It might be trivial or not a good cover. Or is that not how covers work?
Yes, it's an open cover.
Neither (-1,2) nor (-1,3) are covers for the closed interval [1,2] since neither contain the point 1 in [1,2].
@@DanielChanMaths -1
Sorry, and thanks for spotting my mistake! Let me correct what I said, (1,2) and (1,3) are not covers for [1,2] since they don't cover 1. The way I defined things, (-1,3) is not a cover for [1,2], but some people define covers differently where it is.
@@DanielChanMaths I believe it is an open cover by your definition. If X = [1,2] is your top space, then (-1,3) = X in X and so covers X and is open.
I love you (no homo)