Haven't even finished the video yet, but I have learned more so far than the past 8 weeks in my Analysis class. As a college student, I say thank you and greatly appreciate the video!!!!
As far as I understand, the given definition of the accumulation point, as a point such that set X\x still has other elements, is incorrect. The definition I came across (in a book on set theory by Ralf Schindler) is: x is X's accumulation point iff for every a
striderpsv I would just write it as \doubleN. Actually I watched the video and the number line I drew made it look like I was talking about integers, \doubleZ. If you had to use set notation you would say {1,2,3,...} for naturals.
My question is that you have to expand the set of Natural numbers to include the Real numbers in order for you to define an open set containing n element of N that contains no other N. Is this justified?
This makes no sense.. Say the open interval is I, then if x is isolated we should find that I(intersect)S={a} but your I:=(2,3) doesnt contain a=1.. So wtf.. Does the open interval have to be in S??
Hi, What if I take an interval [0,1) U (1,2]. 1 is clearly not isolated. But is 1 an accumulation point. Beaches in the below comment you mentioned that every point has to be either isolated or accumulation point. But I don't think 1 is accumulation point.
Yes, you caught it! Also, I believe 1 would be considered closure point, and also a boundary point that is not included in the set. I should go back and extend this play list. Thank you for bringing my attention back to it! And good catch on your own comment!!!
Ibrahim Chalhoub Good one. My instinct is to say yes. My reasoning being that if you are an accumulation/limit point then there should be a point with a nonzero minimal distance from you. Any radius smaller than that minimal distance would witness your isolation. There are some weird pathological examples sometimes though! Especially in weird topologies that are not R. I think you are right but I will think on it a little bit more, just in case!
Ibrahim Chalhoub If x is isolated, then take a nbhd of x that intersects your set at only x. Then to show that x is a boundary point, let U be any nbhd of that point x. You would need to show that this nbhd U intersects both your set, and the complement of your set. It is pretty clear that the intersection with the set is nonempty (x is in there) and using the radius of the nbhd that witnessed isolation you can find a point that is in U but is not in your set!
Thank you for the video! This really clears things up a lot for me. I'm having a hard time with the terminology for closed, open, not closed, not open, and neither. Do you have any information on that?
striderpsv haha, and clopen, right? The sets that are simultaneously open and closed? I have a video in the playlist for closed somewhere, I will see if I can find it and link it here.
Maybe try this, it is really short, but gives an example of all four cases again in R (so really just intervals). If you have something more specific just let me know! ruclips.net/video/97vXMfhtWd0/видео.html
Haven't even finished the video yet, but I have learned more so far than the past 8 weeks in my Analysis class. As a college student, I say thank you and greatly appreciate the video!!!!
You learn this in college?I learned this in my 3rd year of highschool
As far as I understand, the given definition of the accumulation point, as a point such that set X\x still has other elements, is incorrect. The definition I came across (in a book on set theory by Ralf Schindler) is: x is X's accumulation point iff for every a
How could you construct a set like (n-1/2, n+1/2), and still call it an open subset of N?
Let's extend this idea then!
Does it follow that, if a set S has an isolated point and equipped with the appropriate topology, then S is Hausdorff??
Also, for your example of isolated points, in set notation, how would you write that set?
striderpsv I would just write it as \doubleN. Actually I watched the video and the number line I drew made it look like I was talking about integers, \doubleZ. If you had to use set notation you would say {1,2,3,...} for naturals.
Thank you very much! Really had trouble grasping accumulation points!
My question is that you have to expand the set of Natural numbers to include the Real numbers in order for you to define an open set containing n element of N that contains no other N. Is this justified?
This makes no sense.. Say the open interval is I, then if x is isolated we should find that I(intersect)S={a} but your I:=(2,3) doesnt contain a=1.. So wtf.. Does the open interval have to be in S??
no, the open interval does not have to be contained in S
Hi, What if I take an interval [0,1) U (1,2]. 1 is clearly not isolated. But is 1 an accumulation point. Beaches in the below comment you mentioned that every point has to be either isolated or accumulation point. But I don't think 1 is accumulation point.
Got that...every point in the set is either isolated or accumulation point. 1 is not a part of the set. Sorry. My bad...
Yes, you caught it! Also, I believe 1 would be considered closure point, and also a boundary point that is not included in the set. I should go back and extend this play list. Thank you for bringing my attention back to it! And good catch on your own comment!!!
Sir...what are the limit points of the set {cosn: n is any natural number}?
the limit points of {cosn: n is any natural number} is [-1, 1].
Thanks a bunch💗
So every point in a set S is either an isolation point or an accumulation point right? Or is that incorrect to say?
Ibrahim Chalhoub Good one. My instinct is to say yes. My reasoning being that if you are an accumulation/limit point then there should be a point with a nonzero minimal distance from you. Any radius smaller than that minimal distance would witness your isolation.
There are some weird pathological examples sometimes though! Especially in weird topologies that are not R.
I think you are right but I will think on it a little bit more, just in case!
Great. One question: I know that if x is an isolation point, then x is a boundary point... but how would I go about proving that?
Ibrahim Chalhoub If x is isolated, then take a nbhd of x that intersects your set at only x. Then to show that x is a boundary point, let U be any nbhd of that point x. You would need to show that this nbhd U intersects both your set, and the complement of your set. It is pretty clear that the intersection with the set is nonempty (x is in there) and using the radius of the nbhd that witnessed isolation you can find a point that is in U but is not in your set!
Joshua Helston do you mind if I start an email conversation with you on the topic of topology?
Joshua Helston and thank you for the reply!
Thank you sir. Its beginning to make sense
Thank you for the video! This really clears things up a lot for me. I'm having a hard time with the terminology for closed, open, not closed, not open, and neither. Do you have any information on that?
striderpsv haha, and clopen, right? The sets that are simultaneously open and closed? I have a video in the playlist for closed somewhere, I will see if I can find it and link it here.
Maybe try this, it is really short, but gives an example of all four cases again in R (so really just intervals). If you have something more specific just let me know! ruclips.net/video/97vXMfhtWd0/видео.html
thanks a lot , just the volume , its very low
Awesome explanation!! Thank you
No problem, glad you found it useful!
Awesome Explanation!
Nice explanation
Thanks. This was helpful.
Thank you, I am glad you found it useful!
consise and helpful, thank you
This was helpful, thank you!
Now I am try to understand this. It's hard because of my English level is B1 or less :')
Thank you.
Wow.....that's IT?!???!?!?!??!