Real Analysis Ep 15: Closure of a set

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Episode 15 of my videos for my undergraduate Real Analysis course at Fairfield University. This is a recording of a live class.
This episode is about the closure of a set.
Class webpage: cstaecker.fairf...
Chris Staecker webpage: faculty.fairfie...

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14 окт 2024

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Комментарии : 18

@quant-prep2843 3 года назад

day 15, I am still here... I cant believe i am in high school and able to follow this lecture. thanks to Chris staecker!!

@chaolin3187 2 года назад

I find it is interesting to prove a closure set A_closure is closed. Here is my attempt: I want to show that for any limit point of A_closure, it is also the limit point of A. Since A_closure contains all the limit points A, A_closure contains all of it s limit points. Then the proof is done. Assume x is a limit point of A_closure. Then for any r > 0, the (x - r, x + r) intersects A_closure is not empty. I want to show (x - r, x + r) intersects A is not empty. For any point y belongs to the intersection of (x - r, x + r) and A_closure, If y belongs to A, then we are good. If y doesn't belong to A, then y is the limit of A. First since y belongs to (x - r, x + r), there exists r' such that (y - r', y + r') is a subset of (x - r, x + r). Second, since y is a limit point of A, (y - r', y + r') intersects A is not empty. (x - r, x + r) contains an element of A. Therefore (x - r, x + r) intersects A is not empty. x is the limit point of A. shewn. It may be a little verbose. 😆

@algorithmo134 3 года назад

I look forward for more real analysis video like contraction principle, implicit function theorem and inverse function theorem

@ChrisStaecker 3 года назад

Thanks! But I'm not taking requests at the moment- these videos were for a class I was teaching. Maybe someday I'll teach on these topics, but not anytime soon.

@TheTacticalDood 3 года назад

For showing that a closed interval is indeed a closed set, can we proceed by contradiction? Here is the proof's layout. For the sake of contradiction, assume x is a limit point of the closed interval L =[a, b] and x is not in L. Since x is a limit point of L, then by the definition of a limit point, the intersection of the epsilon-neighborhood of x with L contains points of L other than x for any epsilon > 0. This means that the intersection of the epsilon-neighborhood of x with L can be written as the union of {x} and some other set, say A, that contains the other points of the intersection. Thus, x is an element of L. Here we reach a contradiction, and so our initial assumption of x not being in L is incorrect. Is my reasoning correct? Thanks in advance!

@ChrisStaecker 3 года назад

I'm not sure this argument works- it is immediately suspicious because you don't seem to have used the fact that L is a closed interval anywhere. When you say "This means that the intersection...", this part doesn't follow. That intersection may not include x itself. (The epsilon-neighborhood does include x, but once we intersect with L, it may not anymore.)

@oriyo2248 Год назад

Wouldn't the complement of N be (-infinity, 1) union (1,2) union (2,3) and so one because none of the negative numbers are included in natural numbers (based on the example at 38:56)

@ChrisStaecker Год назад

Yes you're right! What I wrote is the complement of Z (the integers). The complement of N is still a countable union of open intervals, just not exactly the ones I said.

@oriyo2248 10 месяцев назад

@@ChrisStaecker Thanks for your reply! I doubt you'll see this comment again, but I just wanted to thank you for getting me through Real Analysis this semester, i kept going back to your playlist throughout it, and even during the summer to prepare because of how digestible you make these topics. I finished my final on Monday and felt pretty confident thanks to you!

@tudormarginean4776 2 года назад

Let's say we have an open interval O= (0,1). Then consider an open subset I = (0.5, 0.6). The complement of I must be closed, but that complement is (0, 0.5] U [0.6, 1), which doesn't contain all its limit points, namely 0 and 1. Maybe there is something I miss, but I really don't understand how this counterexample can be rejected...

@ChrisStaecker 2 года назад

This is a great question- as I'm using it in this course, the complement always means "complement in all of R". So in your case the complement of (0.5,0.6) is the union (-∞,0.5] U [0.6,∞), and this set is closed. It seems like you are trying to think of (0,1) as if it were the full universe of discourse in this example. If we take that view, then the complement would indeed be (0, 0.5] U [0.6, 1) like you said, but this actually WOULD be closed. It's closed because 0 and 1 are not actually limit points because they are not part of the universe of discourse- so we must act as if there is no such thing at all as 0 and 1. (Just like in a typical example where the universe is all of R, the "endpoints" of ∞ and -∞ are not limit points because they are not part of R.) These ideas have led to the idea of the "subspace topology"- see wikipedia if you're interested.

@tudormarginean4776 2 года назад

@@ChrisStaecker Thank you a lot, this answer is illuminating for me! The answer I gave to myself was something around those lines, but I wasn't sure such a solution would pe possible. So is it true that when we consider complements we always define them over the entire Universe of discourse?

@tudormarginean4776 2 года назад

I mean, if in my example the interval O is contained in a larger interval, (-1, 2), does the same answer apply? In this case 0 and 1 do exist, so the complement of (0.5, 0.6) would be (0, 0.5] U [0.6, 1), which is not closed

@ChrisStaecker 2 года назад

@@tudormarginean4776 In this class, yes we always assume that the universe is R, and when we do the complement we mean the complement in R. In more technical settings, the word "complement" has a slight ambiguity, because it requires the universe to be specified. In this class I wrote X^c for the complement of X (this is how the textbook does it). If I were doing this stuff in a more general setting, I would probably write the complement as (R-X), which makes it clear that I am taking R to be the universe.

@ChrisStaecker 2 года назад

@@tudormarginean4776 I would say in that case that the complement is (-1,0.5] U [0.6,2), which is closed in (-1,2) for the same reasons as before.

@mostafaalkady6556 3 года назад

At 33:08, how do we know that L is a subset of F? Maybe some of the limit points of A are located outside F itself. How can we be sure?

@ChrisStaecker 3 года назад

Thanks for this question- I didn't really spell this out in full detail. Since A is a subset of F, this means that every limit point of A is also a limit point of F. So every point in L is a limit point of F, and since F is closed, it must contain all points of L.

@hunterroy8485 3 года назад

I much rather have taken real analysis and discrete mathematics in my undergraduate career. I was forced to take two general chemistry classes and two laborious labs....the injustice!