Probably the good place to stop of this video disappeared in the black hole from the video about isolated points 😂

17:55 that was not a good place to stop :(

@abdoulkarenzo3138

3 жыл бұрын

İf u anderstand the lesson u can keep doing it yrself

During quarantine , Michael penn really got savage and worked harder to teach every single thing of pure mathematics. Thanks Michael. Keep it up. With love from India

Don't leave us hanging!

!!!! Intuition !!!! Nicely - nay, masterfully - done. You rock.

This is a very good lecture. Easy to take notes, and not miss anything. The economy of his presentation is awesome. I am self teaching a look at Modern Analysis using Simmons as lectures to the more comprehensive, but drier Bachmann and Narici, so these lectures by Prof Penn are priceless to me. Remember folks, do the math!

Great video delivered by a great professor . Thank you Professor !

Oh, what happened? Does this compensate the extra minutes from the previous video? 😂 love this RA course

Thanks so much for this great video .

Very clear explanation !

Just for fun and completeness, the last proof could have finished as: let x in V_e(a) if x=a, because a is in A then x is in A if x =/= a then x is in V_e(a) \ {a} then x is not in A^c (if it where, then (V_e(a) \ {a}) intersect A^c would not be empty) then x is in A and thus V_e(a) is completely inside A

@thesecondderivative8967

Жыл бұрын

The proof ended at "The intersection of V_e(a) and A complement - the singleton a is empty. Thus for all x in V_e(a), x is not in A complement. That means x is in A. All elements of V_e(a) are in A. I.e. V_e(a) is a subset of A. That finishes the proof. That's how I proved the ending.

Please do videos about elliptic integrals

Analysis is too much rigorous but interesting. Thanks for these lectures Mr. Penn.

@RandomBurfness

3 жыл бұрын

Analysis isn't very rigorous at all, lol. If you want to see real rigour, just turn yourself to abstract algebra. Analysis is a street fight, whereas abstract algebra is like going to court.

@harrywang6792

Жыл бұрын

@@RandomBurfness I remember trying to prove why -0 = 0, or 0(a) = 0, or -a = (-1)a, god that was so annoying

@jeffrey8770

Жыл бұрын

@@RandomBurfness that's just not true

is limit point and boundary point same thing or different?

You want to prove that [b, c] is closed. According to your definition, you don't need to prove that x in [b, c] implies x is a limit point. Your definition only requires: x is a limit point of [b, c] implies x in [b, c].

@Zeegoner

3 жыл бұрын

@VeryEvilPettingZoo But you don't need to *construct* all limit points. You can just do the second proof (9:34), showing "if x is not in [b, c], x is not a l.p." The contrapositive is: "x is a limit point implies x is in [b, c]." Since x is an arbitrary limit point, this statement is automatically true for *all* limit points. I think the first part is a good exercise, but not necessary. To be clear, I agree with your comment but you are not appreciating OP is concerned with what the definition strictly requires (for proof).

@hybmnzz2658

3 жыл бұрын

@@Zeegoner yeah thank you this has been kind of confusing. The set [0,1] U {5} would be considered a closed set in the real numbers right? The limit points of that set are just [0,1] so it does contain all of its limit points. Please tell me if this is correct.

@schweinmachtbree1013

3 жыл бұрын

@@hybmnzz2658 Yes that is correct. if you take a course in topology, you would prove this quickly as follows: two standard results about closed sets are that "the union of finitely many closed sets is closed" and "sets containing one element are closed*", and hence [0, 1] U {5} is closed because it is the union of two closed sets (Michael proved that closed intervals are closed sets in one of the previous videos). In fact, there is an even quicker proof because {5} is the closed interval [5, 5], so you don't actually have to use the second result about closed sets I gave. *to be completely precise one should say "sets containing a single real number are closed (in the standard topology of the real numbers)" because if one considers the topology of more complicated sets then there can be sets containing one element which are not closed. of course you shouldn't concern yourself with this if you only care about the topology of the real numbers though.

@Zeegoner

3 жыл бұрын

@VeryEvilPettingZoo That is not OP's question whatsoever. He is not asking about checking limit points vs all limit points. He is asking why point-checking is necessary at all, when the second proof ("applying the definition," the contrapositive of it) does not appear to use the result from the first proof (which is that all points x in [b, c] is a limit point). "Penn was first determining all the limit points so that he *could then apply* the definition." Again, this is wrong. You do not need to determine all limit points so you "can then" apply the definition. "Ordered his point-checking differently" No, "eliminated the point-checking"

@Zeegoner

3 жыл бұрын

@VeryEvilPettingZoo My previous comment was just to eliminate the "might" in the second true/false question. I may have read the whole "maybe-this-maybe-that" in your second paragraph differently than you may have intended, because of the "could then apply" (implying a dependence) and "ordered a little differently" (as opposed to eliminate) in your first paragraph.

I may missing some fundimental part here. Suppose A = { 0 and 1/n | n is positive integer }, from the definition, A is closed, because it contains all limit points, which is only one point "0". But I'm pretty sure A completement is open.

@caladbolg8666

3 жыл бұрын

Correct

@thesecondderivative8967

Жыл бұрын

You're correct. A complement is open

7:52 I can prove that b is in the epsilon neighbourhood of x if epsilon > x-b but not if epsilon >= x-b :(

Not a good place to stop.

pardon, but it appears to me that your proof of the first claim contains some redundancies. if you prove that x is not in [b,c] implies it is not a limit point (which you do) it necessarily follows that the only limit points of [b,c] are contained in [b,c] and you are done. you do not need to show that every point in [b,c] is a limit point even though it is true

@Zeegoner

3 жыл бұрын

Yeah I think the proof by cases he wrote is meant to answer: "Prove that all points in [b, c] are l.p.'s of [b, c]"

Could you talk about p-adic metric I am still struggling with the convergence idea

How is this real analysis if there are no metric spaces? :P

being inactive during quarentine? covid's got nothing on my boy

That was not a good place to stop.

(0,1] is open as it doesn't contain 0, surely the complement is also open as it doesn't contain 1? What am I missing?

@EnteiFire4

3 жыл бұрын

In (0, 1], there is no neighbourhood of 1 that is a subset of (0, 1], so it is not open. It is also not closed IIRC.

@jamescollis7650

3 жыл бұрын

@@EnteiFire4 Thanks 👍

@stephenbeck7222

3 жыл бұрын

A set like (0,1] is neither open nor closed. On the other hand, the entire set of real numbers is both open and closed, i.e. “clopen “.

@gaaraofddarkness

3 жыл бұрын

best and sure way to check if a set is closed is to check if it's complement is open. In this case the complement (-∞,0] U (1,∞) is not open..so the given set is not closed.

Hi, Despite this time I had time enough to eat my cake, I... stay on my hunger 😂

Aaaand that’s a bad place to stop.

Bonus question: Is the finite set {1,2,3} closed? (Ans: Yes; it does not have any limit points)

@thesecondderivative8967

Жыл бұрын

All finite sets have no limit points. Take an arbitrary element a and set A = {a_1, a_2, ... ,a_n} and assuming e > 0 we can set e =1/2 min {a_k - a_(k-1) , a - a_k } for all k between 1 and n. The epsilon neighborhood of a will never touch two neighboring points at the same time. Thus, the only element of set A in the epsilon neighborhood of a is at most a_k. The intersection is empty (except for a_k of course) i.e. a is not a limit point. I hope my proof is correct.

I don't think that your proof takes into account A being the set of real numbers and A compliment being the empty set.

@schweinmachtbree1013

3 жыл бұрын

very good. whenever I use the membership symbol in a proof I like to check that the set is non-empty (perhaps this is too "nitpicky" to do while composing the proof, and it might distract you from the more "important" parts of the proof, but one should definitely do this when one is checking their completed proof - indeed, if you don't check, and one of the sets is empty, then your proof is incorrect, as in this video)

@CousinoMacul

3 жыл бұрын

@VeryEvilPettingZoo It would take a little more than one line because while I believe that he proved that R and the empty set are both open, he only just introduced the definition of closed sets in this video. We would need to prove that both sets are closed. R seems pretty straightforward. Any limit point of R must be a real number and hence contained in R. The empty set takes a little more work but it's also quick. I would probably bwoc assume that the empty set is not closed. Then there exists a limit point of the empty set not contained in the empty set. But then applying the definition of limit points to the empty set quickly leads to there being no limit points of the empty set, which is a contradiction. Hence the empty set is closed, and so is that part of the proof.

@CousinoMacul

3 жыл бұрын

@VeryEvilPettingZoo And here I was expecting a 1-line reply. ;-P Seriously though, thank you for the thorough response and I agree with everything you say. The funny thing is that when I was watching this video, my main thought was that back when I took Introduction to Topology, being the complement of an open set was the definition of a closed set. Of course that's the most general definition and it applies to all topologies, even if they don't have a concept of a limit point.

Show the end😭

@________6295

3 жыл бұрын

Then v_epsilon(a) is contained in A which inplies A is open

@Janox81

3 жыл бұрын

@@________6295 I see. a is in V_eps(a) and definitely in A. Since V_eps(a) minus {a} contains no points of A^c, every point of V_eps(a) minus {a} is in A. Together this is V_eps(a) contained in A.

Mr. Michael a closed set is the set that contains all its boundary points not the limit points

@schweinmachtbree1013

3 жыл бұрын

well he just proved it so it's not exactly up for debate?

@josephhajj1570

3 жыл бұрын

@@schweinmachtbree1013 no I'm sure we just took it

Lol ,why do we need a real analysis to understand the intuition behind closed sets? That's pretty overkill lol.

@sergeiivanov5739

3 жыл бұрын

Either way, it might be said that every set of the type U = [a, b] in R is supposed to be closed if there exists some constant M such that ever element of U does not exceed this. Otherwise, it is open. Understandably, the author ' is more formal. But the idea is the same... Or not?

@billh17

3 жыл бұрын

He is not using real analysis to understand the intuition behind closed sets. Rather, he is showing that the idea of closed sets in real analysis does not contradict what we call closed intervals.

@sergeiivanov5739

3 жыл бұрын

@@billh17 could you account for what you have written in more details?

@billh17

3 жыл бұрын

@@sergeiivanov5739 I am not sure what you want more. In calculus one (or even in high school), the idea of a closed interval [b, c] is introduced. It is the set of real numbers x such that b

@sergeiivanov5739

3 жыл бұрын

@@billh17This is what I requested for. Much obliged!