6:12 I am really confused why X tilda = Union of Intersection of that set. We also have "for all k", so shouldn't it be equal to Intersection of Union of Intersection of the set?
@Problemathic9 күн бұрын
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@yopenzo9 күн бұрын
Thank you very much for your work here. You would deserve much more views and like! (Maybe only a little more volume at your agreeable voice 😉)
@Problemathic9 күн бұрын
@@yopenzo thank you! Your support means a lot to me. Newer videos have better audio quality!
@ivanrioscervino43699 күн бұрын
Thank you!
@Problemathic9 күн бұрын
@@ivanrioscervino4369 I’m glad it’s helpful 😊
@serkandogan376112 күн бұрын
I think the definition of N_r is wrong. It should be the union of two sets which you defined.Not the intersection.
@PauloSantos-uw3wq13 күн бұрын
thank you!
@counting123414 күн бұрын
hmm, not sure about your proof for R^omega in the box topology - I think we do have to show that B is open. Also, I'm only seeing a theorem in Munkres that X is connected if and only if the only subsets of X that are both open and closed in X are the empty set and X itself, but I think in your video you present it without closed being a necessary condition.
@yurigouveawagner943220 күн бұрын
i have a question. G is not a sequence, so what does it mean to say μ(G) --> M?
@Problemathic18 күн бұрын
That’s a good point, what I meant was that the measure of the union of Gn from n=1 up to N tends to M when N tends to inf
@counting123421 күн бұрын
im wondering if you could perhaps make a video on how to identify the triangle inequality in weird situations and apply it? like an "advanced triangle inequality" video. i struggle with recognizing how to use it when it's not directly identifiable, like in p. 122 or p.125 in this chapter of metric spaces in munkres. thank you! (:
@TheNumberCruncher091322 күн бұрын
Nice ASMR
@TheNumberCruncher091322 күн бұрын
And also a good exercise and explanation.
@Problemathic20 күн бұрын
@@TheNumberCruncher0913 thanks 😂
@counting123424 күн бұрын
great video, once again!! thank you!! it takes me such a long time to read munkres and i feel concerned when i don't understand a small detail; you help me identify and comprehend the main storyline.
@Problemathic24 күн бұрын
@@counting1234 thank you! It’s a lot if effort making this videos, it’s great news knowing that they’re actually helping people! 🥰
@counting123426 күн бұрын
i cannot thank you enough for your videos. my professor is extremely unclear, and i find munkres difficult to parse (and am frequently wanting more concrete examples, counterexamples, or images - as an example, i was not sure about what kind of objects his notation represented in his definitions in this section). your videos are a lifesaver in my topology course this fall!!
@Problemathic24 күн бұрын
@@counting1234 Thank you for the kind words, I’m so glad you’re finding it helpful
@weloveyoujeff-uq7bdАй бұрын
I don't think the proof for part 1 is correct, you used the same x in two different variables.
@jaffi730Ай бұрын
Please explain exercises 3.26.1 please help tomorrow is seminar😢
@VictorValdebenitoSilvaАй бұрын
Hola! Pregunta: en el minuto 05:00 ¿no es lo mismo decir que el complemento de tal conjunto B es abierto y eso probaría que B es cerrado? Digo, porque la elección de x0 es arbitraria y siempre vas a poder hallar un abierto Wx que es disjunto de B. Gracias!
@ProblemathicАй бұрын
Si! es básicamente lo mismo que hacemos en el video, solo que dicho con otras palabras 😊
@islamatef2526Ай бұрын
I want to thank you about your great videos I found many things that are not found anywhere else only here ❤
@ProblemathicАй бұрын
Thanks! I'm so glad you find it helpful! 🥰
@islamatef2526Ай бұрын
We can not take the case Ai's are countable except two uncountable sets A1 and A2 since the measure of the union will still 1 but the sum will be 2 so we must take one uncountable set not two .. ok why ? Is the reasons are the formula of the segma algebra & Ai's are disjoint ? I proved this using contapositive l said let Ai's are countable except two sets A1 and A2 where their completens are countable.. Since Ai's are disjoint then A1 is a subset of the completent of the union of Ai's (with A2) And if we take the complement of each side then we will replace each side I mean the union will be subset of the complement of A1 wich is a contradiction ( uncountable subset of countable is a contradiction)... the union is uncountable since it has A2 which is uncountabl and the complement of A1 is a countable from the formula of the segma algebra ..ok is it a logic reason or proof ? So there exist no case which say Ai's are countable except two ? If yes what about if we take all A are uncountable? .. we will have the same problem What about this case ?
@ProblemathicАй бұрын
Hi, exactly as you say in your comment and what I say in the video. There can only be at most one index j for which Aj is uncountable. So the only two possible cases are - All Aj are countable - All but one Aj are countable (and that other one is of course uncountable) So the case that you mention, with two uncountable sets is not a valid option, it never happens!
@islamatef2526Ай бұрын
@Problemathic Thank you so much..lots of love ❤️❤️❤️
@paolorepeto4368Ай бұрын
Hi why in proving countable union of measure le set is measurable you suppose the sets are disjoint?
@deepakagrahari997Ай бұрын
nice
@improlawlАй бұрын
Thank you for the video. I had a question on the first part of the problem. I thought the wording countable implied that we also needed to prove that the basis was a countable one as well. How would you suggest we go about doing that part? Thank you!
@HariwanMSalihАй бұрын
Can you please make a video on extension?
@ProblemathicАй бұрын
@@HariwanMSalih What do you mean by extension?
@HariwanMSalihАй бұрын
I have just found it , I mean the caratheodory theorem ...thanks @@Problemathic
@debmalyaroy76392 ай бұрын
Well described , really that brought a clear idea about dictionary order topology. But I am struggling to understand why the dictionary order topology restricted to I=[0,1] is not same as the subspace topology on I×I obtained from the dictionary order topology on R×R...Can you please explain? I can understand that for inherited subspace topology there may be half-open intervals as it is just the intersection of an interval of R×R with the unit square... Is it for like the basis only consists of open intervals for this type of order topology ? Am I right?
@Problemathic2 ай бұрын
@@debmalyaroy7639 Hi, thanks, I’m glad it’s useful! Regarding your question, this video might be helpful: kzbin.info/www/bejne/qmTCaZaZncqlo9Esi=eKucU1DPDUmOGby-
@爸爸到底-s9x2 ай бұрын
Urysohn’s lemma can also be used to prove the Portmanteau theorem.
@Problemathic2 ай бұрын
@@爸爸到底-s9x Yes, it has lots of applications! It’s an amazing theorem 😍
@HariwanMSalih2 ай бұрын
Sound is very low...thanks
@Problemathic2 ай бұрын
@@HariwanMSalih Yes, sorry about that, newer videos have better sound!
@ayaneshlal77182 ай бұрын
Finally a video where I get the intuition behind how the simple functions are defined. I was never able to understand the intuition till now. Thank you so much!
@Problemathic2 ай бұрын
@@ayaneshlal7718 I’m so happy you find it helpful! Thanks for the kind words! 🥰
@Problemathic2 ай бұрын
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@devkaransingh43782 ай бұрын
in the first question's example...why is Y convex in X??
@fadilawand44342 ай бұрын
Very useful video and nicely put together. Gives the viewer a better intuition of the countability axioms by going through an example.
@Problemathic2 ай бұрын
@@fadilawand4434 Thanks! I’m glad it’s helpful!
@mdred1004 ай бұрын
5:45 why it is inf Lim not the opposite Lim inf¿
@user-uh2ql9og1e4 ай бұрын
@ 4:48 when we are considering other case, why are we doing "there exist for some..." when in first case we just start that E_i are countable for all i? Is it just there to eventually force the form (U E_i)^c is countable? Since, in the first case, we already forced (U E_i) is countable by starting all E_i are countable? Proving two case since for something to be in A: (something is subset of X AND something is countable) OR (smthing is subset of X AND smthing-complement is countable} where, in our case, that Something = (U E_i). Lastly, in your video, are we to assume fancy A is subset of Powerset of X? so Since we claimed E in A earlier. E is also subset of X. So union of subsets of X is still subset of X so (U E_i) is subset of X, satisfying first part of each 2 cases? I ask this because I think you only briefly mentioned in the last video fact A is subset of P(X)... sorry for my English.
@Problemathic4 ай бұрын
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@Problemathic5 ай бұрын
☕ Support the channel by buying us a coffee: ko-fi.com/problemathic
@ilknurkulege46945 ай бұрын
Hello teacher, can you provide proof regarding the real analysis course?
@Problemathic5 ай бұрын
Hi! Which proof are you referring to?
@ilknurkulege46945 ай бұрын
@@Problemathic instagram adress please? Send the Photo
@ilknurkulege46945 ай бұрын
@@Problemathicmeasure convergence and a.e convergence
@Problemathic5 ай бұрын
@@ilknurkulege4694 we uploaded videos about that previous to this one! Check out the measure theory reproduction list?
@ilknurkulege46945 ай бұрын
@@Problemathic I looked at them, teacher, but it was about Egorov's theorem.
@Solo-vp2vn5 ай бұрын
Great work thank you🫶
@Problemathic5 ай бұрын
I’m glad you like it!! 🥰
@swaruppaul44175 ай бұрын
Thanks for the video! This is super helpful! I have learnt this interesting problem! Please do upload more such videos and also request you to upload some interesting problems from functional analysis!
@Problemathic5 ай бұрын
You’re welcome! The idea is to do a functional analysis reproduction list when we finish measure theory! 🤓
@arpanpiano5 ай бұрын
This channel is gold!!!
@Problemathic5 ай бұрын
Thanks for the kind words!! 🥰
@wryanihad5 ай бұрын
The way you produce the vedio is very elegance I think you are french Good for you lady Amazing explanation
@Problemathic5 ай бұрын
Thank you! I’m happy you find it helpful! 🥰 I’m not French😅!
@wryanihad5 ай бұрын
@@Problemathic I guess your granfather may be was french You are looking like frenches Mercy madam
@michaelmo9315 ай бұрын
Such a well explained video❤
@Problemathic5 ай бұрын
Thanks! I’m glad it’s helpful
@SanjaySG-dk1su5 ай бұрын
hello problemathic girl i am sanjay from india i am doing my Bachelors in technology in electronics and communication engineering, i am in senior year now, but i have always had a deep love for theoretical mathematics since my childhood, i am planing to get a masters in mathematics after my engineering degree, your videos are very inspiring and helpful and please keep amazing content your work is great
@Problemathic5 ай бұрын
Thank you!🥰
@Problemathic5 ай бұрын
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@ayzikdig19835 ай бұрын
tnx
@depressedguy94675 ай бұрын
Amazing , just going through it for my phd interview at IISC
@Problemathic5 ай бұрын
Thanks! That’s great, good luck!! 💪🏻
@HarpreetSingh-ek4hv5 ай бұрын
Please keep uploading u r soon going to become a legend in youtube higher mathematics🙌
@Problemathic5 ай бұрын
Thank you!! I'll continue uploading, currently trying to get at least one per week 🥰
@HarpreetSingh-ek4hv5 ай бұрын
Wonderful❤❤
@Problemathic5 ай бұрын
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@dyludylu5 ай бұрын
got lost a bit! when we say x belongs to exactly one N_r, uniquess is proven but what about existence? how do we know N_r is not empty? when proving that uniqueness, how are the inequalities ( r<= x <= 1 or x < r ) used? why does s<>r imply z-r and z-s belong to the same equivalence class?
@dyludylu5 ай бұрын
hey, love the channel. btw the lemma as written says E is a subset of A, shouldn't that be E is a member of A instead?
@Problemathic5 ай бұрын
Hi, thanks! I think subset is correct. Because the idea is that the sigma-algebras A will contain the family of sets E, and so their intersection will give the smallest sigma-algebra that contain the family of sets in E. A cannot have E as a member because both A and E are subsets of P(X). Does this make sense?
@themptytree31455 ай бұрын
I think you missed intervals of the type (a,b] when defining h-intervals. Also could you please explain why we need *disjoint* unions for the algebra?