Limit Points (Sequence and Neighborhood Definition) | Real Analysis

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  • Опубликовано: 13 янв 2025

Комментарии • 42

  • @teffiegurl6253
    @teffiegurl6253 Год назад +8

    Your videos truly help me understand my lesson. Thank you.

  • @BuraRich
    @BuraRich Месяц назад +1

    Thanks math-guy, u help us to remember the concepts by repeating them, thank u for ur patience, we love u and keep going bruh 😍😍😍

  • @punditgi
    @punditgi 2 года назад +6

    Great topic! So important.
    Now, can you tie this discussion to point set topology and open vs closed sets and then to compact sets and the associated theorems for the set of real numbers and then for metric spaces and finally for general topological spaces? That would be a master class on this super important topic and require several videos. Can't wait to see it! 😇

    • @WrathofMath
      @WrathofMath  2 года назад +4

      Thanks Ezra! I'm working my way through the real analysis playlist, so no general topological spaces yet! I'd like to create full playlists for every undergraduate course before moving on to grad topics, but topology is probably near the bottom of the line in that list - partly because I never took a course on it so I need to study it more. But I never took a course on graph theory either, so it's nothing more than a minor inconvenience. But indeed the proof connecting closed sets and limit points is next! Then the Heine-Borel theorem whenever I have time to sit down and record it!

    • @punditgi
      @punditgi 2 года назад +1

      @@WrathofMath Awesome! 👍

  • @Nick-ts1qc
    @Nick-ts1qc Год назад +2

    Thanks for the video. Question -- 4:16 why is 1 a limit point of the set (0, 2)?
    I had interpreted your explanation at 1:20 as "the set cannot contain the limit point", though to your credit, after rewatching that portion I realize you actually said "the *sequence* cannot contain the limit point."
    But if that's the case, and if 1 can be a limit point of a set like (0, 2), can't literally every number in that set be its limit point? Given the notion of the set being "everywhere dense" (such that, for *any point a* in (0, 2) , there is an infinitely dense subset in the 𝛿 neighborhood of a)? E.g., why can't 1.5 also be a limit point, since there's an infinite sequence 1.49, 1.499, 1.499 ... approaching 1.5?
    Perhaps I'm confusing concepts and terminology somewhere here. If so, please let me know. Thanks.

    • @laurenwhite6686
      @laurenwhite6686 Год назад +1

      It’s from my understanding of what he said that every number in the set IS a limit point. So you’re right, 1.5 is a limit point of (0,2). See the video at 4:55.

    • @Nikkikkikkiz
      @Nikkikkikkiz Год назад +1

      1 is a limit point but not a boundary. Boundaries are usually like that

  • @Sarah-mp9lb
    @Sarah-mp9lb 8 месяцев назад +1

    Extremely helpful, thanks!

  • @aimerscien2464
    @aimerscien2464 25 дней назад

    Hi, see the concept at 0:12. You defined A\{x}, so if there's a {1}, there will be an empty set, and hence, there's no sequence of 1,1,1...

  • @TUB_HUB
    @TUB_HUB Год назад +2

    bro ur a gem.,!

  • @rashidap7706
    @rashidap7706 Год назад +5

    So, the limit point dont have be a point in the set?, in your example (0,2),0 is a limit point which is not in (0,2).

  • @wtt274
    @wtt274 2 года назад

    An very important concept which is so well explained . Thank you sir for this video !

  • @tlee146
    @tlee146 Год назад +1

    Excellent video! Can you do a proof by induction of the AM GM inequality? Would be really appreciated!

    • @WrathofMath
      @WrathofMath  Год назад +1

      Thank you! I'd be happy to, but it might be a little while until I have time!

  • @henrywoo1668
    @henrywoo1668 Год назад

    The concept has been well and clearly explained 😊

  • @sitienlieng
    @sitienlieng Год назад +1

    Thanks Wrath of Math, I have seen definitions of limit points defined as “if any neighborhood of x contains a point in S\{x} then x is a limit point of S. I know it is very similar to the definition you provided but it seems much weaker because it only requires 1 point (in S )different than x in every neighborhood of x for x to qualify as a limit point. Could you please clarify this in another video?

    • @WrathofMath
      @WrathofMath  Год назад

      Thanks for watching and the question! The definition you describe appears the same as what I used, you'll have to rephrase it if there is a difference I am missing.
      The definition used in the video is: A point x is a limit point of S if every neighborhood of x intersects A at some point other than x. In other words, every neighborhood of x must contain at least one point that is also in A.
      The definition you describe, as I understand it, is: A point x is a limit point of S if every neighborhood of x contains a point (other than x) that is also in S.
      You used the word "any" at the beginning of your definition, which would mean "at least one", but I assume you meant every.

  • @TranquilSeaOfMath
    @TranquilSeaOfMath 10 месяцев назад

    Nice presentation.

  • @douglasstrother6584
    @douglasstrother6584 4 месяца назад +1

    My experience with Real Analysis is limited to what is covered in Calculus (single-varible through vector) to validate the basic ideas of differentiation and integration. Since Calculus is so geometric (visual), what makes Real Analysis so difficult?

    • @amberjha5974
      @amberjha5974 3 месяца назад +1

      I personally feel Real Analysis is difficult because it is very counter intuitive. In order to use it as a tool to better understand in places where all other tools fail to work, we are forced to learn it in places where it isn't much useful. In Calculus every derivative and integral follows a singular line of logic based on the limit definition which is just expanded and shortcuts are discovered. In Real Analysis, we are forced to learn so much of ground that it makes it super daunting. For an example, trying to find the limit of a function in calculus just makes us check value from left and right side on the graph of the function. In Real Analysis, we are forced to define a lot of terms such as the property of the function, taking an epsilon, comparing the actual value with the epsilon and then mapping the properties derived from the comparison onto the function to get the value. This proof oriented solving makes the entire tool feel like a black box unless, you are completely familiar with the entire process. Entirety of calculus is spent around understanding first, solving later, to master the subject. Real Analysis however asks you to solve first without knowing what is happening and then you are supposed to click in how it worked when you were solving it. Very annoying. Sorry for the rant, this is my first undergraduate course in maths and I love it but I hate it as well.

    • @douglasstrother6584
      @douglasstrother6584 3 месяца назад +1

      @@amberjha5974 Thanks for the detailed answer.
      You'll find a lot of love/hate relationships in Math. I certainly did in Physics.

    • @douglasstrother6584
      @douglasstrother6584 2 месяца назад

      @@amberjha5974 "The Bright Side of Mathematics" has a fun video series on Real Analysis; his presentation is visual and intuitive. Understanding the subtilies of the definitions and having strong proof-writing skills are essential to learning Real Analysis.

  • @krasimirronkov17
    @krasimirronkov17 2 года назад

    Well-explained and visuallised video. Do you know a proof of the fact that if the limit of the sequence a_n+1/ a_n is between 0 and 1 then the sequence a_n converges to 0.

    • @Aman_iitbh
      @Aman_iitbh Год назад

      if all an>0 must be in condition thats ratio test

  • @climitod8524
    @climitod8524 6 месяцев назад

    why is your def so much different than taos doesn't even seem like that's the def at all

  • @PodcastClips910
    @PodcastClips910 9 месяцев назад

    I wish I could double-like this video!

    • @WrathofMath
      @WrathofMath  9 месяцев назад +2

      I wish I could double like this comment!

  • @H3XED_OwO
    @H3XED_OwO 8 месяцев назад

    i'm not sure what "intersects" means but i'm guessing "intersects" means something like this: A intersects B := cardinality of A∩B is bigger than 0

  • @nuraddinvalibayoghlu1511
    @nuraddinvalibayoghlu1511 Год назад

    Thank you for this great video on this topic. I have a question for you.
    There is a convergent sequence defined on the set X. A finite number (for example, 10) of elements of that sequence forms a closed set, which also has an open neighborhood. Is it possible that the limit of this sequence lies outside that closed set and its open neighborhood? Or should it necessarily reside in this closed set?
    I would be very happy if you answer my question.

  • @homayratabassum929
    @homayratabassum929 4 месяца назад

    Thanks

    • @WrathofMath
      @WrathofMath  4 месяца назад +1

      Glad to help, thanks for watching!

  • @oderoivan
    @oderoivan 2 месяца назад

    The work is not well seen

  • @nuraddinvalibayoghlu1511
    @nuraddinvalibayoghlu1511 Год назад

    Thank you for this great video on this topic. I have a question for you.
    There is a convergent sequence defined on the set X. A finite number (for example, 10) of elements of that sequence forms a closed set, which also has an open neighborhood. Is it possible that the limit of this sequence lies outside that closed set and its open neighborhood? Or should it necessarily reside in this closed set?
    I would be very happy if you answer my question.

    • @WrathofMath
      @WrathofMath  Год назад +1

      Thanks for watching! I'm not sure I understand your question, the open neighborhood part in particular. Where is this open neighborhood? It sounds like you're saying the closed set has it, but you also said the closed set consists of a finite number of elements, which can't ever be open (aside from the empty set).

    • @nuraddinvalibayoghlu1511
      @nuraddinvalibayoghlu1511 Год назад

      @@WrathofMath let me ask my question more clearly:
      There is a space X and there exists a sequence in this space. In this space X, there exists a closed set composed of a certain finite number of elements of the sequence, and it also has an open neighborhood. This sequence has a unique limit, but it belongs neither to this closed set nor to its open neighborhood. Can this limit point be separated by a function from that closed set?