Real Analysis | Uniform continuity and compact sets.

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  • เผยแพร่เมื่อ 22 ธ.ค. 2024

ความคิดเห็น • 36

  • @carl3260
    @carl3260 3 ปีที่แล้ว

    Small point, I think the modulus at around 6:45 isn’t necessary. Continuity give the limits f(x), f(y), so adding the convergent series (without modulus) has the limit f(x)-f(x)=0, the modulus then comes back in the next line by definition of limit.

  • @hydraslair4723
    @hydraslair4723 4 ปีที่แล้ว +4

    Michael, any future plans for a series on point-set topology? Would be great to go through it with your intuition about the subject.

    • @get2113
      @get2113 4 ปีที่แล้ว

      I totally agree.

  • @timurpryadilin8830
    @timurpryadilin8830 4 ปีที่แล้ว +2

    Can't wait to see integrals in this course

  • @chessematics
    @chessematics 4 ปีที่แล้ว +2

    Make a playlist on Number Theory, for you have the best content and many videos on it

  • @ruanramon1
    @ruanramon1 4 ปีที่แล้ว +6

    1:30 that's better

  • @chuckaway6580
    @chuckaway6580 4 ปีที่แล้ว +7

    This is actually a very sneaky proof and much more elegant than the one you see in baby Rudin. Few outclass baby Rudin when it comes to proof elegance. Makes me suspect this proof is wrong in a way that I haven't noticed yet.
    Edit:
    I think I figured out why Rudin doesn't go this way with the proof. He tends to prefer epsilonics to sequences in baby Rudin (arguably a flaw with the text, but I think I know why he takes this approach). In particular, he never proves the sequences characterization of uniform continuity. As such, this proof is not really an option for him.
    I should add that this proof method really is just better and there's no reason not to prefer it if you've built up the right machinery. It readily generalizes to general metric spaces, and since that's the most general setting where uniform continuity is defined, that's all you need (okay, we could talk about uniform spaces, but let's please not).

    • @PhotonEcho
      @PhotonEcho 4 ปีที่แล้ว +1

      Rudin was just kind enough to leave this version as an exercise.

    • @get2113
      @get2113 4 ปีที่แล้ว

      I think Bartle is better but to each own.

    • @hydraslair4723
      @hydraslair4723 4 ปีที่แล้ว

      e n t o u r a g e

    • @gustavocardenas6489
      @gustavocardenas6489 4 ปีที่แล้ว

      Baby Rudin ch4 ex 10

    • @SedgeHermit
      @SedgeHermit 4 หลายเดือนก่อน

      I like Rudin's proof better. More direct, directly uses the definition of compactness rather than using a lemma, and not altogether inelegant.

  • @get2113
    @get2113 4 ปีที่แล้ว +1

    The really needed condition is absolute continuity in order to recover function by integrating its derivative. AC is a little stronger than uniformly

  • @daniellin503
    @daniellin503 4 ปีที่แล้ว +1

    Could we change the condition from "K is compact" to "K is bounded"? Because the "closed" part of the definition of the compactness is not used in the proof.

    • @chuckaway6580
      @chuckaway6580 4 ปีที่แล้ว

      You could, but then the proof won't generalize to general metric spaces, as compact = closed and bounded doesn't apply to all metric spaces (in fact, not even all Banach spaces).

    • @ruanramon1
      @ruanramon1 4 ปีที่แล้ว +2

      Maybe theclosed part is used when he says in 4:51 that the subsequence converge to a point in the compact. Think in a bounded open set, it is possible a convergence to a limit point not contained in the open

    • @chuckaway6580
      @chuckaway6580 4 ปีที่แล้ว

      @@ruanramon1 Yes, you need closed to say that it converges to a point in the set, but it's not clear to me that we actually need that in this case. Seems like the proof still works if it converges to a point outside the set, unless you don't assume continuity there, which I guess is possible.

    • @tomkerruish2982
      @tomkerruish2982 4 ปีที่แล้ว

      No. Consider tan x on (-pi/2, pi/2). Choosing x sufficiently close to an endpoint will invalidate any chosen (positive) delta for a given (positive) epsilon.

    • @chuckaway6580
      @chuckaway6580 4 ปีที่แล้ว

      @@tomkerruish2982 There it is. I think I follow, but to build your point out more for others: Consider tan on (-pi/2 , pi/2) U (pi/2, 3pi/2). This is clearly a bounded set on which tan is continuous. Take x_n -> pi/2 from the left and y_n -> pi/2 from the right. Then |x_n - y_n| -> 0, but |f(x_n) - f(y_n)| -> infinity. We need for the limit of the convergent subsequences of x_n and y_n to be in the set or the proof falls apart.
      Kind of subtle, though. It would be easy to write a somewhat convincing fake proof that leaves off closure. The fake proof becomes a real proof if you assume that the function is continuous on some closed set containing the bounded set.

  • @사기꾼진우야내가죽여
    @사기꾼진우야내가죽여 4 ปีที่แล้ว

    I think we also need to show that x which is the limit of the two subsequences of Xn and Yn is contained in K using the fact that x is a limit point of K and K is closed.

    • @armansimonyan5772
      @armansimonyan5772 8 หลายเดือนก่อน

      By Bolzano Weierstrass the subsequential limit exists if the sequence belongs to some compact set A and belongs to the same set A

  • @goodplacetostop2973
    @goodplacetostop2973 4 ปีที่แล้ว +2

    8:43

    • @pbj4184
      @pbj4184 4 ปีที่แล้ว

      And that's a good place-

    • @chessematics
      @chessematics 4 ปีที่แล้ว +2

      @@pbj4184 the stop stopped the TO STOP

  • @pyy6433
    @pyy6433 4 ปีที่แล้ว

    😮 I like your lecture

  • @arvindsrinivasan424
    @arvindsrinivasan424 4 ปีที่แล้ว +1

    🔥🔥🔥

  • @melissaneacsu4543
    @melissaneacsu4543 4 ปีที่แล้ว

    Thank you so much for these videos!