Real Analysis | Open subsets of ℝ.

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  • Опубліковано 3 жов 2024
  • We give the standard definition of an open subset of the real numbers, give a few examples, and prove some classic results.
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КОМЕНТАРІ • 50

  • @cletushumphrey9163
    @cletushumphrey9163 4 роки тому +34

    I believe the mistake at 6:10 is putting a minus sign in front of the parentheses instead of a plus

    • @izumiasmr
      @izumiasmr 11 місяців тому

      I wonder however do we need 1/2 coefficient when picking ε, just getting the minimum seems to be enough 🤔

  • @sanjursan
    @sanjursan 2 роки тому +2

    Michael Penn and the three "C's" of Real Analysis. Clear, Concise, and Comprehensive.

  • @lucascaique2943
    @lucascaique2943 4 роки тому +29

    This is such a great playlist that I'm actually looking forward for any new video.

    • @loopingdope
      @loopingdope 4 роки тому +5

      Like a netflix series, but better

  • @Manuel-pd9kf
    @Manuel-pd9kf 4 роки тому +6

    This deadass is a great playlist, keep it up!

  • @e.m.winter500
    @e.m.winter500 Рік тому

    Very much appreciated all the editing done to save a few seconds here and there. Must have been a lot of work but it makes the video much more smooth. Beautifully done thank you.

  • @murielfang755
    @murielfang755 3 роки тому

    life saver playlist. Clear explanation and nice speed.

  • @jonaskoelker
    @jonaskoelker 2 роки тому +1

    Note that for the finite intersection of open sets, it's enough to show that the intersection of two open sets is open. [This principle holds for any class of objects closed under some combination of two of them: from two you get any finite quantity.]
    If it holds for two, it follows that (((U_1 * U_2) * U_3) * U_4) is open, and so on by induction [where * is the combination of two objects, here the intersection of two sets].
    This would probably make for a slightly easier proof, at least with respect to notation.

    • @izumiasmr
      @izumiasmr 11 місяців тому

      Thanks nice remark! I guess Michael's approach might be in a way instructive sort of low level approach, and then it comes as a bummer what you said, sort of automating via induction

  • @willyh.r.1216
    @willyh.r.1216 4 роки тому +1

    What a beautiful refresher for me Michael. Thank you. This reminds me my first college math back in 1987, french math curricula. Dedekind's approach is also very interesting for the construction of R. R is bounded and complete (with Bolzano-Weirstrass theorem). All Cauchy sequences are convergent. And we can also prove that any real number is a limit of a rational sequence. Meaning, the set of rational numbers denses in R. Those key results came back spontaneously to my mind while watching your video. Thank you.

    • @spencerpencer
      @spencerpencer 4 роки тому +2

      R is most certainly not compact my friend

    • @willyh.r.1216
      @willyh.r.1216 4 роки тому +1

      Thank you for correcting me. It's been a long time I did this real analysis.

    • @bobajaj4224
      @bobajaj4224 4 роки тому +1

      @@spencerpencer that's true, that's why we used the Alexandroff's extension

    • @bobajaj4224
      @bobajaj4224 4 роки тому

      and the limit of rational means that Q is dense in R

    • @willyh.r.1216
      @willyh.r.1216 4 роки тому

      @@bobajaj4224 Alexandroff compactidication of R. I have recollection of that.

  • @abhijitharakali
    @abhijitharakali 4 роки тому

    Thanks Prof. Penn. I'm glad you are making these videos. They are valuable for us and I hope you'll continue to post such videos.

  • @thesecondderivative8967
    @thesecondderivative8967 Рік тому

    12:32 I believe we can make epsilon as small as we want since the intersection of open intervals is open.

  • @xoppa09
    @xoppa09 Рік тому

    This guy is brilliant and his proofs are unassailable ( i cant find any mistake). He is the like jesus of math , saving undergrad math majors from failing. :P I don't even have to speed up the video because he moves so quickly through the proofs. Never a dull moment.

  • @PMBUNESA-wj3li
    @PMBUNESA-wj3li 11 місяців тому +1

    Why are you take the epsilon as the half of the minimum? Not just the minimum?

  • @hopegarden7636
    @hopegarden7636 4 роки тому +1

    Great video as usual also them gains are showing itself

  • @BlueRobair
    @BlueRobair 2 роки тому

    Thank you !

  • @goodplacetostop2973
    @goodplacetostop2973 4 роки тому +5

    18:32

  • @tomatrix7525
    @tomatrix7525 3 роки тому

    These are really good, thanks!

  • @__hannibaalbarca__
    @__hannibaalbarca__ 4 роки тому

    I love General Topology; and A Counterexample in GT

  • @GKinWor
    @GKinWor 2 роки тому

    so helpful

  • @vardaandua3585
    @vardaandua3585 4 роки тому

    Sir please tell how to find the integral of (x/(xsinx +cosx ))^2 without using integration by parts

  • @romeoaubrey4119
    @romeoaubrey4119 Рік тому

    Im confused. I thought the empty set is equivalent to the complement of R and so since R is open then it's complement which is the empty set is closed. Am I wrong?

    • @Lucashallal
      @Lucashallal Рік тому +1

      Yes, the empty set is closed

  • @rafael7696
    @rafael7696 4 роки тому +1

    It's a very simple concept

    • @rafael7696
      @rafael7696 4 роки тому

      @@mr.knight8967 very easy too

  • @sthetatos
    @sthetatos 4 роки тому

    Take a=Pi and epsilon = sqrt of 2. How to calculate (a-epsilon, a+epsilon)? Is this neighborhood well defined? Thanks.

  • @matsnordstrom8584
    @matsnordstrom8584 4 роки тому

    Is this course or playlist following Rudin's "principles of mathematical analysis "? Great work. Will follow!

    • @spicyy812
      @spicyy812 Рік тому +1

      late reply, but its following Abbotts understanding analysis.

    • @izumiasmr
      @izumiasmr 11 місяців тому

      ​@@spicyy812thanks 🙏

  • @ishaangoud3180
    @ishaangoud3180 2 роки тому

    Is this topic connected to Metric Spaces?

  • @freddyfozzyfilms2688
    @freddyfozzyfilms2688 3 роки тому

    when take an epsilon from each set in the finite intersection, does this step require the axiom of choice? Since there could be an uncountable number of epsilons

    • @사기꾼진우야내가죽여
      @사기꾼진우야내가죽여 3 роки тому

      Although there exist uncointably many choices of epsilon for each set, there are only finitely many sets from which we choose epsilon, so we don't need the axiom of choice .
      I think the choice of epsilons from each of finitely many sets can be done by mathematical induction.

    • @freddyfozzyfilms2688
      @freddyfozzyfilms2688 3 роки тому

      @@사기꾼진우야내가죽여 The fact that each set is open means that the epsilon has been chosen for us right?

  • @moorsyjam
    @moorsyjam 4 роки тому

    For the proof of the finite insection of open intervals, doesn't that only hold if the intersection is non-empty?

    • @griffine6111
      @griffine6111 4 роки тому

      If the intersection of a finite number of open sets is the empty set, that still works since the empty set is open!
      If you want to talk about the "empty intersection" which is when you are intersecting no sets, this is the whole set, which is also open. (This is like multiplying no copies of a number together and getting 1. Written usually as x^0 =1.)

    • @moorsyjam
      @moorsyjam 4 роки тому

      @@griffine6111 I get that the theorem still holds, since he showed the empty set is open. He's just taking an element of the intersection for the proof, which is kinda difficult if it's the empty set.

    • @lucascaique2943
      @lucascaique2943 4 роки тому

      It's implied that the intersection is non-empty, since we know the empty set is open.

    • @thesecondderivative8967
      @thesecondderivative8967 Рік тому

      I believe the proof implies that the intersection is non-empty. If the intersection were empty, then we use the fact that the empty set is open.

  • @tahafakhech7712
    @tahafakhech7712 4 роки тому +2

    This will be the greastest playlist of all time

  • @arvindsrinivasan424
    @arvindsrinivasan424 4 роки тому +1

    🔥🔥🔥

  • @athelstanrex
    @athelstanrex 4 роки тому +1

    im first

  • @mrrashedali
    @mrrashedali 4 роки тому

    And I'm second