f implies continuous

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  • Опубліковано 22 гру 2024

КОМЕНТАРІ • 27

  • @ProfOmarMath
    @ProfOmarMath 3 роки тому +24

    As soon as I saw the thumbnail I thought "discrete metric"

    • @athelstanrex
      @athelstanrex 3 роки тому +2

      The power of metrics... Also i love your channel, Professor Omar!!

    • @drpeyam
      @drpeyam  3 роки тому +3

      Nice!!!

  • @seanfraser3125
    @seanfraser3125 3 роки тому +16

    This is what a mathematician thinks clickbait is

  • @EpicMathTime
    @EpicMathTime 3 роки тому +17

    The flip side of this coin: every function is also continuous when we put the _indiscrete_ topology on the target space.

    • @ProfOmarMath
      @ProfOmarMath 3 роки тому +4

      Hey Epic. Your video on the group of matrices with all entries equal was really nice. It's a great example to really focus on understand the definition of a group. Good stuff.

    • @EpicMathTime
      @EpicMathTime 3 роки тому +4

      @@ProfOmarMath Thanks, glad you liked it!

    • @curiosityzero2151
      @curiosityzero2151 3 роки тому

      Love ur videos man

  • @imnotarobot6927
    @imnotarobot6927 3 роки тому +4

    Love the beginning :D
    "Every function is continuous"
    *stares*
    "ᵖʳᵒᵛᶦᵈᵉᵈ ᵗʰᵃᵗ ʸᵒᵘ ᵘˢᵉ ʷʰᵃᵗˢ ᶜᵃˡˡᵉᵈ ᵃ ᵈᶦˢᶜʳᵉᵗᵉ ᵐᵉᵗʳᶦᶜ"

  • @curiosityzero2151
    @curiosityzero2151 3 роки тому +8

    Yes thats what I have been always advocating for

  • @iabervon
    @iabervon 3 роки тому +1

    As for the combination of both spaces having the discrete metric, the thing is that all functions whose domain uses that metric are uninterestingly locally constant, since they only locally have one input.

  • @MrNygiz
    @MrNygiz 3 роки тому +1

    let f map to a discrete space and assume f is continuous and not locally constant.
    Then there exists a point s in S such that f is not constant on any U-open containing s.
    Now take the set U=f^-(f(s)) this must be open by continuity of f and contains s. as f is also constant on this set we have a contradiction

  • @abdonecbishop
    @abdonecbishop 3 роки тому +1

    Thank you Dr. Peyam

  • @factsheet4930
    @factsheet4930 3 роки тому +2

    The real numbers are a countable union of countable sets...
    Provided you forget about the axiom of choice 👍🏻

  • @Jaylooker
    @Jaylooker 3 роки тому

    f^-1 (U) -> U reminds of covering spaces

  • @f5673-t1h
    @f5673-t1h 3 роки тому +1

    Every function is continuous!!!... provided they are to an INdiscrete space

  • @txikitofandango
    @txikitofandango 3 роки тому +2

    This proof is like the cracks in the edifice of real numbers

  • @mathadventuress
    @mathadventuress 3 роки тому +1

    Even 1/x at 0?

    • @drpeyam
      @drpeyam  3 роки тому +4

      You can’t strictly speaking talk about continuity at points where it’s undefined

    • @toaj868
      @toaj868 3 роки тому

      The definition of a function requires it to be defined for all elements in the domain so for 1/x to be a function we would have to exclude 0 from its domain.

  • @pythoncake2708
    @pythoncake2708 3 роки тому

    i remember we had an exercise to prove that in Analysis I

  • @albertaraujo6304
    @albertaraujo6304 3 роки тому

    YAY, I never have to prove continuity again! Every function is continuous. lol. Way over my head! :)

  • @dgrandlapinblanc
    @dgrandlapinblanc 2 роки тому

    Weird. Thank you very much.

  • @mikhailmikhailov8781
    @mikhailmikhailov8781 3 роки тому

    Every subset is open, hence every function is continuous.
    meh, its a stupid example. i guess for n point discrete metric spwces you can isometrically embed them to space to R^n, so there is that. It would be interesting if you can get infinite dimensional "simplices" in some Hilbert space somehow and embed R or Z with the discrete topology in them.

  • @mrmathman202
    @mrmathman202 3 роки тому

    You're number one you're number one everyone's number one

  • @Tristan.211
    @Tristan.211 3 роки тому

    Me failing my calc 1 exam