• wisha@lemmy.ml
      link
      fedilink
      English
      arrow-up
      10
      arrow-down
      1
      ·
      edit-2
      6 months ago

      It might sound trivial but it is not! Imagine there is a lever at every point on the real number line; easy enough right? you might pick the lever at 0 as your “first” lever. Now imagine in another cluster I remove all the integer levers. You might say, pick the lever at 0.5. Now I remove all rational levers. You say, pick sqrt(2). Now I remove all algebraic numbers. On and on…

      If we keep playing this game, can you keep coming up with which lever to pick indefinitely (as long as I haven’t removed all the levers)? If you think you can, that means you believe in the Axiom of Countable Choice.

      Believing the axiom of countable choice is still not sufficient for this meme. Because now there are uncountably many clusters, meaning we can’t simply play the pick-a-lever game step-by-step; you have to pick levers continuously at every instant in time.

      • finestnothing@lemmy.world
        link
        fedilink
        English
        arrow-up
        10
        ·
        6 months ago

        This would apply if I had to pick based on the set of levers in each group. By picking the first one I see I get out of the muck of pure math, I don’t care about the set as a whole, I pick the first lever I see, lever x. Doesn’t matter if it’s levers -10 to 10 real numbers only, my lever x could be lever -7, the set could be some crazy specific set of numbers, doesn’t matter I still pick the first one I see regardless of all the others in the set.

        Pure math is super fun, but reality is a very big loophole

        • wisha@lemmy.ml
          link
          fedilink
          English
          arrow-up
          1
          arrow-down
          1
          ·
          6 months ago

          But look at the picture: the levers are not all the same size- they get progressively smaller until (I assume from the ellipsis) they become infinitesimally small. If a cluster has this dense side facing you, then you won’t “see” a lever at all. You would only see a uniform sea of gray or whatever color the levers are. You now have to choose where to zoom in to see your first lever.

          • finestnothing@lemmy.world
            link
            fedilink
            English
            arrow-up
            6
            ·
            6 months ago

            They get smaller to show that they’re further away in the background not that they get infinitely small. If they were actually getting smaller, then sure, I grab an electron microscope, look at a field of levers, zoom until I see one, and pick that one, then somehow throw an electron sized lever, move to the next, smaller, physics defying lever group and just wait for quantum mechanics to do it’s thing I guess

            • wisha@lemmy.ml
              link
              fedilink
              English
              arrow-up
              2
              arrow-down
              1
              ·
              6 months ago

              They have to get smaller to fit the problem statement- if all levers are the same size or have some nonzero minimum size then the full set of levers would be countable!

              Now we play the game again 🤓. I start by removing the levers in the field/scale of view of your microscope’s default orientation.

              • finestnothing@lemmy.world
                link
                fedilink
                English
                arrow-up
                3
                ·
                6 months ago

                Then I moved the microscope until it finds at least one, pick the first one from the new lever group, and my power takes care of throwing that first found/seen lever in the same instant as me throwing it in a normal set of levers

      • XPost3000@lemmy.ml
        link
        fedilink
        English
        arrow-up
        2
        ·
        6 months ago

        “You have to pick levers continuously at every instant in time”

        Supertasks: 🗿

    • i_love_FFT@lemmy.ml
      link
      fedilink
      English
      arrow-up
      2
      ·
      6 months ago

      Yeah but then like that person said, they will disassemble the trolley in a weird way and put back together two trolleys, one on each track.

  • RagingHungryPanda@lemm.ee
    link
    fedilink
    English
    arrow-up
    9
    ·
    6 months ago

    I know you can’t enumerate them all, but you just have to enumerate them faster than the trolly. and live forever

  • Leate_Wonceslace@lemmy.dbzer0.com
    link
    fedilink
    English
    arrow-up
    7
    ·
    6 months ago

    The image suggests that a closest element of each cluster exists, but a furthest element does not, so I will pull the closest lever in each cluster.

  • mcz@lemmy.world
    link
    fedilink
    English
    arrow-up
    5
    ·
    6 months ago

    Help me, I assumed that it’s possible but then two men appeared to decompose the train and put the parts back together into two copies of the original train

    • pyre@lemmy.world
      link
      fedilink
      English
      arrow-up
      3
      ·
      edit-2
      6 months ago

      that’s what i thought. I’m sure something’s going way over my head but my first thought was “how is this a tough choice or even a question”

  • Zkuld@lemmy.world
    link
    fedilink
    English
    arrow-up
    2
    ·
    6 months ago

    I invoke the axiom of choice and hope for the best. because if it doesn’t work we have bigger problems then 4 dead people