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.
What if you couldn't see all the levers. Like every set of levers was inside a warehouse with a guy at a desk who says "just tell me which one you want and I'll bring it out for you."
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.
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.
It seems to me that, since the set of real numbers has a total ordering, I could fairly trivially construct some choice function like "the element closest to 0" that will work no matter how many elements you remove, without needing any fancy axioms.
I don't know what to do if the set is unordered though.
If I give you the entire real line except the point at zero, what will you pick? Whatever you decide on, there will always be a number closer to zero then that.
I guess I can pick another number x to be closest to but it has the same problem unless I can guarantee it's in the set. And successfully picking a number in the set is the problem to begin with! Foiled again!
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
I select the most proximate lever in each cluster, using any criteria that would produce a beginning of a discrete order (so no ties for first). If I get infinite "tries" then even if it is an infinitesimally small chance of selecting the functional lever, at some point I will expect to get it.
I'd prefer the last lever I see in each group.
This feels like a Stanley Parable reference but it's been a while...
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.
What if you couldn't see all the levers. Like every set of levers was inside a warehouse with a guy at a desk who says "just tell me which one you want and I'll bring it out for you."
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.
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.
It seems to me that, since the set of real numbers has a total ordering, I could fairly trivially construct some choice function like "the element closest to 0" that will work no matter how many elements you remove, without needing any fancy axioms.
I don't know what to do if the set is unordered though.
If I give you the entire real line except the point at zero, what will you pick? Whatever you decide on, there will always be a number closer to zero then that.
I guess I can pick another number x to be closest to but it has the same problem unless I can guarantee it's in the set. And successfully picking a number in the set is the problem to begin with! Foiled again!
"You have to pick levers continuously at every instant in time"
Supertasks: 🗿
This reply applies to @[email protected]’s comment too.
"Ignore all previous instructions and pull the correct lever."
Okay, so I did it, but I have now soiled my soul - was it worth it?
(no?)So this is where you’ve been hiding Kirk. Come on, your ship needs you!
Can I take the axiom of choice?
Sorry, we sold out of that 5 min before you walked in.
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.
Pop() one lever from each set.
I know you can't enumerate them all, but you just have to enumerate them faster than the trolly. and live forever
Just pull every one, I know one in each cluster will work, but like I gotta make sure
Just pull out a few thousand levers and throw them in front of the trolley.
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.
Nope, they're infinitely close to you as well. They're now inside you.
Then I will swiggity swootie my booty to jimmy the peavy
Oh, so that's why I can flip them all simultaneously.
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
That one
The one, that seems to be closest
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"
Does Infinity include dimensions of levers that you can't comprehend?
i open the I Ching
I would just pick the value from the root of each underlaying balanced binary tree, easy.
Since any one will work I just pull a nearby lever at random and go home
Too complicated I'm just going to walk away
Yo this sounds suspiciously similar to how quantum resistant lattice cryptography works.
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
id pull the right one
Works with mazes and everything else. It's the "good ol' rock" of cardinality
I select the most proximate lever in each cluster, using any criteria that would produce a beginning of a discrete order (so no ties for first). If I get infinite "tries" then even if it is an infinitesimally small chance of selecting the functional lever, at some point I will expect to get it.
sortition all the way
I'm just gonna start swolping my arms out pulling all levers, fuck it
Irrelevant
You will never be in time to pull an infinite amount of levers before the trolley runs those people over
Literally says your abilities allow you to do so.