You must log in or register to comment.
The first lever I see in each group.
I’d say whichever lever in each group feels the most like a “Steve” to me.
I’d prefer the last lever I see in each group.
This feels like a Stanley Parable reference but it’s been a while…
The first one I encounter.
Did I do it? Or did I fundamentally misunderstand the question
https://en.m.wikipedia.org/wiki/Axiom_of_choice
The axiom of choice asserts that it is possible to pick an arbitrary element from every set. Most of mathametics accepts this. However constructivist math does not.
Not as a general principle. That doesn’t mean that constructivists say that there can’t be sets for which the operation is valid. In particular enumeration is not a precondition for a thing to be pickable.
Now they say that the levers are indistinguishable, which means that their difference actually does not lie in their identities, but their relationship to the space they’re in (or everything would collapse into itself), thus I don’t have to look at the levers I can look at the space. They say that “I can’t enumerate them all” but that means that there’s at least a countably infinite number of them.
So the solution is easy: I take the space, throw away all of it that doesn’t hold a that countably infinite subset, observe that the result is now isomorphic to the naturals, then cut it down to six, and throw a dice. There, not just arbitrary but even (a bit) random.
Really, only ultrafinitists would have trouble with this… but then they’d turn it around and challenge you to actually construct that infinite number of levers for real, not just in the abstract, and untie everyone while you’ve stopped the tram due to being caught in an endless loop.
The problem is “indistinguishable” levers.
In the strict sense, if there was a lever you could see first, they would not be indistinguishable. They should not be distinguishable by any property including location
“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!
Just pull every one, I know one in each cluster will work, but like I gotta make sure
Can I take the axiom of choice?
Sorry, we sold out of that 5 min before you walked in.
I know you can’t enumerate them all, but you just have to enumerate them faster than the trolly. and live forever
Just pull out a few thousand levers and throw them in front of the trolley.
Pop() one lever from each set.
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
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.
That one
Since any one will work I just pull a nearby lever at random and go home
Yo this sounds suspiciously similar to how quantum resistant lattice cryptography works.
Too complicated I’m just going to walk away
Murderer…
I would just pick the value from the root of each underlaying balanced binary tree, easy.
