I didn’t follow your argument here, particular the part under “If P returns K=S:”
Ah sorry, I was somewhat sloppy with notation. Let me attempt to explain in a somewhat cleaned up form.
For a given statespace S (that is, S is a set of all possible states in a particular problem), you’re saying there exists a deterministic predictor PS that fulfills certain properties:
First, some auxiliary definitions:
TS⊆S is the subset of the statespace S where the ‘true’ answer is ‘YES’.
FS⊂S is the subset of the statespace S where the ‘true’ answer is ‘NO’
By definition from your question, TS∪FS=S and TS∩FS=∅
Then PS is a deterministic function:
PS(K)={Done,iff K=TS,Rotherwise.
where:
K⊆R⊆TS
(And both K and R are otherwise unconstrained.)
Hopefully you follow thus far.
So.
I choose a statespace, S={(0,0),(1,0),(0,1),(1,1)}
I assume there exists some deterministic predictor PS for this statespace.
I choose a particular problem: TaS={(0,0),(1,0),(0,1)}
(That is, in this particular instance of the problem (1,1) is the only ‘NO’ state)
I run PS({(0,0),(1,0),(0,1)}) and get a result Ra
That is, with the parameter K={(0,0),(1,0),(0,1)}
K⊆TaS, so this is correct of me to do.
There are three possibilities for Ra:
Ra=Done
I choose a different problem: TbS={(0,0),(1,0),(0,1),(1,1)}
K⊆TbS , so this is correct of me to do.
I run PS({(0,0),(1,0),(0,1)}) and get a result Rb
That is, with the parameter K={(0,0),(1,0),(0,1)}
K⊆TbS, so this is correct of me to do.
There are three possibilities for Rb:
Rb=Done
K≠TbS, so PS did not fulfill the contract
Hence contradiction.
Rb=S={(0,0),(1,0),(0,1),(1,1)}
In this case, PS(K)≠PS(K), as Ra≠Rb.
Hence, PS is not deterministic.
Hence PS did not fulfill the contract.
Hence contradiction.
Rb=<any other value>
In this case PS did not fulfill the contract
Hence contradiction.
Ra=S={(0,0),(1,0),(0,1),(1,1)}
In this case, (1,1)∈Ra but (1,1)∉TaS
Hence Ra⊈TaS
Hence PS did not fulfill the contract
Hence contradiction.
Ra=<any other value>
In this case PS did not fulfill the contract
Hence contradiction.
All paths lead to contradiction, hence PS cannot exist. An analogous argument works for any statespace that isn’t the null set.
(The flaw in your argument is that multiple different TS sets—multiple different sets of ground-truths can be compatible with the same set K of observations thus far. You can try to argue that TS in practice is a complex enough set that it uniquely identifies a single possible world, but that is a very different argument than the flat statement you appear to be making.)
Ah sorry, I was somewhat sloppy with notation. Let me attempt to explain in a somewhat cleaned up form.
For a given statespace S (that is, S is a set of all possible states in a particular problem), you’re saying there exists a deterministic predictor PS that fulfills certain properties:
First, some auxiliary definitions:
TS⊆S is the subset of the statespace S where the ‘true’ answer is ‘YES’.
FS⊂S is the subset of the statespace S where the ‘true’ answer is ‘NO’
By definition from your question, TS∪FS=S and TS∩FS=∅
Then PS is a deterministic function:
PS(K)={Done,iff K=TS,Rotherwise.
where:
K⊆R⊆TS
(And both K and R are otherwise unconstrained.)
Hopefully you follow thus far.
So.
I choose a statespace, S={(0,0),(1,0),(0,1),(1,1)}
I assume there exists some deterministic predictor PS for this statespace.
I choose a particular problem: TaS={(0,0),(1,0),(0,1)}
(That is, in this particular instance of the problem (1,1) is the only ‘NO’ state)
I run PS({(0,0),(1,0),(0,1)}) and get a result Ra
That is, with the parameter K={(0,0),(1,0),(0,1)}
K⊆TaS, so this is correct of me to do.
There are three possibilities for Ra:
Ra=Done
I choose a different problem: TbS={(0,0),(1,0),(0,1),(1,1)}
K⊆TbS , so this is correct of me to do.
I run PS({(0,0),(1,0),(0,1)}) and get a result Rb
That is, with the parameter K={(0,0),(1,0),(0,1)}
K⊆TbS, so this is correct of me to do.
There are three possibilities for Rb:
Rb=Done
K≠TbS, so PS did not fulfill the contract
Hence contradiction.
Rb=S={(0,0),(1,0),(0,1),(1,1)}
In this case, PS(K)≠PS(K), as Ra≠Rb.
Hence, PS is not deterministic.
Hence PS did not fulfill the contract.
Hence contradiction.
Rb=<any other value>
In this case PS did not fulfill the contract
Hence contradiction.
Ra=S={(0,0),(1,0),(0,1),(1,1)}
In this case, (1,1)∈Ra but (1,1)∉TaS
Hence Ra⊈TaS
Hence PS did not fulfill the contract
Hence contradiction.
Ra=<any other value>
In this case PS did not fulfill the contract
Hence contradiction.
All paths lead to contradiction, hence PS cannot exist. An analogous argument works for any statespace that isn’t the null set.
(The flaw in your argument is that multiple different TS sets—multiple different sets of ground-truths can be compatible with the same set K of observations thus far. You can try to argue that TS in practice is a complex enough set that it uniquely identifies a single possible world, but that is a very different argument than the flat statement you appear to be making.)