3 ms·

I really don't understand your objection, mainly because I can't identify the referent of "the a_n < a < b_n part of the proof" in "Does that interact with the

I really don't understand your objection, mainly because I can't identify the referent of "the a_n < a < b_n part of the proof" in "Does that interact with the a_n < α < b_n part of the proof in strange ways?". I can't make "I don't think it is supported that a_n < a < b_n for all n still forces convergence out of S" be a meaningful sentence - it's the case that a_n < a < b_n, but certainly that doesn't force $a$ to lie outside S. If you restrict to the case when S is countable, then the thing that forces $a$ to lie outside S is rather that Bob has enumerated S and made certain that Alice's $a$ is bounded some concrete distance away from every s_n; in fact, for every $s_n$ he could tell you why $a$ is not equal to $s_n$, by telling you "$a$ is at least this much away from $s_n$" for each $n$. We don't even care what $a$ is; we just know that it's far away from every member of $S$.

roenxi 6y ago
Thanks for your time and attention on the matter; I've enjoyed the thread. I think you might have convinced me on that point; certainly my attempts to follow up were incoherent to the point where even I could tell. > I really don't understand your objection Bob's strategy is bunk, we've started by assuming something that is false. Therefore, the assumption that the reals are countable might already invalidate the proof that a bounded but monotonically increasing sequence converges. If that assumption is invalid, then the argument in this proof might already fail even before the contradiction we later identify. So it is obvious to me we're going to have a contradiction. But the contradiction might be elsewhere than in the true facts that the proof gives. So the proof might not be correct. I suppose I accept the proof will find a contradiction, but I'm not sure that the contradiction it found was the first one to arise. I can imagine that this might be a proof that Bob has a winning strategy without further argument: * Reals are countable * Bounded monotonic increasing sequence converges to a Real * These are inconsistent assumptions so I can prove anything * Therefore Bob has a winning strategy So that means all the additional logic the proof adds in has to be re-verified from first principles to ensure that everything is still valid and internally consistent as far as it went after assuming that the reals are countable. I'm not good enough at maths to know that this specific proof of a winning strategy doesn't contain subtle flaws, so I don't trust it (maybe proof by contradiction just isn't for me). I struggle to accept it as simpler - the Cantor proof constructs a direct counterexample and is very satisfying to me.
roenxi 6y ago
I had a few days to think about it and looked up the Wikipedia article for the monotonic limit theorum family and figured out what my confusion was about. I don't accept that theorem applies; and I don't think this game converges to a real number. Real numbers have a least upper bound [0]. This game doesn't describe something with a least upper bound (assume the least upper bound is U, Bob will then pick a number < U as the upper bound, this contradicts - in the limit there is no upper bound unless the players agree to cooperate). I don't think the game describes a Real number - I think it describes some other mathematical object. [0] https://en.wikipedia.org/wiki/Least-upper-bound_property https://en.wikipedia.org/wiki/Least-upper-bound_property

roenxi 6y ago

Thanks for your time and attention on the matter; I've enjoyed the thread. I think you might have convinced me on that point; certainly my attempts to follow up were incoherent to the point where even I could tell. > I really don't understand your objection Bob's strategy is bunk, we've started by assuming something that is false. Therefore, the assumption that the reals are countable might already invalidate the proof that a bounded but monotonically increasing sequence converges. If that assumption is invalid, then the argument in this proof might already fail even before the contradiction we later identify. So it is obvious to me we're going to have a contradiction. But the contradiction might be elsewhere than in the true facts that the proof gives. So the proof might not be correct. I suppose I accept the proof will find a contradiction, but I'm not sure that the contradiction it found was the first one to arise. I can imagine that this might be a proof that Bob has a winning strategy without further argument: * Reals are countable * Bounded monotonic increasing sequence converges to a Real * These are inconsistent assumptions so I can prove anything * Therefore Bob has a winning strategy So that means all the additional logic the proof adds in has to be re-verified from first principles to ensure that everything is still valid and internally consistent as far as it went after assuming that the reals are countable. I'm not good enough at maths to know that this specific proof of a winning strategy doesn't contain subtle flaws, so I don't trust it (maybe proof by contradiction just isn't for me). I struggle to accept it as simpler - the Cantor proof constructs a direct counterexample and is very satisfying to me.

roenxi 6y ago

I had a few days to think about it and looked up the Wikipedia article for the monotonic limit theorum family and figured out what my confusion was about. I don't accept that theorem applies; and I don't think this game converges to a real number. Real numbers have a least upper bound [0]. This game doesn't describe something with a least upper bound (assume the least upper bound is U, Bob will then pick a number < U as the upper bound, this contradicts - in the limit there is no upper bound unless the players agree to cooperate). I don't think the game describes a Real number - I think it describes some other mathematical object. [0] https://en.wikipedia.org/wiki/Least-upper-bound_property https://en.wikipedia.org/wiki/Least-upper-bound_property