Open thread, Oct. 12 - Oct. 18, 2015

MrMind

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If it's worth saying, but not worth its own post (even in Discussion), then it goes here.

Notes for future OT posters:

1. Please add the 'open_thread' tag.

2. Check if there is an active Open Thread before posting a new one. (Immediately before; refresh the list-of-threads page before posting.)

3. Open Threads should be posted in Discussion, and not Main.

4. Open Threads should start on Monday, and end on Sunday.

So the sequence of distributions that you gave, or my example of the sequence 0,1,2,3,4,... both don't converge. Calling them a supertask doesn't change that fact.

I don't understand you.

Thomas, please read and understand query's response above. In attempting to dismantle a concept you don't like, you've lost precision. Formalize your questions and concerns rigorously and then see if a seeming contradiction is still there.

12query11y

Your question of "after finishing the supertask, what is the probability that 0 stays in place" doesn't yet parse as a question in ZFC, because you haven't specified what is meant by "after finishing the supertask". You need to formalize this notion before we can say anything about it. If you're saying that there is no formalization you know of that makes sense in ZFC, then that's fine, but that's not necessarily a strike against ZFC unless you have a competitive alternative you're offering. The problem could just be that it's an ill-defined concept to begin with, or you just haven't found a good formalization. Just because your brain says "that sounds like it make sense", doesn't mean it actually makes sense. To show that ZFC is inconsistent, you would need to display a formal contradiction deduced from the ZFC axioms. "I can't write down a formalization of this natural sounding concept" isn't a formal contradiction; the failure is at the modeling step, not inside the logical calculus.

6twanvl11y

Define the sequence S by S(0) = 0 S(n+1) = 1 + S(n) This is a sequence of natural numbers. This sequence does not converge, which means that the limit as n goes to infinite of S(n) is not a natural number (nor a real number for that matter). You could try to write it as a function of time, S'(t) such that S'(1-0.5^n) = S(n). That is, S'(0)=0, S'(0.5)=1, S'(0.75)=2, etc. A possible formula is S'(t) = -log_2(1-t). You could then ask what is S'(1). The answer is that this is the same as the limit S(infinity), or as log(0), which are both not defined. So in fact S' is not a function from numbers between 0 and 1 inclusive to natural or real numbers, since the domain excludes 1. You can similarly define a sequence of distributions over the natural numbers by T(0) = {i -> 0.5 * 0.5^i} T(n+1) = the same as T(n) except two values swapped This is the example that you gave above. The sequence T(n) doesn't converge (I haven't checked, but the discussion above suggests that it doesn't), meaning that the limit "lim_{n->inf} T(n)" is not defined.