“If a gate has particles at both switch-wire inputs, this formulation allows some successor states that have two particles at the same position. However, that does not occur in any of the examples here.”

So apparently two particles can just occupy the same placeat the same time.

Incidentally, one way to think of this theorem is as a generalization of the fundamental theorem of algebra since in the single variable case all polynomials are surjective. Similarly, one gets from Picard’s theorem that all injective analytic functions are surjective (consider f composed with itself).

Thank you for a very interesting post. Unfortunately my set theory is very poor so I got lost near the beginning:

Can someone explain how we know that if a series converges to 0 for all {1/n | n c N}, it must converge to 0 at 0? I don’t understand why we know it “converges to 0 on, e.g. (-1,0) U (1,0)”: how do we know that this is true on the infinitely many points of the form x/y (1<x<y,x is coprime to y)?

Thanks!

No, externally we know there are infinitely many numbers less than ω, but the model doesn’t know it. The model thinks there are finitely many– namely, ω many (which the model thinks is a finite amount).

Second, you say: “construct Z and Q from N as usual”. Could you clarify? How do we extend an arbitrary primitive recursive function from N to Z to Q? Especially if it’s defined using recursion, which leads to an infinite loop if given anything besides a natural…

This is surprising, considering that the ordinal notations do not form an arithmetical, or even hyperarithmetical set. Would you mind explaining how it can be done?

Your example with G(f,x) = 0 indicates that you don’t understand the solution. An important point to the solution is that the representative g that is chosen is the same regardless of what x is. Otherwise we could not assert that the guesser is wrong for only finitely many x because g(x) differs from f(x) at only finitely many places. In your example, you would have g(x) be identically zero, which obviously does not match f(x) at all but finitely many places.

Your probability argument doesn’t hold because G(f,x) is not a random variable.

This is not correct because you are only asking what his response to the question is. Assume the knight is holding up 1 finger and you ask “Will you answer ‘no’ if I ask you if you are holding one finger up?”
His response to THAT question would be “No(I will answer ‘yes’(when you ask that question))” and he would not have lied.
Ask the knight “Will you answer ‘no’ if I ask you if you are holding two fingers up?” he would respond “Yes(I WILL say ‘no’ (when you ask THAT question))”
By asking “What will your response to Q be?”, you are asking only that, and not question Q itself

thanks

LOL!!!

“The Mathematical Undecidability within Quantum Physics: The Origin of Indeterminacy and Mechanism of Decision at Measurement”

To get a copy click on the following link:
http://steviefaulkner.files.wordpress.com/2010/04/undecidability-in-qm_1034.pdf

Suppose that f is white noise with a non-zero variance. Each f(x) is random, and f(x) and f(x’) are independent of one another if x’ =/= x. Since G(f,x) is a deterministic mapping, and since it uses information about f everywhere except at x, G(f,x) and f(x) are independent. By independence, the variance of the error VAR[G(f,x) - f(x)] for a given x is at least VAR[f(x)]. It follows that the probability of a correct guess (even over random x) is zero.

The article’s proof seems to be based in the statement “the set {x | g(x) differs from f(x)} has measure 0,” which is why a random selection of x helps. However, I believe it is implicit in the proof is that f is fully known to the guesser — it’s fixed at the beginning of the problem and used to construct a “nice” G(f,x) that yields G(f,x) = f(x) except possibly at a finite number of points.

If the guesser didn’t know f, G(f,x) couldn’t be chosen so nicely. Suppose f is non-zero except at finitely many points. For any given f and x, choose G(f,x) = 0. The function g with g=f except at x where we define g(x) = 0 certainly belongs to the same class as f, so we can make it the representative function for that class. However, fix f and define g(x) = G(f,x). Now, g=0, which means that g and f differ at infinitely many points. The probability of guessing f(x) is now zero.

Is there a subtlety in the axioms that I’m misunderstanding?

- max

http://www.math.wisc.edu/~keisler/calc.html

To AlefSin: Looking at your link, it looks like Wolpert has disproved something like the existence of a prediction machine existing within the universe itself, but the kind of determinism that Drescher is defending does not require that.

hmmm, but if I understand correctly it contradicts Worlpert’s results (Wolpert, D. H. Physica D 237, 1257–1281 (2008)). It’s a bit technical so here is a simpler version: http://www.astro.uhh.hawaii.edu/documents/Binder_nv-toae.pdf

Basically, Wolpert uses Cantor diagonalization to disprove Laplace’s demon.

I’ll need another read through to properly get it though, probably after I’ve had some more sleep sometime

It looks as though you are not considering field variables but am I right in believing that under the Field Axioms, formulae proposing the existence of non-rational numbers are undecidable due to combined effects of Soundness and Completeness Theorems? If so, you might find this an interesting connection to your work.

However, that probably says more about my lack of breadth of knowledge than it does about the applications of the context. I’ve seen the book “Symmetric Properties of real functions” referenced as a analytical basis for studying these types of things, and you can see a preview at:
http://books.google.com/books?id=BMWk0X8rl_YC&lpg=PP1&ots=SF0LuDF6P8&dq=thomson%20symmetric%20properties%20of%20real%20functions&pg=PR11

In any case, I could easily be wrong, but as far as I know, this result is only entertainment; I don’t know that it tells anything fundamentally new either about cardinalities or real functions.