| > Stage 4 of the CDA outline can be critiqued on several grounds. The first is that if we have all the numbers in [0,1] in our table a priori, then using the diagonal process to create a number that is not in the original set does precisely that: It creates a number that is not in [0,1]. So why does the constructed anti-diagonal number seem to have a form that puts it within [0,1]? Could it be that ignoring the zero ahead of the decimal point is a bit sneaky? You wrote all this and you don't seem to understand how proof by contradiction works? > If, instead, we construct the anti-diagonal number by counting down the table by positions before each digit selection we find that any anti-diagonal number constructed (from the rectangle) is now within the set we have counted through, and therefore not outside the counted set. This Stretched Diagonal counterargument remains true as d→ ∞, irrespective of any ordering of number representations in the table. This is not how anything works. You can't take a proof where they create an example that leads to contradictions and say, well, if you choose a different example, it doesn't lead to contradictions, so the proof doesn't work. Your "proof" in Appendix B is also just gibberish from a mathematical perspective. The infinity and cardinalities aren't numbers - this isn't a mathematically rigorous statement: lim(n -> inf) log(n) = inf = aleph_0 - it's nonsense. What you're saying here is that as n increases without bound, 2^n also increases without bound. That doesn't prove anything about the cardinality of infinite sets. Also, since you seem to accept that |R| is equivalent 2^|N|, it's also not hard to prove that 2^|S| > |S| for any S. 2^|S| < |S| is trivially untrue and if |S| = 2^|S], there must be at least one bijection F such that F: S => P(S) and inverse F^-1: P(S) => S Now define S' to be a subset of S such that s is in S if & only if it's in S, but not in F(s). Now, consider F^-1(S') - it must be in S, due to how it's defined. Let's consider whether it's also in S': 1. If F^-1(S') is not in S' (= F(F^-1(S')), by definition, it must be in S' 2. If F^-1(S') is in S', by definition it must not be in S' Since both these lead to contradictions, the assumption that there must be a bijection between P(S) and S mut be false. On the whole, you seem to be generally confused - R is not a real thing - it's a mathematical abstraction. |