| I am not remotely convinced by this argument. The first flaw I see is that the author is imprecise by commingling probabilities (0%, 100%) with absolutes (possible, impossible, none, never, etc). > After all, probability-zero events do happen. Not a problem! Just pick two new real numbers! And if this fails, pick again! Probability-zero events happen all the time. The probability of getting any specific value selected uniformly at random from the unit interval (say, 0.232829) is zero. Probability-zero events should not be conflated with properties that exist nowhere. > We can now state that for any such mapping, none of the three reals is in the countable set assigned to the others. And this entails that we can prove that |(ω)| > |ω2|! In other words, we can prove that there are at least TWO cardinalities in between the reals and the naturals! That's... not how cardinalities work. Just because you have two sets with different elements does not mean they have different cardinalities. For instance, consider the set of integers {..., -1, 0, 1, 2, ...} vs the set of half-integers {..., -1/2, 1/2, 3/2, 5/2, ...}. These clearly have different elements, but you can easily construct a bijection between the two (just add 1/2 to each element in your set of half-integers), so you can demonstrate that they have the same cardinality. > We define f(x) to be {y | y ≤ x} Um, no. This demonstrates the existence of one such mapping. It does not demonstrate that the set of such mappings covers any substantial portion of the entire space of possible mappings. |