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| Re: Only two alephs?
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In article <00002098+00007b3c@msn.com>, James Wanless <grokltd@msn.com> wrote: >I don't see that Cantor's diagonal proof of the existence of aleph1 >extends to proving the existence of aleph2... As I see it, it relies >on the base aleph's being countable. Can anyone show me otherwise? You need to read a real book about set theory, instead of whatever you have been reading. Here's a proof that will work on any set: Given a set S, it constructs a larger set, called P(S). For concreteness, let's consider your question, which is about sets larger than the set of real numbers. Let's call the set of real numbers R. We will construct a set that is larger than R. Consider all the sets that contain only real numbers; that is, consider the subsets of R. The collection of all such sets is called P(R), the `power set of R', and we will show that it is larger than R. Now, if R and P(R) were the same size, we would be able to find a one-to-one correspondence between their elements. Let's call such a correspondence f. For every real number r there is a set f(r) of real numbers that corresponds to it, and vice versa. f(r) might be the set of real numbers that are between 3 and pi, or it might be the set of real numbers that are roots of eleventh-degree polynomial equations, or it might be the empty set, or it might be something else. Now, since r is a real number and f(r) is a set of real numbers, r might happen be a member of the set f(r). Either r is a member of f(r) or it isn't. If r is a member of f(r), paint it blue; if not, paint it green. Ever real number r is now painted green or blue, but not both. Let G be the set of green numbers. Since f is a one-to-one correspondence, there is some real number g that corresponds to G. That is, such that f(g) = G. Every real number is blue or green. What color is g? Is it green? If it is green, then we painted it green because g is not a member of f(g). But f(g) = G, and G contains all the green numbers, including g. This is a contradiction. Well, then, maybe g is blue. If g is blue, then that is because g is a member of f(g). But f(g) = G, and everything in G is green. g can't be in f(g), because g is blue, not green. Another contradiction! The conclusion is that there is no such corresopndence. P(R) and R are not the same size. On the other hand, it's easy to see that P(R) is at least as big as R, by placing elements of R in one-to-one correspondence with a proper subset of P(R). (That's an exercise for you.) So there you go. P(R), the set of all subsets of R, is bigger than R itself. This proof works just fine for any set at all, so if you want a set bigger than P(R), just do it again; you get P(P(R)). -------------------------------- You should note that aleph-1 is *not* necessarily the size of R. Aleph-1 is defined to be the smallest infinite cardinal number larger than aleph-0. It might be the size of R, or it might the size of some smaller set that is still larger than the set of integers. The conjecture that aleph-1 is in fact the size of R, and that there are no infinite sets larger than Z and smaller than P(Z), is called the `continuum hypothesis.' The C.H. turns out to be independent of the usual axioms of set theory, which means that our intuition about sets is not strong enough to resolve it one way or the other: After we write down a list of all the properties we want sets to have, the C.H. could go either way and still not contradict any of them. -- Mark-Jason Dominus mjd@plover.com
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