Cantor Confusion
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From: David Marcus on 12 Jan 2007 16:22 Andy Smith wrote: > Yes, sorry speaking too loosely. Of course no problem > with an infinite set having a max and a min. But if you > can systematically order the set between max and min > that puts its elements in correspondence with the natrural > numbers, but has a first and a last element (and there > exists to greatest last natural number, trivially from > Peano axiom). Here is a question for you: Suppose you have a countably infinite ordered set X. So there is a bijection f:N -> X. Does it have to be true that m < n implies that f(m) < f(n)? -- David Marcus
From: Virgil on 12 Jan 2007 17:09 In article <MPG.2011c51e73b463e1989aef(a)news.rcn.com>, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: > Dave Seaman wrote: > > On Fri, 12 Jan 2007 11:48:38 EST, Andy Smith wrote: > > > > > that you can have an ordered > > > infinite sequence which has a finite start and end point. > > > For infinite, read "countably infinite". > > > > Of course. The rationals in [0,1], for example. > > Normally, a "sequence" doesn't have a last element. So, the rationals > are an example of a countably infinite set (not sequence) that has a > finite start and end point (where "start" and "end" mean smallest and > largest). If you mean "those" rationals, the ones in [0,1], yes. But "the" rationals could be misinterpreted to mean all rationals, particularly if the lead in sentence gets snipped. > > Andy seems to be confusing sequence with countably infinite. Not sure > why he is doing this or how to make him stop doing it.
From: cbrown on 12 Jan 2007 17:32 David Marcus wrote: > cbrown(a)cbrownsystems.com wrote: > > Consider the set of all ordered pairs of natural numbers, (m, n). > > > > Now define the ordering: > > > > (m, n) <= (x,y) if > > n < y > > OR > > n=y and m<=x. > > > > Then you should be able to see that: > > > > This is a total order: > > (m,n) < (x,y) or (m,n) = (x,y) or (m,n) > (x,y) > > > > There is a smallest element: > > (0,0) <= (m,n) for all m,n > > > > The ordering is a well-order: > > If S is any set of such elements {(x,y)}, there exists (m,n) in S such > > that for all (x,y) in S, (m,n) <= (x,y). > > > > It contains a subset whose ordering is the same as the ordering of the > > naturals: > > (m, 0) <= (n,0) iff m <= n > > > > There is no largest element /in/ that /subset/: > > (m, 0) < (m+1,0) for all m > > > > There is an element which is "larger than any natural": > > (m,0) <= (0,1) for all m > > > > In some sense, (0,1) is "like" the ordinal w (roughly - for > > visualization purposes only!) Think of w+1 as (1,1), w+2 as (2,1), w + > > w = 2*w as (0,2), etc. > > That's a good example. It wouldn't be hard to make your "like" precise, > i.e., as the two being order isomorphic. > Sure; just define U((m,n)) = {(u,v) : (u,v) < (m,n)}. Let U = {U((m,n))}. Then there is a bijection between the elements of U and the ordinals less than w^2 that preserves ordering under inclusion. We could extend this to w^2, w^3, etc. by applying a similar lexicographic ordering to the set of all finite length sequences (a_0, a_1, ..., a_n, 0, 0, ...) where each a_i is in N. But when we "get to" w^w, even I get confused! I suppose then we can consider U x N to be the "next step"... ? Cheers - Chas
From: MoeBlee on 12 Jan 2007 18:01 Andy Smith wrote: > As I understand it, ordinals are constructed abstractly as a set of successive elements - then one can say, oh, and there is the set of ALL of them, then another with that and a successor, and so on. First, 0 is proven to exist from the axiom schema of separation (and, as generally for sets, uniqueness is proven from the axiom of extensionality). Each finite successor ordinal can be proven to exist using the proven existence of 0 and then, primarily, the union and pairing axioms, even without the axiom of infinity. Then, omega, is proven to exist from the axiom of infinity. Then successor ordinals after omega are proven to exist by using the proven existence of omega and again using union and pairing axioms . And to prove the existence of limit ordinals after omega, we use the axiom schema of replacement. For example, the first limit ordinal after omega is proven to exist by applying the axiom schema of replacement. > Cardinality relates to demonstrating a > correspondence (or not) between elements of infinite sets. The cardinality of a set x is (in the usual method) the least ordinal that has a 1-1 correspondence with x. Then it follows that sets x and y then have the same cardinality iff there is a 1-1 correspondence between x and y. But equinumerosity is usually even earlier in the definitions than cardinality, since we don't need to mention the cardinality of sets just to say that there is or is not a 1-1 correspondence between them. > But how would one visualise the abstract concept of a > transfinite number - in a comparable way to {1/2,1/4,1/8, ...}? By visualizing order structures. The best way to approach that is to read about them in a textbook on set theory. MoeBlee
From: Andy Smith on 12 Jan 2007 09:18
> > Here is a question for you: Suppose you have a > countably infinite > ordered set X. So there is a bijection f:N -> X. Does > it have to be true > that m < n implies that f(m) < f(n)? > At the risk of appearing an idiot, I would say yes. Because X is ordered such that x(n) > x(m) for all n > m. Am I missing something? |