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Help with ordinal numbers

  1. May 16, 2012 #1
    Why is there no number class containing all the ordinal numbers?
     
  2. jcsd
  3. May 16, 2012 #2

    micromass

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    What is a "number class"??
     
  4. May 16, 2012 #3
    If there were such an ordinal, then it would be the largest ordinal. This is impossible, the easy answer being given any cardinal number (pick the one corresponding to the largest ordinal) you can create a larger cardinal by the powerset operation. This contradicts the assumption of a largest ordinal.

    This problem is related to the impossibility of a set of all sets.

    I don't know a lot of this subject, but I do know the Princeton Companion to Mathematics edited by Timothy Gowers has a wonderful discussion of ordinal numbers. This is truly a fantastic book, one every math major should look through (just don't be surprised if you have difficulty putting it down once you start reading!).
     
  5. May 17, 2012 #4
    I'm totally new to ordinal numbers, I'll have to check out that book.

    It's seems you're saying the following. An ordinal number corresponds with a certain cardinality and these correspond with the number classes. For example, the set of all finite ordinals (each corresponding to the different finite cardinalities -- some natural number?) makes up the first number class. Then the countable ordinals make up the second number class. Every ordinal number A has the same cardinality as some ordinal number B if the set of all ordinal numbers less than A or B share the same cardinality. But, then A and B would also be in the same number class. So, if there was a number class that contained all ordinal numbers, it would contain the largest ordinal number -- or it would be the number class OF the largest cardinal (?). But this doesn't work as you showed. There will always be a next cardinality, and therefore a next ordinal number and a next number class.

    Is that right?
     
  6. May 17, 2012 #5
    That sounds right to me. I think we mean the same thing by number class and cardinal. A cardinal is an equivalence class of ordinals (hence your terminology). I am using some hand waving when I say pick an ordinal corresponding to the cardinal, but I'm not sure if there is a way to avoid choice.
     
  7. May 17, 2012 #6

    micromass

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    Choice?? I don't see how choice should be involved here??
     
  8. May 17, 2012 #7

    jgens

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    That the class of all ordinals is a proper class has a simpler proof than this. If the class of all ordinals were not a proper class, then it would be an ordinal, which is a contradiction.
     
  9. May 17, 2012 #8

    micromass

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    A contradiction only if [itex]x\in x[/itex] is not allowed. If you allow that, then you need to look at x+1.
     
  10. May 17, 2012 #9

    jgens

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    True enough. I always assume ZF/ZFC unless the OP states otherwise, but I guess you can still do a lot of interesting math without the axiom of regularity. So that is probably a better proof since it works in both cases.
     
  11. May 17, 2012 #10

    micromass

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    I was just being annoying :biggrin:

    For some reason I never see regularity as a proper axiom of ZFC. I don't know why I don't want to accept it...
     
  12. May 17, 2012 #11
    I didn't know that a cardinal number is an equivalence class of ordinals. I'm not entirely sure what this means. For example, you have some cardinal number, say, aleph_1 and then you have some set or class of all countable ordinals. Now, all I know is that the set of all countable ordinals makes up Cantor's second number class and has cardinality aleph_1. I'm not sure where "equivalence classes" comes in or how to use the term. Is an equivalence class a kind of number class? I guess I'm not to sure on what a number class is either, other than what different "kinds" of ordinals (finite, countable,...) are "members" of. But, I'm still unclear about what "equivalence classes" does here?
     
  13. May 17, 2012 #12
    As the name suggests, you use equivalence classes when you want, for all intents and purposes, to call two (technically) distinct objects the same.

    Take for example modular arithmetic. Lets assume we are working mod 12. Then we write such things as [itex]3 +4\equiv 7, 8+8\equiv 4[/itex]. It's as easy as clockwork (pun intended). The thing is, in one sense (integers) 16 and 4 are different numbers. Mod 12, they are the same. We make up some terminology, and say anything that should equal 4 (mod 12) is in the equivalence class of 4.

    So an equivalence classes can be thought of in a set theoretic way where two elements are in the same class if they behave like equality. What are the properties of the equal sign? Well,
    1. [itex]x = x[/itex] for every element
    2. [itex]x = y \Rightarrow y = x [/itex]
    3. [itex]x = y \mathrm{\ and\ } y = z \Rightarrow x = z [/itex]

    In the previous example, the "modular" equals is our equivalence relation (denoted [itex]\equiv[/itex]. There are 12 equivalence classes. Numbers relating to 0,1,2,...,11.

    Cardinality are the equivalence classes of ordinal numbers. So take the first infinite ordinal [itex]\omega[/itex]. Then [itex]\omega + 1[/itex] is also an ordinal. These two are in the same equivalence class, with the class as a whole being called the cardinal.

    Edit: Thanks jgens, for the correction.
     
    Last edited: May 18, 2012
  14. May 17, 2012 #13

    jgens

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    I have no argument with the mathematical content of your post, just a friendly FYI that the correct phrase is "for all intents and purposes" and not "for all intensive purposes"

    You should specify the equivalence relation (I know bijection gives you the relation you want, but the OP might not).
     
  15. May 18, 2012 #14
    Well, it may be because, contrary to popular belief, regularity is not what prevents Russell's paradox from going through in ZFC. Indeed, if the other axioms allowed Russel's paradox to go through, then regularity would be powerless to stop it. Rather, it is the replacement of the axiom of unrestricted comprehension with the axioms of restricted comprehension, infinity, and replacement that allows the universe of sets to have a model. After that, the rest of the axioms (of ZF, not choice) just constitute conservative extensions, meaning they just narrow the class of models, not prevent there from being one.
     
  16. May 20, 2012 #15
    Sweet, I didn't know that either.
    Nice. I didn't get it at all at first, as I had no idea what mod arithmetic was. But I got it (both, really) shortly after. So, if we are "working" mod 4, 2 + 3 = 1?
    So, this equivalence is with respect to something right? They act as if they have the same cardinality, but certainly not as if they have the same order type. So, they are equivalence classes "with respect to" one-to-one correspondences, right? I think this is what jgens was saying....
     
  17. May 21, 2012 #16
    Actually, it was shown by Gödel that consistency of ZFC follows from consistency of ZF. So you can add another axiom to your list :smile:
     
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