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Math Help - how to prove that any limit ordinal is really a limit?

  1. #1
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    how to prove that any countable limit ordinal is really a limit?

    If ordinal number \alpha=\beta+1, then we call \alpha a successor ordinal. An ordinal is a limit ordinal if it is not a successor. My question is: how to prove that for any countable limit ordinal \alpha=\{\beta|\beta<\alpha\}, |\alpha|=\aleph_0, there is a sequence in \alpha converging to \alpha? Here we define a sequence in X to be a function on the set of all natural numbers \mathbb N and having values in X, and a sequence of ordinals (\alpha_n) converging to ordinal \alpha if for any \beta<\alpha, there is a natural number N such that \beta<\alpha_n\leq\alpha for all n\geq{N}.
    For example, sequence \alpha_n=\omega+n\rightarrow \omega\cdot2; \beta_n=\omega^n\rightarrow \omega^\omega, .... But these are specific sequences, I need a proof in general situation.
    Thanks!
    Last edited by zzzhhh; August 8th 2009 at 06:05 AM. Reason: Add the omitting "countable" condition
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  2. #2
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    Hi

    You can't prove that; for instance for \aleph_1 there is no such sequence. (And that makes sense: if there is such sequence, for more convenience assume it strictly increases. Think of the "distance" (sorry for my approximative english, by "distance" between a and b I mean card(\{c\ ;\ a<c<b\}) ) between two successive ordinals in that sequence: if at a moment, it exceeds \aleph_0, then the limit will be greater than \aleph_1. But if all "distances" are at most countable, since a countable reunion of countable sets is countable, then you cannot reach \aleph_1 ).

    You can find answers with cofinality (Cofinality - Wikipedia, the free encyclopedia )
    Last edited by clic-clac; August 7th 2009 at 01:23 PM. Reason: parenthesis forgotten
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    Thank you for pointing out the mistake. I forgot to add the restriction "countable" in front of "limit ordinal". Without this, the sequence is like a car heading for a place far far away but carrying insufficient gasoline; it must stop somewhere in the middle without the least hope of arriving. I'm very sorry for omitting such an important condition. Now I have edited the post and wonder if you could give me some ideas. Thanks!
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    Hi

    Countable is a sufficient condition for the existence of such a sequence, but not a necessary one (as you said in your exemples, you can reach \omega^{\omega} which is not countable) This is important, great limit ordinals can be reached with not so much gasoline ( \aleph_\omega for instance)

    Now for a countable limit ordinal \alpha. Well in fact \alpha itself gives you a sequence as you wanted..!
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    Thank you but ... \omega^\omega IS a countable ordinal, that is, there is a 1-1 correspondence between \omega^\omega and \omega_0, but there is no such one between \omega^\omega and \omega_1. See the following picture extracted from "Introduction to Set Theory", 3ed, Revised and Expanded, written by Karel Hrbacek and Thomas Jech, P122(available from gigapedia). I used this textbook to study axiomatic set theory by myself.

    And, what is the sequence I want that countable limit ordinal \alpha can give? Could you write down explicitly the general term \alpha_n or prove the existence of this sequence by contrapositive? Thanks.
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  6. #6
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    Quote Originally Posted by zzzhhh View Post
    If ordinal number \alpha=\beta+1, then we call \alpha a successor ordinal. An ordinal is a limit ordinal if it is not a successor. My question is: how to prove that for any countable limit ordinal \alpha=\{\beta|\beta<\alpha\}, |\alpha|=\aleph_0, there is a sequence in \alpha converging to \alpha? Here we define a sequence in X to be a function on the set of all natural numbers \mathbb N and having values in X, and a sequence of ordinals (\alpha_n) converging to ordinal \alpha if for any \beta<\alpha, there is a natural number N such that \beta<\alpha_n\leq\alpha for all n\geq{N}.
    For example, sequence \alpha_n=\omega+n\rightarrow \omega\cdot2; \beta_n=\omega^n\rightarrow \omega^\omega, .... But these are specific sequences, I need a proof in general situation.
    Thanks!
    If \alpha = \{\beta:\beta<\alpha\} is countable then there exists an enumeration \alpha = \{\beta_1,\beta_2,\beta_3,\ldots\} of \alpha. Construct a subsequence of (\beta_n) by omitting all elements less than the previous element of the subsequence. More precisely, let n_1=1, and for k\geqslant1 let n_{k+1} = \min\{n:\beta_n>\beta_{n_k}\}. Then (\beta_{n_k}) is an increasing sequence of ordinals less than \alpha that is eventually larger than any ordinal less than \alpha. Therefore it converges to \alpha.
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    Yes! This is exactly what I want, thank you, Opalg!
    Fill two minor gaps:
    (1)Feasibility of the construction: During the recursive process, when k is given, since n_k is a finite number, the set \{\beta_1, \beta_2,..., \beta_{n_k}\} is finite; so there must be some \beta_i in the enumeration and not in this finite set that is greater than \beta_{n_k} because limit ordinal has no greatest element. So the set \{\beta_i|i>n_k\wedge\beta_i>\beta_{n_k}\} is nonempty and thus the least element of this set exists due to well-ordering. Set this least element to be the next element \beta_{n_{k+1}} of the subsequence.
    (2)Convergence of the subsequence: After the subsequence (\beta_{n_k}) is constructed, we can prove by induction the following property: For all i<n_k for some k, \beta_i<\beta_{n_k}. Now given any \beta<\alpha, it has an index i in the enumeration which is a finite natural number due to the 1-1 correspondence with \mathbb{N}; the index of the subsequence must be able to exceed it at, say, the N-th position, that is, i<n_N. According to the above property, when k\geq N, \beta_{n_k}\geq\beta_{n_N}>\beta_i=\beta always holds, as desired.
    Last edited by zzzhhh; August 9th 2009 at 09:57 AM.
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