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Thread: 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 $\displaystyle \alpha=\beta+1$, then we call $\displaystyle \alpha$ 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 $\displaystyle \alpha=\{\beta|\beta<\alpha\}, |\alpha|=\aleph_0$, there is a sequence in $\displaystyle \alpha$ converging to $\displaystyle \alpha$? Here we define a sequence in $\displaystyle X$ to be a function on the set of all natural numbers $\displaystyle \mathbb N$ and having values in $\displaystyle X$, and a sequence of ordinals $\displaystyle (\alpha_n)$ converging to ordinal $\displaystyle \alpha$ if for any $\displaystyle \beta<\alpha$, there is a natural number $\displaystyle N$ such that $\displaystyle \beta<\alpha_n\leq\alpha$ for all $\displaystyle n\geq{N}$.
    For example, sequence $\displaystyle \alpha_n=\omega+n\rightarrow \omega\cdot2; \beta_n=\omega^n\rightarrow \omega^\omega$, .... But these are specific sequences, I need a proof in general situation.
    Thanks!
    Last edited by zzzhhh; Aug 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 $\displaystyle \aleph_1$ 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 $\displaystyle a$ and $\displaystyle b$ I mean $\displaystyle card(\{c\ ;\ a<c<b\})$ ) between two successive ordinals in that sequence: if at a moment, it exceeds $\displaystyle \aleph_0$, then the limit will be greater than $\displaystyle \aleph_1$. But if all "distances" are at most countable, since a countable reunion of countable sets is countable, then you cannot reach $\displaystyle \aleph_1$ ).

    You can find answers with cofinality (Cofinality - Wikipedia, the free encyclopedia )
    Last edited by clic-clac; Aug 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 $\displaystyle \omega^{\omega}$ which is not countable) This is important, great limit ordinals can be reached with not so much gasoline ($\displaystyle \aleph_\omega$ for instance)

    Now for a countable limit ordinal $\displaystyle \alpha$. Well in fact $\displaystyle \alpha$ itself gives you a sequence as you wanted..!
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    Thank you but ... $\displaystyle \omega^\omega$ IS a countable ordinal, that is, there is a 1-1 correspondence between $\displaystyle \omega^\omega$ and $\displaystyle \omega_0$, but there is no such one between $\displaystyle \omega^\omega$ and $\displaystyle \omega_1$. 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 $\displaystyle \alpha$ can give? Could you write down explicitly the general term $\displaystyle \alpha_n$ 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 $\displaystyle \alpha=\beta+1$, then we call $\displaystyle \alpha$ 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 $\displaystyle \alpha=\{\beta|\beta<\alpha\}, |\alpha|=\aleph_0$, there is a sequence in $\displaystyle \alpha$ converging to $\displaystyle \alpha$? Here we define a sequence in $\displaystyle X$ to be a function on the set of all natural numbers $\displaystyle \mathbb N$ and having values in $\displaystyle X$, and a sequence of ordinals $\displaystyle (\alpha_n)$ converging to ordinal $\displaystyle \alpha$ if for any $\displaystyle \beta<\alpha$, there is a natural number $\displaystyle N$ such that $\displaystyle \beta<\alpha_n\leq\alpha$ for all $\displaystyle n\geq{N}$.
    For example, sequence $\displaystyle \alpha_n=\omega+n\rightarrow \omega\cdot2; \beta_n=\omega^n\rightarrow \omega^\omega$, .... But these are specific sequences, I need a proof in general situation.
    Thanks!
    If $\displaystyle \alpha = \{\beta:\beta<\alpha\}$ is countable then there exists an enumeration $\displaystyle \alpha = \{\beta_1,\beta_2,\beta_3,\ldots\}$ of $\displaystyle \alpha$. Construct a subsequence of $\displaystyle (\beta_n)$ by omitting all elements less than the previous element of the subsequence. More precisely, let $\displaystyle n_1=1$, and for $\displaystyle k\geqslant1$ let $\displaystyle n_{k+1} = \min\{n:\beta_n>\beta_{n_k}\}$. Then $\displaystyle (\beta_{n_k})$ is an increasing sequence of ordinals less than $\displaystyle \alpha$ that is eventually larger than any ordinal less than $\displaystyle \alpha$. Therefore it converges to $\displaystyle \alpha$.
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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 $\displaystyle k$ is given, since $\displaystyle n_k$ is a finite number, the set $\displaystyle \{\beta_1, \beta_2,..., \beta_{n_k}\}$ is finite; so there must be some $\displaystyle \beta_i$ in the enumeration and not in this finite set that is greater than $\displaystyle \beta_{n_k}$ because limit ordinal has no greatest element. So the set $\displaystyle \{\beta_i|i>n_k\wedge\beta_i>\beta_{n_k}\}$ is nonempty and thus the least element of this set exists due to well-ordering. Set this least element to be the next element $\displaystyle \beta_{n_{k+1}}$ of the subsequence.
    (2)Convergence of the subsequence: After the subsequence $\displaystyle (\beta_{n_k})$ is constructed, we can prove by induction the following property: For all $\displaystyle i<n_k$ for some $\displaystyle k$, $\displaystyle \beta_i<\beta_{n_k}$. Now given any $\displaystyle \beta<\alpha$, it has an index $\displaystyle i$ in the enumeration which is a finite natural number due to the 1-1 correspondence with $\displaystyle \mathbb{N}$; the index of the subsequence must be able to exceed it at, say, the $\displaystyle N$-th position, that is, $\displaystyle i<n_N$. According to the above property, when $\displaystyle k\geq N, \beta_{n_k}\geq\beta_{n_N}>\beta_i=\beta$ always holds, as desired.
    Last edited by zzzhhh; Aug 9th 2009 at 09:57 AM.
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