how to prove that any limit ordinal is really a limit?

**zzzhhh** · August 4th 2009, 09:32 AM

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!

**clic-clac** · August 7th 2009, 01:22 PM

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 )

**zzzhhh** · August 8th 2009, 06:00 AM

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!

**clic-clac** · August 8th 2009, 08:27 AM

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..!

**zzzhhh** · August 8th 2009, 11:18 AM

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.

**Opalg** · August 8th 2009, 12:29 PM

Originally Posted by zzzhhh

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 $\alpha$ .

**zzzhhh** · August 9th 2009, 01:25 AM

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.

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