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Given an well-ordered class $(A,\leq)$ we want to proof that every element $a \in A$ can be reached by applying the successor relation on a the smallest element of $A$ or on an limes element of $A$ finitely often.

I tryed to construct an infinite decreasing sequence $a_{n+1} \leq a_n$ with $a_i \in A$ and no $m$ such that $a_m = a_{m+1}$. But my problem is that I cannot define a limes element $b \in A$ with $b < a$ and no limes element $c$ with $b < c < a$. Any hints?

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2 Answers 2

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You do it by induction.

Suppose that for all $\beta<\alpha$ we have that there is a minimal or limit $\gamma<\beta$ and a finite sequence $\gamma=\beta_0\leq\ldots\beta_n=\beta$, where $\beta_{i+1}$ is the successor of $\beta_i$.

If $\alpha$ is a limit point then we are done. It is itself a limit. Otherwise $\alpha$ is the successor of $\beta$, and simply add to the sequence from $\gamma$ that we already have the new point, $\alpha$.

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HINT: Say that an element of $A$ is bad if it cannot be reached by applying the successor operation a finite number of times to $\min A$ or a limit element of $A$, and let $B=\{a\in A:a\text{ is bad}\}$. If $B\ne\varnothing$, let $b=\min B$, and derive a contradiction.

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How is $G$ defined? –  joachim Nov 11 '12 at 17:39
1  
@joachim: Sorry, that’s left over from an earlier version. It should be $B$; I’ll fix it in a moment. –  Brian M. Scott Nov 11 '12 at 17:43

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