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Let $\alpha$ be an ordinal and suppose this ordinal has a cofinal subset. Let the cofinality of $\alpha$ be $\beta$. Thus $\beta$ is isomorphic with $B$, cofinal subset of $\alpha$. Then $\beta$ is a cofinal subset of itself. Now, let the cofinality of $\beta$ be $\gamma$.

Here, how to prove that $\alpha$ has a cofinal subset similar to $\gamma$?

I see this is completely different from proving that cofinal subset of cofinal subset of $A$ is a cofinal subset of $A$, since $\beta$ described above is not a cofinal subset of $\alpha$. (Actually if $\alpha ≠ \beta$, then $\beta$ is not a cofinal subset of $\alpha$.)

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Oh.. I'm really sorry. I proved it. Both the statement '$\beta$ is a cofinal subset of $\alpha$' and the negation of this imply that $\alpha$ has a cofinal subset similar to $\gamma$. –  Katlus Jun 5 '12 at 5:11

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Since the cofinality of $\alpha$ is $\beta$, there is an increasing $\beta$ sequence $(a_{\eta})_{\eta < \beta}$ where each $a_\eta < \alpha$ and $\alpha = \lim_{\eta < \beta} a_\eta$. Since the cofinality of $\beta$ is $\gamma$, there is a increasing $\gamma$-sequence $(b_\zeta)_{\zeta < \gamma}$ such that each $b_\zeta < \beta$ and $\lim_{\zeta < \gamma} b_\zeta = \beta$.

Then $a_{b_{\zeta}}$ is a $\gamma$ sequence of elements in $\alpha$ such that $\lim_{\zeta < \gamma} a_{b_{\zeta}} = \alpha$.

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