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Let $\alpha$ and $\beta$ be two ordinals with $\beta \leq \alpha$. Define

$$ X:= \{\gamma \in \alpha^{+} : \beta + \gamma \leq \alpha\}.$$

I have shown this is an ordinal. Now I need to show it isn't a limit ordinal, but I'm stuck.

Many thanks for your help.

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  • $\begingroup$ What is $\alpha^+$ for you? (Because in many contexts, that denotes the successor cardinal of $|\alpha|$.) $\endgroup$ – Asaf Karagila May 19 '15 at 20:37
  • $\begingroup$ It is the successor of $\alpha$, namely $\alpha \cup \{ \alpha\}$. $\endgroup$ – Frank May 19 '15 at 20:37
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HINT: Assume to the contrary, and use the definition of $\beta+\delta$ when $\delta$ is a limit ordinal, to show that $\beta+\delta\leq\alpha$ as well to obtain a contradiction.

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  • $\begingroup$ What is otp? Thanks. $\endgroup$ – Frank May 19 '15 at 20:39
  • $\begingroup$ Order type, the unique ordinal isomorphic to the well-ordered set given. $\endgroup$ – Asaf Karagila May 19 '15 at 20:40
  • $\begingroup$ Ok thanks. If this is what the writer of the exercise intended I would be surprised, as the next part uses this $\gamma$ as 'another' way of doing the problem. $\endgroup$ – Frank May 19 '15 at 20:41
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    $\begingroup$ Ok thanks you've been very helpful. $\endgroup$ – Frank May 19 '15 at 20:54
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    $\begingroup$ Well, it beats a hard kick in the face! $\endgroup$ – Asaf Karagila May 19 '15 at 20:55

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