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Question: If $\Theta$ is the order type of an uncountable set, then show that for $\alpha < \omega_1$, $\alpha \preceq \Theta$ or $\alpha^* \preceq \Theta.$ Where $\preceq$ is an ordering of order types.

This makes sense to me conceptually. For every order type less than the first uncountable order type, this order type must be less than any uncountable order type. Is this way of thinking correct?

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    I'm not sure what do you mean by $\preceq^*$ and by "$\preceq$ is an ordering of order types". Also is $\Theta$ a linear order type?2011-07-14
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    Ohhh, the $\preceq^*$ means that $\alpha^*$ (i.e. the inverse order of $\alpha$) embeds into $\Theta$. Am I correct in my reading?2011-07-14
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    @Asaf: well I hope so because that's what I assumed in my answer. @furs: I forgot to answer the question about the concept of the second paragraph. That way of thinking is not correct, since $\omega_1$ isn't the "first uncountable order type." It makes sense for wellorders, but for general linear orders there are lots of others to worry about (the reals, for instance) that don't have any clear relation to $\omega_1$.2011-07-14
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    @Asaf: Yes, that it what I had intended to write. @ccc: I see, thank you.2011-07-14

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