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This is based on Ex. 6.4.6 in Stillwell's "Real Numbers."

Using previous exercises, it was established that one can construct for any countable ordinal $\gamma$ disjoint half-open intervals $[a_{\alpha}, a_{\alpha+1})$ for all $\alpha\lt\gamma$, with the properties $a_{\alpha}\lt a_{\beta}$ iff $\alpha \lt \beta$ and $\bigcup_{\alpha \lt \gamma} [a_{\alpha},a_{\alpha+1})=[0,1)$.

What was termed a "$\gamma$-line," $[0,1)\times \gamma$ was defined as containing a copy $[0,1)\times\{{\alpha}\}$ of $[0,1)$ for each $\alpha\lt \gamma$.

As a preliminary question: Is the notation $\times \{\alpha\}$, just an index?

I can see how this $\gamma$-line is homeomorphic and order isomorphic to $[0,1)$ for each countable ordinal $\gamma$.

My main question is, I would appreciate help to:

Similarly define the $\omega_1$-line and explain why it is not homeomorphic and not order isomorphic to $[0,1)$.

I am not sure what to use instead of $\times\{\alpha\}$ as an index be to be less than $\omega_1$ as $\alpha$ was less than $\gamma$- do I use countable ordinals?

And the explanation asked for probably hinges on the uncountability of $\omega_1$, yet I don't know how to explicitly show it.

As a subsidiary question, Brian Scott gave an example of homeomorphic but not order isomorphic here: The relation between order isomorphism and homeomorphism.

Can there be order isomorphism without homeomorphism?

Thanks

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Since in both case the topology is generated by the class of subsets sets of the form $\{x\mid a<x<b\}$ or $\{x\mid x<b\}$, order isomorphism implies homeomorphism.

To see that $[0,1)$ is not even homeomorphic to the long line, one fairly direct approach would be to show that there can't even be a continuous surjection from $[0,1)$ to the long line. Namely, in each of the $\omega_1$ copies of $[0,1)$ in the long line, there must be a point that is the image of a rational, but there are not enough rationals to hit all of them!

(Or, perhaps slicker: $[0,1)$ is separable, but the long line isn't!)

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    $\begingroup$ Or, in other words, only one of these orders has an uncountable well-ordered subset. $\endgroup$ – Asaf Karagila Jul 12 '17 at 15:32
  • $\begingroup$ Thanks - Then in the example of the $\gamma$-line above, each copy of $[0,1)$ has a point that is the image of a rational as you say. What do the copies of $[0,1)$ look like in the $\omega_1$? Can one even define the $\omega_1$ line in a similar fashion as the $\gamma$ line? With regards, $\endgroup$ – user12802 Jul 12 '17 at 15:50
  • $\begingroup$ @Andrew: I don't know exactly how the definition of a "$\gamma$-line" you're working with looks like -- but the one I can imagine would generalize directly to setting $\gamma=\omega_1$. $\endgroup$ – Henning Makholm Jul 12 '17 at 15:58
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    $\begingroup$ Actually, hmm, I see ... "containing a copy $[0,1)\times\{{\alpha}\}$ of $[0,1)$ for each $\alpha\lt \gamma$" doesn't work very well as a definition in my opinion. It would be clearer to say $[0,1)\times \gamma = \{(x,\alpha)\mid 0\le x<1, \alpha\in\gamma\}$ with the reverse lexicographic order. $\endgroup$ – Henning Makholm Jul 12 '17 at 16:00
  • $\begingroup$ Yes, that's elegant. Sorry to trouble you further. Can one say anything similar for the $\omega_1$ line? More specifically, can you even have such copies of $[0,1)$ and where you have $\alpha\in\gamma$ can you have some element, $\delta$, which is $\in\omega_1$? What would such an element be? $\endgroup$ – user12802 Jul 12 '17 at 16:21

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