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?



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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