# Long ray: Proof of being locally euclidean

Consider the topological space $X=\omega_1 \times [0,1)\setminus (0,0)$ equipped with the order topology that arises from the lexicographical order. I want to show that this space is locally euclidean. Clearly, for points $(\alpha,x)$ with $x>0$ I can construct a suitable homeomorphism. That's also no problem for points $(\beta,0)=(\alpha+1,0)$ where $\beta$ is a successor ordinal. But in the general case $(\lambda,0)$ where $\lambda$ is a limit ordinal, I have no idea. Nevertheless, for special limit ordinals like $\omega_0$ I worked out a proof but I don't know how to generalize it to all limit ordinals less than $\omega_1$.

Here is my homeomorphism for $p=(\omega_0,0)$:

We take $U=\{x\in X \mid x<(\omega_0+1,0)\}$ as open neighborhood of $p$.

$$f:U\to(1,3)$$ $$(n,x)\mapsto x\cdot 2^{-n-1}+\sum_{i=0}^n 2^{-i} \quad\text{for}\quad n\in\mathbb{N}$$ $$(\omega_0,x)\mapsto 2+x \quad\text{else}$$

I think, that map should do it.

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The key remark is that $\omega_1$ is the first uncountable ordinal. Therefore, any $\omega<\omega_1$ is countable. You can adapt your proof as follows. Chose $\omega$ and for any $\alpha<\omega$ chose a positive number $\epsilon_\alpha$ such that the $\sum \epsilon_\alpha=1$. This is possible because $\omega$ is countable. Now, build your local homeo, as you did for $\omega_0$, by concatenating intervals of lengths $\epsilon_\alpha$:

$$(\alpha,x)\mapsto x\epsilon_\alpha+\sum_{\beta<\alpha}\epsilon_\beta$$

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That's easy, thanks. I wanted to do it that way but have $\epsilon_\alpha > \epsilon_\beta$ for $\alpha < \beta$. But now I realize that it isn't necessary to choose the $\epsilon_\alpha$ that way. An arbitrary positive number will do it, given the sum of all $\epsilon_\alpha$ is finite. – principal-ideal-domain Mar 5 '14 at 12:32

A useful fact is the following:

Given any countable ordinal $\beta < \omega_1$ there is an order preserving injection $f : [ 0 , \beta ] \to \mathbb{R}$.

So if you are worried about the point $( \lambda , 0 )$ with $\lambda$ a limit ordinal, fix an order preserving injection $f : [ 0 , \lambda + 1 ] \to \mathbb{R}$. For each $\alpha < \lambda$ there is a linear mapping $f_\alpha : [0,1] \to [\;f(\alpha) , f(\alpha+1)\;]$. We then "paste" these linear mappings together to define a homeomorphism between $U = \{ \langle \alpha , x \rangle \in X : \alpha < \lambda+1 \}$ and the open interval $(f(0),f(\lambda+1))$.

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How do you show the existence of this order preserving injection? By the sum of user126154. – principal-ideal-domain Mar 5 '14 at 12:37
@principal-ideal-domain: This answer is the way I usually think of this. – inactive... for now Mar 5 '14 at 12:39
Thanks, that's a very nice proof using transfinite induction. – principal-ideal-domain Mar 5 '14 at 12:48