How to prove the not-so-long rays are homeomorphic to the reals?

Question

The long ray (half of the long line) is an interesting topological space. It is defined as the order topology on $\omega_1$$\times [0, 1)$ with lexicographic order. Basically, it is an uncountable number of half open intervals glued together.

My question is about not-so-long rays, where you take a countable ordinal $\alpha$, and make the space $\alpha \times [0,1)$. This is claimed to be homeomorphic to $[0,1)$. How does one define this homeomorphism? I see how this can be defined for $\omega$, $\omega^2$, and others, but I don't know how to do it in general.

This property is used to prove, for example, that every interval on the long line is homeomorphic to an interval on the real line.

Answer 1

Well, that's actually pretty easy.

Just note that the orders are isomorphic, since they are both complete orders with dense countable subsets and no maximum. There is an isomorphism between the countable dense subsets, and it extends uniquely to the rest of the order.

More specifically, you need to pick an order embedding of $\alpha$ into $\Bbb R$, or $[0,1)$, and then just look at the intervals defined by $[j(\gamma),j(\gamma+1))$ where $j$ is the embedding.

Answer 2

One way to embed any $\alpha$ in $\omega_1$ into $R$ is by transfinite induction. Suppose $\beta$ can be embedded by some $f$ into $[x,y]$, whenever $x,y$ are in R with $x<y$, for any $\beta<\alpha$.The case $\alpha=\phi$ (the ordinal $0$) is trivial. For the case $\alpha=\gamma+1$, let $f$ embed $\gamma$ into $[x,(x+y)/2]$ and let $f(\gamma)=y$ if $\gamma$ is a successor ordinal or is $0$,otherwise let $f(\gamma)$ be the l.u.b. of the image of $\gamma$ . For the case of a non-zero limit ordinal $\alpha$, let $\beta(n) : n=0,1,2,...$be a strictly increasing sequence of members of $\alpha$ with l.u.b. $\alpha$, and let $x(n) : n=0,1,2,...$ be a strictly increasing real sequence with l.u.b. $y$, and with $x(0)=x$. Let $f(0)$ embed $\beta(0)+1$ into $[x(0),x(1)]$ and let $f(n+1)$ embed $(\beta(n+1)+1)/(\beta(n)+1)$ into $[x(n),x(n+1)]$. This is possible because $(\beta(n+1)+1)/(\beta(n)+1)$ is order-isomorphic to an ordinal less than $\alpha$.Now let $f$ be the union of all of the $f(n)$. There is another way but I think this is long enough.