Prove that for every countable ordinal $\alpha$, $\alpha\times[0,1)$ is order isomoprhic to $[0,1)$

Question

I need to prove that for every countable ordinal $\alpha$, $X_{\alpha} = \alpha\times[0,1)$ is order isomorphic to $[0,1)$, where the order on $X_{\alpha}$ is the lexicographical order. I wanted to go by induction on the countable ordinals. I was able to manually construct a function $f:X_{\omega}\to[0,1)$ that is an order isomorphism, so the base case I have.

So, suppose by induction that for some countable ordinal $\alpha$, for all $\beta < \alpha$, $X_{\alpha}\cong [0,1)$. I need to show that $X_{\alpha+1}\cong[0,1)$. So my attempt was to use the fact that $X_{\alpha + 1} = \left(\alpha\cup\{\alpha\}\right)\times[0,1)$ and to use the induction hypothesis to get an order isomorphism $f'$ between $\alpha\times[0,1)$ and $[0,1)$, and to use the trivial isomorphism $f''$ between $\{\alpha\}\times[0,1)$ and $[0,1)$, and then take the union of these functions $f = f'\cup f''$. But then a problem arises when trying to compare $f(\alpha,r)$ and $f(\gamma,q)$ where $\gamma < \alpha$, since nothing promises me that $f'(\gamma,q) < f''(\alpha,r)$.

I assume after I figure out for successor ordinal, the generalization to a limit ordinal will not be difficult?

Answer 1

So, for $\alpha = 0$ this is obviously false. If $0 < \alpha$ is finite, fix a strictly increasing sequence $\{x_0 = 0 < x_1 < \ldots < x_n = 1 \}$. For each $k < n$ fix an order isomorphism $$f_k \colon \{k\} \times [0,1) \to [x_{k}, x_{k+1}).$$

Then $f := \bigcup \{f_0, \ldots, f_{n-1}\}$ is as desired.

This provides an idea how to solve the general case as well: Let $0 < \alpha < \omega_1$. Fix a strictly increasing sequence $\{x_\beta \mid \beta < \alpha\} \subseteq [0,1)$ such that $x_0 = 0$ and $\sup_{\beta < \alpha} x_\beta = 1$. For each $\beta < \alpha$ fix an order isomorphism $$ f_\beta \colon \{ \beta \} \times [0,1) \to [x_\beta, x_{\beta+1}). $$

Then $f := \bigcup \{ f_\beta \mid \beta < \alpha \}$ is as desired.

It only remains to be seen that we can in fact choose a set $\{ x_\beta \mid \beta < \alpha \} \subseteq [0,1)$ as above. This is (obviously? <- feel free to ask for clarification) equivalent to the fact that there is an order embedding $$\pi \colon (\alpha, \in) \to ([0,1), <).$$ This we prove by induction on $\alpha$. For finite $\alpha$ this is trivial. Now suppose that $\alpha \ge \omega$ and that the claim holds for all $\beta < \alpha$. For each $\beta < \alpha$ fix an order embedding $$ \pi_\beta \colon (\beta, \in) \to ([0,1), <). $$ Pick a strictly increasing, cofinal sequence $(\beta_n \mid n < \omega)$ in $\alpha$ and fix a strictly increasing sequence $(x_n \mid n < \omega$) in $[0,1)$. (This is possible, because $\alpha$ is countable.) Now let, for each $n < \omega$, $$ \pi_{\beta_n}^* \colon (\beta_n, \in) \to ([x_n, x_{n+1}), <), \gamma \mapsto x_n + (x_{n+1}-x_{n}) \cdot \pi_{\beta_n}(\gamma). $$

Then $\pi := \bigcup \{ \pi_{\beta_n}^* \restriction [\beta_{n-1}, \beta_n) \mid n < \omega\}$ is as desired.

Answer 2

PART 1. Claim: For $a\in \omega_1$ let $a$ have the $\epsilon$-order topology. Then for $b,c \in \mathbb R$ with $b<c$ and for $a\ne 0$ there is an order-isomorphic homeomorphism $f:a\to d\subset [b,c]$ such that $f(0)=b$ and $\sup \{f(x):x\in a\}=b.$

For brevity I will write OIH for order-isomorphic homeomorphism.

Proof of Claim:(By transfinite induction).Assume it holds for all $a'<a.$

(i). $a<\omega.$ Trivial.

(ii). $\omega <a=a'+1=a''+2.$ Let $f:a'$ be an OIH to $d\subset [b,(b+c)/2]$ with $f(0)=b$ and $f(a'')=(b+c)/2.$ Extend $f$ to the domain $a$ by letting $f(a')=c$.

(iii). $\omega<a=a'+1$ and $a'=\sup a'$. Let $f:a'\to d\subset [b,c]$ be an OIH with $f(0)=b$ and $\sup \{f(x):x\in a'\}=c$. Extend $f$ to the domain $a$ by letting $f(a')=c$.

(iv). $0<a=\sup a<\omega.$ Let $a=\sup \{a_n:n\in \omega\}$ where $(a_n)_{n\in \omega}$ is a strictly increasing sequence, and $a_0=0.$ Let $(y_n)_{n\in \mathbb N}$ be a strictly increasing sequence in $[b,c]$ with $y_1=b$ and $\lim_{n\to \infty}y_n=c$.

Now for $n\in \omega$ there is an order-isomorphism $g_n:(a_{n+1}$ \ $a_n)\to e_n$ to a (unique) ordinal $e_n<a$.The $\epsilon$-order topology on $a_{n+1}$ \ $a_n$ equals its subspace topology as a subspace of the ordered space $a$ because it is an interval of $a$. So $g_n$ is an OIH.

So choose an OIH $f_n:e_n\to d_n\subset [x_n,x_{n+1}]$ with $f_n(a_n)=x_n,$ such that if $e_n$ is a limit ordinal then $\sup \{f_n(x):x\in e_n\}=x_{n+1}$, or if $e_{n+1}=e'_{n+1}+1$ then $f_n(e'_{n+1})<x_{n+1}$. And we also require $f_0(0)=b$.

Now let $f=\cup_{n\in \omega} g_n(f_n).$

PART 2. For $0<a=a'+1<\omega_1$ let $f:a\to d\subset [0,1/2]$ be an OIH with $f(0)=0$ and $f(a')=1/2$. For $ 0<a=\sup a <\omega_1$ let $f:a\to d\subset [0,1)$ be an OIH with $f(0)=0$ and $\sup \{f(x):x\in a\}=1.$

For $a'\in a$ let $h:\{a'\}\times [0,1)\to [f(a'),f(a'+1))$ be an OIH. Then $h:a\times [0,1)$ is an OIH.

Remark: In Part 1 we want an OIH $f:a\to d$ and not merely an order-isomorphic embedding, otherwise in Part 2 we might end up with something like $a=\omega+\omega$ with $\sup \{f(n):n\in \omega\} <f(\omega)$. That is, in Part 1 , when $0<x=\sup x<a$ we need $f(x)=\sup \{f(y):y<x\}.$

Answer 3

Use the theorem that every linear ordinal order embeds into a subset or the rationals of the unit interval to embed alpha into the unit interval. Between each image an element of alpha and its successor insert a copy of the half open unit interval. If alpha has a maximum, append another copy of the half open unit interval.