Countable ordinals in the reals I'm trying to show that for any countable ordinal $\alpha$, there is a subset of $(\mathbb{R},<)$ that has order type $\alpha$. In this post I'm not asking for a solution, but instead for a proof-check. If it's wrong I'll go back to the drawing board.
I tried using induction: assume that every $\beta<\alpha$ admits a subset of $(\mathbb{R},<)$ with order type $\beta$. Since $\alpha$ is assumed countable $\{\beta\in \text{Ord}:\beta<\alpha\}$ is countable, so we can enumerate as $\beta_0,\beta_1\dots$. Then map $f_k:\beta_k\to [k,k+1)\subset \mathbb{R}$ where $f_k$ order preserving and continuous, and $\sup f( \beta_k) = k+1$. Then $\bigcup \beta_k$ is a well-order with order type $\alpha$.
Does this work? If so, is it ok to claim the existence of such $f_k$'s?
 A: Here is an alternative way of showing that every countable ordinal is order-isomorphic to a subset of the real line (or a subset of $[0,1]$, below).
For finite ordinals this is clear (just take a finite subset of $[0,1]$ with the same number of points).
Now let $\nu=\{\gamma:\gamma<\nu\}$ be an infinite countable ordinal. Fix a bijection $f:\{\gamma:\gamma<\nu\}\to\Bbb N_+$ (where $\Bbb N_+$ denotes the set of all positive integers),  and map each $\gamma<\nu$ to $\psi(\gamma):=\sum\limits_{\delta<\gamma}\frac{1}{2^{f(\delta)}}$.
Note that this maps $\gamma=0$ to $0\in[0,1]$ (i.e. $\psi(0)=0\ $) because
$\sum\limits_{\delta<0}\frac{1}{2^{f(\delta)}}$ is a sum with an empty index set, and by convention it sums up to $0$.
Then, $\psi(1)=\frac{1}{2^{f(0)}}$.
Also
$\psi(2)=\frac{1}{2^{f(0)}}+\frac{1}{2^{f(1)}}\ $, etc.
Clearly if $\beta<\gamma<\nu$ then $\psi(\beta)<\psi(\gamma)$. Indeed, $\psi(\beta)+\frac{1}{2^{f(\beta)}}\le\psi(\gamma)$.
Note also that if $A=\{\psi(\gamma):\gamma<\nu\}$, then $\inf(A)=\min(A)=0$ and $\sup(A)\le1$.
We have $\sup(A)=\max(A)<1$ only when $\nu$ is a successor ordinal (assuming $\nu$ is countably infinite, to begin with).
An alternative definition would be
$\varphi(\gamma):=\sum\limits_{\delta\le\gamma}\frac{1}{2^{f(\delta)}}$.
Then, $\varphi(0)=\frac{1}{2^{f(0)}}$.
Also
$\varphi(1)=\frac{1}{2^{f(0)}}+\frac{1}{2^{f(1)}}\ $, etc.
But, $\psi$ might be a better choice, since $\psi(\gamma)=\sup\limits_{\delta<\gamma}\psi(\delta)$ (in $[0,1]$) exactly when $\gamma$ is limit, i.e.
$\gamma=\sup\limits_{\delta<\gamma}\delta$.
The above is not properly an answer to the original question, as the OP did not ask for a different approach, but only asked  for a verification of the approach presented in the question. I put, nevertheless, this different approach here, for reference. It is slick (though not completely self-contained since it presumes background knowledge of sum of an infinite series), but it is also simple and, I would think, good-to-know.
The notation  $\sum\limits_{\delta<\gamma}\frac{1}{2^{f(\delta)}}$ may need some extra explanation since one usually sees something like
$\sum\limits_{n=1}^\infty$ rather than $\sum\limits_{\delta<\gamma}$, but the latter is ok too, since all terms are non-negative (so the order in which they appear won't matter) so
$\sum\limits_{\delta<\gamma}\frac{1}{2^{f(\delta)}}$ could be defined as $\sup$ of all possible sums of the form $\frac{1}{2^{f(\delta_1)}}+\cdots+\frac{1}{2^{f(\delta_k)}}$ where $k\ge1$ and $0\le\delta_1<\dots<\delta_k<\gamma$.
This $\sup$ cannot exceed $\sum\limits_{n=1}^\infty\frac{1}{2^n}=1$.
Yet another way to explain this, if $g=f^{-1}$ is the inverse bijection, $g:\Bbb N_+\to\{\gamma:\gamma<\nu\}$, then for each $\gamma<\nu$ and all $n\ge1$, one may define
$a_{n,\gamma}= \begin{cases} 
      \frac{1}{2^n}\ ,\ {\mathrm{ if }}\ g(n)<\gamma\\
      0\ ,\ {\mathrm{ if }}\ \gamma\le g(n)<\nu 
   \end{cases}\ ,$ and then $\sum\limits_{\delta<\gamma}\frac{1}{2^{f(\delta)}}:=\sum\limits_{n=1}^\infty a_{n,\gamma}\ $.
Clearly $0\le\psi(\gamma)=\sum\limits_{n=1}^\infty a_{n,\gamma}< 
\sum\limits_{n=1}^\infty\frac{1}{2^n}=1$, for each $\gamma<\nu$. Also, if $\beta<\gamma<\nu$ then $0\le a_{n,\beta}\le a_{n,\gamma}$ for all $n$, and if $m=f(\beta)$ then
$a_{m,\beta}=0<\frac{1}{2^m}=a_{m,\gamma}$
so $\sum\limits_{n=1}^\infty a_{n,\beta}<\frac{1}{2^m}+\sum\limits_{n=1}^\infty a_{n,\beta}\le\sum\limits_{n=1}^\infty a_{n,\gamma}\ .$
A: The proof is incorrect.  I think you meant to say that $\cup_kS_k$  has order-type $\alpha$ where $S_k$ is the image of $\beta_k$ under $f_k.$ But it's still wrong. For example if $\alpha=\omega + \omega=\omega \cdot 2$ then for some $k',k'',k'''$ we have $\beta_k'=\omega,\; \beta_k''=\omega +1, \beta_k'''=\omega + 2.$ Then each of the  3 intervals $[k',k'+1),\;[k'',k''+1),\;[k''',k'''+1)$ contains a subset  of $\cup_kS_k$ that is order-isomorphic to $\omega,$ so the order-type of $\cup_kS_k$ is at least $\omega \cdot 3.$
For real numbers $a,b$ with $a<b$ let $g_{a,b}:[0,\infty) \to [a,b)$ be an order-isomorphic bijection.  
For countable ordinal $\alpha ,$ suppose that for all $b<\alpha$ there exists an order-embedding $f_b$ of $b$ into $[0,1] .$  We show there exists an order-embedding of $\alpha$ into $[0,1].$
(i). If $\alpha =0 $ then $\alpha = \emptyset.$ Let $f_0=\emptyset.$ (Trivial special case.)
(ii). If  $\alpha =c+1,$ let $f_c:c\to [0,1]$ be an order-embedding of $c$ into $[0,1] .$ Define $f_{\alpha} (x)=g_{0,1}(f_c(x))$ for $x <c,$ and $f_{\alpha }(c)=1.$ Then $f_{\alpha }$ is an order-embedding of $\alpha$  into $[0,1].$
(iii). If $0\ne \alpha =\cup \alpha,$ take $\{a_n: n\in \omega\}\subset  \alpha$ with $a_n<a_{n+1}<\alpha$ for  each $n\in \omega,$ and $\cup S=\alpha ,$ and $a_0=0.$ 
For each $n\in \omega$ the set $a_{n+1}$ \ $a_n$ is order-isomorphic to a (unique) ordinal $b_n,$ with $b_n<\alpha.$  So let $f_n$ be an order-embedding of $a_{n+1}$ \ $a_n$  into $[0,1].$ 
We have $\cup_{n\in \omega} (a_{n+1}$ \ $a_n)=\alpha$ (because $a_0=0$). For each $x\in \alpha$ there is a unique $n\in \omega$ such that $x\in a_{n+1}$ \ $a_n.$
$$\text {For } x\in a_{n+1} \backslash a_n \text { let }\quad h_{\alpha}(x)=g_{n,n+1}(f_n(x))$$ $$\text {and }\quad f_{\alpha}(x)=g_{0,1}(h_{\alpha}(x)).$$  Then $f_{\alpha}:\alpha\to [0,1]$ is an order-embedding.  
