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I'm reading through this and I'd like to define an injective function from the set of countable ordinals $\Omega$ into $\mathbb{R}$ using transfinite induction (or maybe transfinite recursion?).

Clearly, $\emptyset \in \Omega$ will be defined to map to zero: $f(\emptyset) := 0$. Next one would probably make a distinction between limit ordinals and successor ordinals.

For successor ordinals $\beta$ one would probably assume that if $f(\alpha)$ is defined for all $\alpha < \beta$ and if $\beta = \alpha + 1$ then one can define $f(\beta) := f(\alpha) + 1$.

I'm not sure this is right but I'm even less sure about the case for limit ordinals. Now let $\beta$ be a limit ordinal and assume $f(\alpha)$ is defined for all $\alpha < \beta$. We have $\beta = \sup \{ \alpha \mid \alpha < \beta \}$ by the definition of limit ordinal and because $\beta$ is countable we can write this as $\beta = \sup \{ \alpha_i \mid \alpha_i < \beta , i \in \mathbb{N} \}$. Now I was thinking that maybe $f$ could be defined as $ f(\beta) := \sum_{i=0}^\infty 10^{-i} f(\alpha_i)$.

To show that this is injective it's probably enough to show that different limit ordinals map to different reals.

What do you think of this? Am I on the right track? Thanks for your help.

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    If you're allowed to use Axiom of Choice, you can choose $f(\alpha)$ as an arbitrary element of $\mathbb R\setminus \{f(\gamma); \gamma<\alpha\}$. (This set is non-empty, it is even uncountable.)2012-01-16
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    See also [this question](http://math.stackexchange.com/questions/123969/embedding-ordinals-in-mathbbq), [this question](http://math.stackexchange.com/questions/408300/countable-ordinals-are-embeddable-in-the-rationals-bbb-q-proofs-and-their) and other posts linked there.2016-05-08

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