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I know that there is no increasing function $f: \aleph_1 \to \mathbb{R}$, so it seems like for $\alpha < \aleph_1$, there should exists a function $f: \alpha \rightarrow \mathbb{R}$ that is strictly increasing.

I can't think of how this could be shown/constructed. Any ideas?

Andrés E. Caicedo
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user2640211
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  • Well, $\alpha < \aleph_1$ implies that $\alpha$ is countable... – Crostul Mar 22 '16 at 20:55
  • @Crostul Your comments gives a reason to believe that it could be true, but I guess it does not help much when you have to find an increasing function from $\epsilon_0$ to $\mathbb{R}$, where $$\epsilon_0=\sup {\omega,\omega^\omega,\omega^{\omega^\omega},\ldots}$$ – Darío G Mar 22 '16 at 20:58
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    @user26440211 Try to prove it by induction on the countable ordinals, and have in mind that any time you have an strictly increasing function $f:\alpha\to \mathbb{R}$ you can transform it to an strictly increasing function from $g:\alpha\to (0,1)$. This last remark will help you when doing the inductive step and the limit step. – Darío G Mar 22 '16 at 21:07
  • Moreover, you can arrange that the embedding is continuous and it's range is contained in $\mathbb Q $. – Andrés E. Caicedo Mar 22 '16 at 21:35

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