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Let $<$ be the usual ordering on $\mathbb{R}.$ If $\gamma$ is an ordinal, then $f:\gamma \to \mathbb{R}$ is ordering preserving if $\forall \alpha \in\gamma \forall \beta \in \gamma [\alpha \in \beta \implies f(\alpha) < f(\beta)].$

Let $\gamma $ be an ordinal. Prove there is an ordering preserving $f:\gamma \to \mathbb{R}$ iff $\gamma < \omega_1.$

Could anyone advise on this problem?

Given $\exists$ order-preserving $f: \gamma \to \mathbb{R},$ suppose $\gamma \geq \omega_1.$ We know that there is a rational number between any two real numbers, so how I derive a contradiction from here?

Martin Sleziak
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Alexy Vincenzo
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1 Answers1

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Hint: If $f : \omega_1 \to \mathbb{R}$ is order preserving, consider the following family of open intervals: $$\{ (\; f ( \alpha ) , f ( \alpha + 1 )\; ) : \alpha \in \omega_1 \}.$$ (Recall that $\mathbb{R}$ is separable.)

Given $\gamma < \omega_1$, to construct an order-preserving function $f : \gamma \to \mathbb{R}$ it is easier to proceed by induction on $\gamma$, and moreover show that such an $f$ can be taken to have bounded range (i.e., $|f(\alpha)| \leq M$ for all $\alpha \in \gamma$ for some $M \in \mathbb{R}$).

user642796
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  • Thank you! For the 1st part, instead of proving by contradiction, could we let $X ={ (; f ( \alpha ) , f ( \alpha + 1 ); ) : \alpha \in \gamma }$ and define function $F: X \to \mathbb{Q}$ by $F(A) = \frac{\left \lfloor {f(\alpha)} \right \rfloor + \left \lfloor {f(\alpha +1)} \right \rfloor}{2},$ where $A = (f(\alpha), f(\alpha + 1)) \in X.$ Then $|X| \leq \omega$ and hence $\gamma \leq \omega ?$ – Alexy Vincenzo Nov 14 '13 at 13:03
  • @Alexy: You proposed idea would not work, since we haven't ruled out the possibility that the function $\alpha \mapsto \lfloor f(\alpha) \rfloor$ is constant. (What if $0 \leq f(\alpha) < 1$ for all $\alpha$?) – user642796 Nov 14 '13 at 13:10
  • Thank you for the correction. May I trouble you to elaborate on why would it work for $\gamma \geq \omega_1? $ – Alexy Vincenzo Nov 14 '13 at 13:16
  • Can we consider ${ (; f ( \alpha ) , f ( \alpha + 1 ); ) : \alpha \in \gamma }$ and given $(f(\alpha), f(\alpha + 1)),$ choose a rational $r_ {\alpha} \in (f(\alpha), f(\alpha + 1))?$ Now, define $F:{ (; f ( \alpha ) , f ( \alpha + 1 ); ) : \alpha \in \gamma } \to \mathbb{Q}$ by $F(A) = r_{\alpha},$ where $A=(f(\alpha), f(\alpha + 1)).$ Then, $|{ (; f ( \alpha ) , f ( \alpha + 1 ); ) : \alpha \in \gamma }| \leq \omega? $ – Alexy Vincenzo Nov 14 '13 at 14:46
  • @Alexy: As soon as there is no order preserving function $\omega_1 \to \mathbb{R}$ there cannot be an order preserving function $\gamma \to \mathbb{R}$ for any $\gamma > \omega_1$ (since such an order preserving function restricted to $\omega_1$ would give an order preserving function $\omega_1 \to \mathbb{R}$). But you are correct, the same argument will apply to any $\gamma > \omega_1$. – user642796 Nov 14 '13 at 17:36
  • Sir, thank you for the advice. But I have not assumed $\gamma \geq \omega_1$ and if so, would my proposed proof still work? ( see the my comment just before your latest reply). – Alexy Vincenzo Nov 15 '13 at 07:33
  • @Alexy: Your proof only shows that *if* there is an order-preserving function $f : \gamma \to \mathbb{R}$, then $\gamma < \omega_1$. In order to show the opposite direction you will need to show that for each $\gamma < \omega_1$ *there is* an order preserving function $f : \gamma \to \mathbb{R}$. – user642796 Nov 15 '13 at 07:48
  • I'm actually referring to the "if" part of the statement. Anyways, thanks for your instruction. – Alexy Vincenzo Nov 15 '13 at 13:24