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## Monotone injection of an ordinal into $[0,1]$

This is related to my recent question and would provide a natural positive answer to Question 2. I am sure this must be known to experts.

Question: Is there a monotone injection $(\omega_1,<) \to ([0,1],<)$ ?

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What is the definition of $\omega_1$? There is a monotone injection of every countable ordinal into the unit interval, but no such injection for the first uncountable ordinal. – Jim Conant Oct 9 2010 at 15:03
$\omega_1$ denotes the first uncountable ordinal. – Andreas Thom Oct 9 2010 at 15:08
Thus explaining Alessandro's proof. Thanks. – Jim Conant Oct 9 2010 at 15:10
You might also see mathoverflow.net/questions/25100/… – Robin Chapman Oct 9 2010 at 15:16
(On the other hand, for any countable ordinal $\alpha$, ${\mathbb Q}$ contains subsets order-isomorphic to $\alpha+1$ that are closed as subsets of ${\mathbb R}$.) – Andres Caicedo Oct 12 2010 at 1:31

No, because you could use it to construct an injective map $\omega_1\to\mathbb{Q}$, mapping $\alpha<\omega_1$ to some rational number between $\alpha$ and $\alpha+1$.

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 That was fast. Thank you. – Andreas Thom Oct 9 2010 at 14:46 Glad to be helpful. – Alessandro Sisto Oct 9 2010 at 15:05

Let me add a slightly different argument to Alessandro's quick and clever solution. Assume that $f:(\omega_1,<)\to([0,1],<)$ is an order preserving map. Let $X$ be the range of $f$. Set $a=\sup(X)$, the least upper bound of $X$. Now $X\cap [0,a]$ is uncountable while $X\cap [0,a-\frac{1}{n}]$ is countable for $n=1,2,\dots,$ an impossibility. Question: where did I use that $f$ is o.p.?

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$X\cap [0,a-1/n]$ can be uncountable if $f$ is not o.p. For example, choose an uncountable subset $A$ of $\omega_1$ such that its complement $A^c$ is uncountable as well. Then define $f(A)=1$ and $f(A^c)=1/2$ (this is not injective, but it should not be difficult to construct an injective $f$ using the same idea). – Alessandro Sisto Oct 9 2010 at 15:51

There is a far reaching generalization of this due to Friedman and Shelah. Suppose $X$ is a Borel set in a Polish space and $<$ is a linear order of $X$ that is a Borel subset of $X\times X$. Then there is no order preserving map from $\omega_1$ into $(X,<)$.

The Friedman-Shelah result follows from a later structure theorem of Harrington and Shelah who proved that for any such Borel linear order $(X,<)$ there is a Borel measurable order preserving map into ${\bf R}^\alpha$ for some countable ordinal $\alpha$ where ${\bf R}^\alpha$ is ordered lexicographically. The arguments for $[0,1]$ above can be generalized to show that ${\bf R}^\alpha$ has no $\omega_1$-chains.

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