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

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

2010-10-09T14:33:05Z

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 $(\omega_1,<) \to ([0,1],<)$ ?

Answer by Alessandro Sisto for Monotone injection of an ordinal into $[0,1]$

2010-10-09T14:44:44Z

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 `$\alpha+1$.

Answer by Péter Komjáth for Monotone injection of an ordinal into $[0,1]$

2010-10-09T15:30:51Z

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.?

Answer by Dave Marker for Monotone injection of an ordinal into $[0,1]$

2010-10-12T12:46:32Z

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.