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 $(\omega_1,<) \to ([0,1],<)$ ?
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 $(\omega_1,<) \to ([0,1],<)$ ?
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 $\alpha+1$.
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.
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.?
$X\cap [0,a-1/n]$
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).
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Oct 9, 2010 at 15:51