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

 A: 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. 
A: 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.? 
A: 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 $\alpha+1$.
