5
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ 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. $\endgroup$
    – Jim Conant
    Oct 9, 2010 at 15:03
  • $\begingroup$ $\omega_1$ denotes the first uncountable ordinal. $\endgroup$ Oct 9, 2010 at 15:08
  • $\begingroup$ Thus explaining Alessandro's proof. Thanks. $\endgroup$
    – Jim Conant
    Oct 9, 2010 at 15:10
  • $\begingroup$ You might also see mathoverflow.net/questions/25100/order-types-of-positive-reals/… $\endgroup$ Oct 9, 2010 at 15:16
  • 1
    $\begingroup$ (On the other hand, for any countable ordinal $\alpha$, ${\mathbb Q}$ contains subsets order-isomorphic to $\alpha+1$ that are closed as subsets of ${\mathbb R}$.) $\endgroup$ Oct 12, 2010 at 1:31

3 Answers 3

18
$\begingroup$

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 $\alpha+1$.

$\endgroup$
1
  • $\begingroup$ Glad to be helpful. $\endgroup$ Oct 9, 2010 at 15:05
6
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.?

$\endgroup$
1
  • 1
    $\begingroup$ $X\cap [0,a-1/n]$ 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). $\endgroup$ Oct 9, 2010 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.