Dense sets and Suslin Lines

Question

I'm studying Infinitary Combinatorics and need a hint for proofing that every Suslin Line has a $\omega_1$ dense set. I've tried to do something with the set of all subsets of the line of cardinality $\omega_1$, but had no success. Any hints?

Answer 1

Let $\langle X,\le\rangle$ be a Suslin line. Let $\preceq$ be an arbitrary well-ordering of $X$. For each $x\in X$ let $$\mathscr{I}(x)=\{(a,b):a,b\in X\text{ and }a<x<b\}\;,$$ and let $D=\{x\in X:x=\preceq\text{-}\min I\text{ for some }I\in\mathscr{I}(x)\}$.

• Show that $D$ is dense in $X$.

For $x\in D$ let $I(x)=\bigcup\{I\in\mathscr{I}(x):x=\preceq\text{-}\min I\}$; note that $I(x)$ is an interval containing $x$.

• Show that if $x,y\in D$ and $x\preceq y$, then either $I(x)\cap I(y)=\varnothing$, or $I(y)\subseteq I(x)$.

Suppose that $|D|\ge\omega_2$. Let $$D_0=\left\{x\in D:\forall y\in D\setminus\{x\}\big(I(x)\nsubseteq I(y)\big)\right\}\;;$$

$\{I(x):x\in D_0\}$ is pairwise disjoint, so $|D_0|\le\omega$. Let $E_1=D\setminus D_0$, and let $$D_1=\left\{x\in E_1:\forall y\in E_1\setminus\{x\}\big(I(x)\nsubseteq I(y)\big)\right\}\;;$$ $\{I(x):x\in D_1\}$ is pairwise disjoint, so $|D_1|\le\omega$. If $D_\xi$ has been constructed for $\xi<\eta<\omega_1$, let $E_\eta=D\setminus\bigcup_{\xi<\eta}D_\xi$, and let $$D_\eta=\left\{x\in E_\eta:\forall y\in E_\eta\setminus\{x\}\big(I(x)\nsubseteq I(y)\big)\right\}\;;$$ $\{I(x):x\in D_\eta\}$ is pairwise disjoint, so $|D_\eta|\le\omega$. Thus, $|E_\eta|\ge\omega_2$ for each $\eta<\omega_1$, and we can construct $D_\eta$ for each $\eta<\omega_1$. Moreover, $\left|\bigcup_{\xi<\omega_1}D_\xi\right|\le\omega_1$, so $E_{\omega_1}=D\setminus\bigcup_{\xi<\omega_1}D_\xi\ne\varnothing$. Fix $p\in E_{\omega_1}$.

• Show that for each $\xi<\omega_1$ there is a $p_\xi\in D_\xi$ such that $I(p)\subseteq I(p_\xi)$. Conclude that $I(p_\xi)\supsetneqq I(p_\eta)$ whenever $\xi<\eta<\omega_1$.

For each $\xi<\omega_1$ let $x_\xi\in I(p_\xi)\setminus I(p_{\xi+1})$, and let $A=\{x_\xi:\xi<\omega_1\}$. Let $L=\{x_\xi:x_\xi<p_{\xi+1}\}$ and $R=\{x_\xi:p_{\xi+1}<x_\xi\}$; clearly $A=L\cup R$, and $|A|=\omega_1$, so at least one of $L$ and $R$ is uncountable..

• Show that if $x_\xi\in L$, and $\xi<\eta<\omega_1$, then $x_\xi<x_\eta$.
• Similarly, show that if $x_\xi\in R$, and $\xi<\eta<\omega_1$, then $x_\eta<x_\xi$.
• Conclude that either $\langle L,\le\rangle$ is well-ordered in type $\omega_1$, or $\langle R,\le\rangle$ is inversely well-ordered in type $\omega_1$. In either case show that $X$ contains an uncountable family of pairwise disjoint open intervals.

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The argument that I sketched above is actually a special case of the proof that if $X$ is a linearly ordered topological space whose density is at least $\kappa^+$, then $X$ contains a family of at least $\kappa$ pairwise disjoint open intervals. It follows that for a LOTS $X$, $d(X)\le c(X)^+$.

Answer 2

Hint: Construct a tree $\langle I_{x} : x \in 2^{< \omega_1}\rangle$ of intervals in $L$. It should only have countably many non degenerate interval at each level and it should not have an uncountable branch.