Why are ordered spaces normal? [collecting proofs]

Question

Greets

This is a problem I wanted to solve for a long time, and finally did some days ago. So I want to ask people here at MSE to show as many different answers to this problem as possible. I will offer a Bounty in two days, depending on the interest in the problem, and eventually increase it as it gets more voted. Of course, I will show my answer to this question to know whether it is correct.

Thanks

Answer 1

$\newcommand{\cl}{\operatorname{cl}}$ Theorem. Let $\langle X,\tau,\le\rangle$ be a LOTS; then $X$ is $T_5$.

Proof. Let $H$ and $K$ be separated subsets of $X$. For each $x\in H$ there is a convex $V_x\in\tau$ such that $x\in V_x\subseteq X\setminus K$, and for each $x\in K$ there is a convex $V_x\in\tau$ such that $x\in V_x\subseteq X\setminus H$. Let $V_H=\bigcup_{x\in H}V_x$ and $V_K=\bigcup_{x\in K}V_x$; clearly $H\subseteq V_H$, $K\subseteq V_K$, and $V_H\cap V_K\subseteq X\setminus(H\cup K)$.

Let $V=V_H\cap V_K$. If $V=\varnothing$, we’re done, so suppose that $V\ne\varnothing$. Define a relation $\sim$ on $V$ by $p\sim q$ iff $\left[\min\{p,q\},\max\{p,q\}\right]\subseteq V$; it’s easily verified that $\sim$ is an equivalence relation whose equivalence classes are the order-components of $V$.

Let $T\subseteq V$ contain exactly one point of each $\sim$-class. Suppose that $x\in H$, and $p,q\in V_x\cap T$ with $p<q$; I’ll show that $p<x<q$. Suppose that $x<p$. Since $p\in T\subseteq V$, there is a $y\in K$ such that $p\in V_y$; $[x,q]\subseteq V_x\subseteq X\setminus K$, so $y\notin[x,q]$. If $y<x$, then $x\in[y,p]\subseteq V_y\cap H=\varnothing$, so $x<p<q<y$. But then $[p,q]\subseteq V_x\cap V_y\subseteq V$, so $p\sim q$, contradicting the choice of $p$ and $q$ and showing that $p<x$. A similar argument shows that $x<q$. Similarly, if $x\in K$ and $p,q\in V_x\cap T$ with $p<q$, then $p<x<q$. Note that it follows immediately that $|V_x\cap T|\le 2$ for all $x\in H\cup K$.

Now fix $p\in T$. Let $H_p=\{x\in H:p\in V_x\}$ and $K_p=\{x\in K:p\in V_x\}$; $H_p\ne\varnothing\ne K_p$, since $p\in V$. Suppose that $x<p$ for some $x\in H_p$. If $y\in K_p$ and $y<p$, then either $x<y$ and $y\in V_x$, or $y<x$ and $x\in V_y$, since the sets $V_x$ and $V_y$ are convex; neither is possible, so $p<y$, and since $y\in K_p$ was arbitrary, $p<K_p$. A similar argument then shows that $H_p<p$ and hence $H_p<p<K_p$. If instead $p<x$ for some $x\in H_p$, it follows similarly that $K_p<p<H_p$.

For each $x\in H\cup K$ define $W_x\in\tau$ as follows: $$W_x=\begin{cases}V_x,&\text{if }V_x\cap T=\varnothing\\V_x\cap(p,\to),&\text{if }V_x\cap T=\{p\}\text{ and }p<x\\V_x\cap(\leftarrow,p),&\text{if }V_x\cap T=\{p\}\text{ and }x<p\\V_x\cap(p,q),&\text{if }V_x\cap T=\{p,q\}\text{ and }p<x<q\;.\end{cases}$$ Let $$W_H=\bigcup_{x\in H}W_x\qquad\text{and}\qquad W_K=\bigcup_{x\in K}W_x\;;$$ clearly $W_H$ and $W_K$ are open, $H\subseteq W_H$, and $K\subseteq W_K$, and I claim that $W_H\cap W_K=\varnothing$.

Suppose not; then there are $x\in H$ and $y\in K$ such that $W_x\cap W_y\ne\varnothing$; without loss of generality suppose that $x<y$. Fix $q\in W_x\cap W_y$; it’s not hard to see that $x<q<y$, since $W_x$ and $W_y$ are convex. Moreover, $q\in V$, so $q\sim p$ for a unique $p\in T$. Let $I$ be the closed interval with endpoints $p$ and $q$. Then $I\subseteq V\subseteq X\setminus(H\cup K)$, so $x,y\notin I$, and therefore $x<p<y$. If $p\le q$, then $p\in W_x\cap T\subseteq V_x\cap T$, and by construction $W_x\subseteq(\leftarrow,p)$, and $p\notin W_x$, a contradiction. If, on the other hand, $q\le p$, then $p\in W_y\cap T\subseteq V_y\cap T$, so that $W_y\subseteq(p,\to)$, and $p\notin W_y$, which is again a contradiction, and it follows that $W_H\cap W_K=\varnothing$. $\dashv$

Despite the sometimes finicky details, the idea of the argument is very simple. $V$ is the intersection of the open nbhds $V_H$ and $V_K$ of $H$ and $K$, respectively. It’s open, so we partition it into its open order-components. If $C$ is one of these components, let $H_C=\{x\in H:V_x\cap C\ne\varnothing\}$ and $K_C=\{x\in K:V_x\cap C\ne\varnothing\}$; we show that either $H_C<C<K_C$ or $K_C<C<H_C$. Then we pick a point $p$ in $C$ and contract the intervals $V_x$ that meet $C$ by intersecting them with $(\leftarrow,p)$ for $x<C$ and with $(p,\to)$ for $C<x$. This contraction removes $C$ from the intersection of the nbhds of $H$ and $K$, and since we do it simultaneously for all components $C$, we end up with disjoint nbhds $W_H$ and $W_K$ of $H$ and $K$.

As an immediate consequence we get that every LOTS $X$ is hereditarily normal: disjoint closed sets in a subspace of $X$ are separated sets in $X$. Thus, we have for free that every GO-space (generalized ordered space) is hereditarily normal.

Moreover, the argument is very easily modified to show that if $\mathscr{F}$ is a separated family of subsets of $X$, meaning that $F\cap\cl\bigcup(\mathscr{F}\setminus\{F\})=\varnothing$ for each $F\in\mathscr{F}$, then there is a pairwise disjoint family $\mathscr{U}=\{U_F:F\in\mathscr{F}\}$ of open sets in $X$ such that $F\subseteq U_F$ for each $F\in\mathscr{F}$, i.e., that $X$ is hereditarily collectionwise normal.

Answer 2

Proposition 1. If $\langle X,<\rangle$ is a connected linear order with endpoints, then the ordered space $X$ is compact.

Proof. Note that under these hypothesis every nonempty subset of $X$ has a supremum. Hence this property is proved using the idea in the classic proof that $[0,1]$ is compact; which is done considering an open covering $\Lambda$ of $[0,1]$ and showing that $\sup\{x\in [0,1]:[0,x]$ can be covered by a finite subset of $\Lambda\}$ equals $1$; using the completeness of $[0,1]$.

Proposition 2. If $\langle X,< \rangle$ is a linear order, then $\langle X,< \rangle$ can be embedded in a connected order without endpoints.

Proof. First embed $\langle X,< \rangle$ in a dense linear ordering $\langle X',< \rangle$ without endpoints; which is pretty easy, and then using Dedekind cuts, $\langle X',< \rangle$ can be embedded in a connected linear order without endpoitns $\langle X'',< \rangle.$

Proposition 3. There exists a family $M$ of closed intervals of $(X,<)$ such that the intersection of any two elements of $M$ is at most a point and $\bigcup M=X$.

Fix $x\in X$ not an endpoint. Put $x_0=x$, and let $x_0\in X$ be such that $x<x_0$. Suppose that for some ordinal $\alpha$ an increasing $\alpha$-sequence $\langle x_{\beta}:\beta < \alpha \rangle$ such that $x_{\gamma}<x_{\beta}$ whenever $\gamma<\beta$ has been constructed. Let $x_{\alpha}\in X$ be such that $x_{\alpha}$ is greater than all elements of $\langle x_{\beta}:\beta < \alpha \rangle$, if such $x$ exists. But $X$ is a set, so there exists some ordinal $\alpha$ such that the construction cannot be continued at step $\alpha$. Then we have $[x,\infty)=\bigcup_{\beta<\alpha}[x_{\beta},x_{\beta+1}]$, similarly there is a family of closed intervals $\mathfrak{I}$ such that the intersection of any two elements of $\mathfrak{I}$ is at most a point and such that $\bigcup \mathfrak{I}=(\infty,x],$ then put $M=\{[x_{\beta},x_{\beta+1}]:\beta<\alpha\}\cup \mathfrak{I}$.

Proposition 4. If $\langle X,<\rangle$ is connected, then $\langle X,< \rangle$ is normal.

Proof. Let $M$ be a family of closed intervals with the property of the previous proposition. Then each $I\in M$ is compact by Proposition 1, but also each $I\in M$ is Hausdorff, hence each $I\in M$ is a normal subspace of $\langle X,<\rangle$. Now let $A,B$ be closed disjoint subsets of $\langle X,<\rangle$. For each $I\in M$ let $A_I$ and $B_I$ be disjoint open subsets of $I$ with $A\cap I\subseteq A_I$ and $B\cap I\subseteq B_I$. Furthermore the sets $A_I,B_I$ can be chosen so that if $A$ does not contain an endpoint,$x$, of $I$, then $x\notin A_I$, and $B_I$ can also be chosen so that this holds, this ensures that for distinct $I,J\in M$, $A_I\cap B_J=\emptyset$; since $A\cap B=\emptyset$ and $I$ and $J$ have only at most one point in common. Hence if $A'=\bigcup_{I\in M}A_I$ and $B'=\bigcup_{I\in M}B_I$, $A'\cap B'=\emptyset$, and both $A'$ and $B'$ are open subsets of $X$ separating $A$ and $B$; since $\bigcup M=X$. Therefore $\langle X,<\rangle$ is normal.

Proposition 5. Every ordered space is normal.

Proof. Let $\langle X,<\rangle$ be a linear order, and let $\langle X',<\rangle$ be connected linear order without endpoints exteding $\langle X,<\rangle$. Let $A,B\subset X$ be disjoint and closed, then if $Y=X'-(\operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B))$, $A,B\subset Y$ since $X\cap \operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B)=\emptyset$, because of the hypothesis, and so $X\subseteq Y$. But $Y$ is open in $X'$, hence $Y$ is the union of a disjoint family of open intervals, each being connected, hence by Proposition 4, $Y$ is a normal subspace of $X$. But $\operatorname{Cl}_{Y}(A)\cap \operatorname{Cl}_{Y}(B)=\operatorname{Cl}_{X'}(A)\cap \operatorname{Cl}_{X'}(B)\cap Y=\emptyset$, thus there are disjoint open subsets $A',B'$ of $Y$ such that $\operatorname{Cl}_{Y}(A)\subseteq A'$ and $\operatorname{Cl}_{Y}(B)\subseteq B'$. Therefore $A'\cap X$ and $B'\cap X$ are disjoint open subsets of $X$ separating $A$ and $B$, respectively. Thus $\langle X,<\rangle$ is normal.

Answer 3

(Generalised) ordered spaces (GO-spaces) are monotonically normal, and monotonically normal spaces are hereditarily normal. I wrote proofs here.