When is the complement of the diagonal of a LOTS functionally countable?

Let $x,y\in X$ such that $x\neq y$, we need to find an open set $W\subset X^{2}$ such that $\langle x,y\rangle\in W$ and $\lvert\{U\in\mathcal{U}\colon(I_{U}\times J_{U})\cap W\neq\emptyset\}\rvert\leq 1$.

First, consider the case when $y<x$. Let $W_{0}$ and $W_{1}$ be open subsets of $X$ such that $y\in W_{0}\subset(\leftarrow,x)$, $x\in W_{1}\subset(y,\rightarrow)$ and $W_{0}\cap W_{1}=\emptyset$. Then $W=W_{0}\times W_{1}$ is such that $\langle x,y\rangle\in W$ and $(I_{U}\times J_{U})\cap W=\emptyset$ for all $U\in\mathcal{U}$.

So assume that $x<y$. We divide our analysis in four cases:

By our hypothesis we may assume that $U_{0}<_{s}U_{1}$. Let $p\in(I_{U_{0}}\cup J_{U_{0}})\cap(x,y)$ and $q\in(I_{U_{1}}\cup J_{U_{1}})\cap(x,y)$. Notice that $x<p<q<y$. Let $W=(\leftarrow,p)\times(q,\rightarrow)$. Then $W$ is an open subset of $X^{2}$ such that $\langle x,y\rangle\in W$; we claim that $W\cap(I_{V}\times J_{V})=\emptyset$ for every $V\in\mathcal{U}$.

Let $V\in\mathcal{U}$ and $\langle a,b\rangle\in W$. If $V=U_{0}$ or $V<_{s}U_{0}$, since $b>q$ and $q\in I_{U_{1}}\cup J_{U_{1}}$, $b\notin J_{V}$; thus, $\langle a,b\rangle\notin I_{V}\times J_{V}$. If $U_{0}<_{s}V$, since $a<p$ and $p\in I_{U_{0}}\cup J_{U_{0}}$, $a\notin I_{V}$; thus, $\langle a,b\rangle\notin I_{V}\times J_{V}$.

Let $z\in(I_{U}\cup J_{U})\cap(x,y)$ and $W=(\leftarrow,z)\times(z,\rightarrow)$. Clearly, $\langle x,y\rangle\in W$; we claim that $W\cap(I_{V}\times J_{V})=\emptyset$ for every $V\in\mathcal{U}\setminus\{U\}$.

So let $V\in\mathcal{U}$ with $U\neq V$ and $\langle a,b\rangle\in W$. If $V<_{s}U$, since $z<b$ and $z\in I_{U}\cup J_{U}$, we conclude that $b\notin J_{V}$; thus, $\langle a,b\rangle\notin I_{V}\times J_{V}$. If $U<_{s}V$, since $a<z$ and $z\in I_{U}\cup J_{U}$, we convince ourselves that $a\notin I_{V}$; thus, $\langle a,b\rangle\notin I_{V}\times J_{V}$.

Let $W=(\leftarrow,y)\times(x,\rightarrow)$, notice that $\langle x,y\rangle\in W$. We claim that $W\cap(I_{V}\times J_{V})=\emptyset$ for every $V\in\mathcal{U}\setminus\{U\}$.

So let $V\in\mathcal{U}$ with $U\neq V$ and $\langle a,b\rangle\in W$. If $V<_{s}U$, since $J_{V}\subset(\leftarrow,x]$ and $x<b$, $b\notin J_{V}$; therefore, $\langle a,b\rangle\notin I_{V}\times J_{V}$. If $U<_{s}V$, since $I_{V}\subset[y,\rightarrow)$ and $a<y$, $a\notin I_{V}$; therefore, $\langle a,b\rangle\notin I_{V}\times J_{V}$.

In this case we also take $W=(\leftarrow,y)\times(x,\rightarrow)$. Again, $\langle x,y\rangle\in W$. We claim that $W\cap(I_{U}\times J_{U})=\emptyset$ for every $U\in\mathcal{U}$.

Let $U\in\mathcal{U}$ and $\langle a,b\rangle\in W$. If $I_{U}\cup J_{U}\subset(\leftarrow,x]$, since $x<b$, $b\notin J_{U}$ and thus, $\langle a,b\rangle\notin I_{U}\times J_{U}$. If $I_{U}\cup J_{U}\subset[y,\rightarrow)$, since $a<y$, $a\notin I_{U}$ and thus, $\langle a,b\rangle\notin I_{U}\times J_{U}$.

Since these are all the possible cases, we conclude that $\{I_{U}\times J_{U}\colon U\in\mathcal{U}\}$ is discrete. ∎

Assume that $X$ contains an ordered copy of $\omega_{1}$. So there is $\{x_{\alpha}\colon\alpha<\omega_{1}\}\subset X$ such that $\alpha<\beta<\omega_{1}$ implies $x_{\alpha}<x_{\beta}$.

Given $\alpha<\omega_{1}$ there are unique ordinals $\beta<\omega_{1}$ that is a limit or $0$ and $n<\omega$ such that $\alpha=\beta+n$; define $U_{\alpha}=(x_{\beta+3n},x_{\beta+3n+3})$. Then it is easy to see that the family $\mathcal{U}=\{U_{\alpha}\colon\alpha<\omega_{1}\}$ satisfies the hypothesis of Lemma 3. Thus, $X$ does not satisfy the DCCC.

If $X$ contains an ordered copy of $\omega_{1}^{\ast}$, we may follow an analogous argument to conclude that $X$ does not satisfy the DCCC. ∎

We prove the contrapositive implication. Let $X$ be a LOTS and assume that there is an uncountable $Y\subset X$ such that there is an order embedding $e\colon Y\to\mathbb{R}$; call $e[Y]=Z$. Without loss of generality we may assume that $Z$ is bounded in $\mathbb{R}$. We will find an uncountable $A\subset Y$ and define a continuous function $f\colon X\to\mathbb{R}$ such that $f\restriction A$ is injective.

First, let $\mathcal{I}$ be the collection of all open intervals $I$ of $\mathbb{R}$ of the form $(a,b)$, where $a,b$ are rational numbers and $a<b$, such that $I\cap Z$ is countable. Let $Z_{0}=Z\cap(\bigcup\mathcal{I})$. Then $Z_{0}$ is countable.

Next, let $Z_{1}$ be the set of points $x\in Z\setminus Z_{0}$ such that $x$ is either an immediate succesor or an immediate predecesor of a point in $Z\setminus Z_{0}$. Since $c(\mathbb{R})=\omega$, $Z_{1}$ is countable. Let $p=\inf(Z\setminus Z_{0})$ and $q=\sup(Z\setminus Z_{0})$. Define $Z_{2}=Z\setminus(Z_{0}\cup Z_{1}\cup\{p,q\})$.

Notice that $Z_{2}$ is an uncountable subset without endpoints satisfying the following property: if $a,b\in Z_{2}$ with $a<b$, then $(a,b)\cap Z_{2}$ is uncountable. Since $\mathbb{R}$ is second countable, $Z_{2}$ has a countable, topologically dense set $D\subset Z_{2}$. Then it follows that every time $a,b\in Z_{2}$ and $a<b$, there exists $d\in D$ such that $a<d<b$.

Finally, let $A=e^{\leftarrow}[Z_{2}]$ and $Q=e^{\leftarrow}[D]$. Since $e$ is an order embedding, we conclude the following properties:

1. (i)

   $Q$ is countable, densely ordered and has no endpoints.

2. (ii)

   If $a,b\in A$ and $a<b$, then there are $p,q\in Q$ with $a<p<q<b$.

As it was famously shown by Cantor, condition (i) implies that $Q$ is order-isomorphic to the rationals $\mathbb{Q}$. So we may choose an order isomorphism $i\colon Q\to\mathbb{Q}\cap(0,1)$. Then we define a function $f\colon X\to[0,1]$ by

It is well known (and can be easily checked by the reader) that $f$ is continuous. By condition (ii) above it follows that $f\restriction A$ is injective. Thus, $X$ is not functionally countable. ∎

Let $X=(\mathbb{Q}\times\{0,2\})\cup(\mathbb{R}\times\{1\})$ with the subspace topology of $\mathbb{R}\times 3$ with the lexicographic order. ∎

Let $U$ be the set of lonely points of $X$, as defined in Definition 9, and let $Y=X\setminus U$. Notice that since $Y$ is a closed subspace of $X$, it is a LOTS with the subspace topology. We define $f\colon X\to Y$ in the following way.

1. (i)

   If $x\in Y$, let $f(x)=x$.

2. (ii)

   If $x=\max X$ and $x\in U$, let $f(x)$ be the immediate predecesor of $x$.

3. (iii)

   If $x\in U$ and $x\neq\max X$, let $f(x)$ be the immediate succesor of $x$.

It should be clear that $f$ is continuous and that the fibers of $f$ are convex with cardinality less than or equal to $2$.

First, notice that $Y$ is uncountable. If $U$ is countable, then this is clear. Otherwise, since $f\restriction U$ is clearly injective, $f[U]$ is an uncountable subset of $Y$.

Next, let us prove that $Y$ is not separable. By [8, Proposition 3.2], $X$ is functionally countable. Then by [8, Proposition 3.1(a)] the continuous image $Y$ of $X$ is functionally countable. If $D\subset Y$ is countable, by [9, Theorem 3.10] we conclude that $\overline{D}$ is countable so $D$ is not dense in $Y$.

Now, let us prove that $c(Y)=\omega$ by assuming the opossite. We would like to apply Lemma 3 to conclude that $Y^{2}\setminus\Delta_{Y}$ does not safisfy the DCCC.

Define $A=\{x\in Y\colon\{x\}\textrm{ is open}\}$. Assume first that $A$ is countable. Let $\mathcal{U}$ be a collection of $\omega_{1}$-many open nonempty intervals of $Y$ that are pairwise disjoint. Then there are only countably many elements of $\mathcal{U}$ that intersect $A$; let $\mathcal{V}$ be the collection of all elements of $\mathcal{U}$ that do not intersect $A$. Then $\mathcal{V}$ is of cardinality $\omega_{1}$ and each of its elements is infinite. Thus, $\mathcal{V}$ satisfies the conditions in the claim.

Next, consider the case when $A$ is uncountable. Given $a,b\in A$ we define $a\sim b$ if either $a=b$ or $[\min\{a,b\},\max\{a,b\}]$ is finite. It is easily seen that $\sim$ is an equivalence relation on $A$; let $\mathcal{E}=A/{\sim}$ be the set of equivalence classes. Moreover, every element of $\mathcal{E}$ is convex. By the definition of $Y$ it can be shown that every element of $\mathcal{E}$ is of cardinality at least $2$. It can also be easily shown that every element of $\mathcal{E}$ is order isomorphic to one of the following ordered sets:

1. (1)

   a natural number greater than or equal to $2$,

2. (2)

   the set of natural numbers $\omega$,

3. (3)

   the linearly ordered set $\omega^{\ast}$, which is the reverse order of $\omega$, or

4. (4)

   the integers $\mathbb{Z}$.

Thus, every element of $\mathcal{E}$ is a countable open set with at least two points. Since $A$ is uncountable, $\mathcal{E}$ is uncountable as well. Thus, we may take $\mathcal{V}$ to be any subcollection of $\mathcal{E}$ of size $\omega_{1}$ and the conditions in the claim are satisfied.Thus, we may write $\mathcal{V}=\{V_{\alpha}\colon\alpha<\omega_{1}\}$ and for each $\alpha<\omega_{1}$, $I_{\alpha}$ and $J_{\alpha}$ are nonempty open subsets with $I_{\alpha}\cup J_{\alpha}\subset V_{\alpha}$ and $I_{\alpha}<_{s}J_{\alpha}$. By Lemma 3 we conclude that $\{I_{\alpha}\times J_{\alpha}\colon\alpha<\omega_{1}\}$ is a discrete family in $Y^{2}\setminus\Delta_{Y}$.

Let us define a function $g\colon X^{2}\setminus\Delta_{X}\to Y^{2}$ by $g(\langle x,y\rangle)=\langle f(x),f(y)\rangle$ for all $\langle x,y\rangle\in X^{2}\setminus\Delta$. We will prove that $\{g^{\leftarrow}[I_{\alpha}\times J_{\alpha}]\colon\alpha<\omega_{1}\}$ is a discrete family in $X^{2}\setminus\Delta_{X}$. Let $\langle x_{0},x_{1}\rangle\in X^{2}\setminus\Delta_{X}$.

If $f(x_{0})\neq f(x_{1})$ we know that $g(\langle x_{0},x_{1}\rangle)=\langle f(x_{0}),f(x_{1})\rangle\in Y^{2}\setminus\Delta_{Y}$ so there exists an open set $W\subset Y^{2}\setminus\Delta_{Y}$ with $g(\langle x_{0},x_{1}\rangle)\in W$ and $W$ intersects at most one element of $\{I_{\alpha}\times J_{\alpha}\colon\alpha<\omega_{1}\}$. Thus, $g^{\leftarrow}[W]$ is an open subset of $X^{2}\setminus\Delta_{X}$ with $\langle x_{0},x_{1}\rangle\in g^{\leftarrow}[W]$ and such that $g^{\leftarrow}[W]$ intersects at most one element of $\{g^{\leftarrow}[I_{\alpha}\times J_{\alpha}]\colon\alpha<\omega_{1}\}$.

Now, assume that $f(x_{0})=f(x_{1})$. Then, by the definition of $Y$ and $f$, one of $x_{0}$ or $x_{1}$ must be a lonely point. We will assume that $x_{0}<x_{1}$, that $x_{1}\neq\max X$, and thus, that $x_{0}$ is lonely; the other cases can be treated in a similar way. Since $x_{0}$ is isolated, $W=\{x_{0}\}\times(x_{0},\rightarrow)$ is an open set such that $\langle x_{0},x_{1}\rangle\in W$. If $\alpha<\omega_{1}$ notice that $W\cap g^{\leftarrow}[I_{\alpha}\times J_{\alpha}]\neq\emptyset$ is equivalent to $W\cap(f^{\leftarrow}[I_{\alpha}]\times f^{\leftarrow}[J_{\alpha}])\neq\emptyset$, which implies that $x_{0}\in f^{\leftarrow}[I_{\alpha}]$; that is, $f(x_{0})\in I_{\alpha}$. Since the elements of $\mathcal{V}$ are pairwise disjoint, this can only happen for at most one value of $\alpha$. That is, there is at most one element of $\{g^{\leftarrow}[I_{\alpha}\times J_{\alpha}]\colon\alpha<\omega_{1}\}$ that intersects $W$.

Thus, we have proved that $\{g^{\leftarrow}[I_{\alpha}\times J_{\alpha}]\colon\alpha<\omega_{1}\}$ is an uncountable discrete family of $X^{2}\setminus\Delta_{X}$, which is impossible since $X^{2}\setminus\Delta$ is assumed to be functionally countable. From this contradiction, we conclude that $c(Y)=\omega$ and thus, $Y$ is a Souslin line. ∎

By [5, Exercise III.5.33] we may assume that there is a Souslin line $X$ that is connected with the order topology. Let $a,b\in X$ be such that $a<b$. By Urysohn’s lemma there exists a continuous function $f\colon X\to[0,1]$ such that $f(a)=0$ and $f(b)=1$. Since $X$ is connected, the image of $f$ is a connected subset of $[0,1]$ containing both endpoints $0$ and $1$. Thus, $f$ is surjective and $X$ is not functionally countable. ∎

We prove the contrapositive implication. Assume that $X$ is not first countable. By Lemma 2, we may assume that there exists $\{x_{\alpha}\colon\alpha<\omega_{1}\}\subset X$ such that $\alpha<\beta<\omega_{1}$ implies $x_{\alpha}<x_{\beta}$. For $\alpha<\omega_{1}$, there are unique ordinals $\beta<\omega_{1}$ that is a limit or $0$ and $n<\omega$ such that $\alpha=\beta+n$; let $U_{\alpha}=(x_{\beta},x_{\beta+2})$. Then $\{U_{\alpha}\colon\alpha<\omega_{1}\}$ is an uncountable, pairwise disjoint collection of nonempty open subsets of $X$. Thus, $c(X)>\omega$. ∎

To simplify the proof, let us assume that $X\subset Y$, where $Y$ is the Dedekind completion of $X$. By Lemma 14 and the fact that $X$ is dense in $Y$, $Y$ is first countable.

Let $U\subset X$ be nonempty open and convex. Define $p=\inf U$ and $q=\sup U$; these two points exist in $Y$ but not necessarily in $X$. Let also $x_{0}\in U$.

First, let us argue that $p,q\notin U$. If $p\in U$ then $p=\min{(U)}$. Since $X$ has no endpoints, by definition of the order topology in $X$ there exist $x,y\in X$ with $p\in(x,y)\cap X\subset U$. But since $X$ is densely ordered, $(x,p)\cap X$ is a nonempty subset of $U$, which is a contradiction. We arrive at a similar contradiction if we asssume that $q\in U$.

Let $\{B(p,n)\colon n\in\omega\}$ and $\{B(q,n)\colon n\in\omega\}$ be countable local bases at $p$ and $q$, respectively. Since $Y$ is densely ordered, we may recursively find $\{y_{n}\colon n\in\mathbb{Z}\}\subset(p,q)$ with the following properties:

1. (a)

   $y_{0}=x_{0}$,

2. (b)

   if $n<\omega$ then $y_{-(n+1)}\in(p,y_{-n})\cap U\cap B(p,n)$, and

3. (c)

   if $n<\omega$ then $y_{n+1}\in(y_{n},q)\cap U\cap B(q,n)$.

Then, for each $n\in\omega$, since $X$ is topologically dense in $Y$, let $a_{n}\in(y_{-(n+1)},y_{n})\cap X$ and $b_{n}\in(y_{n},y_{n+1})\cap X$. It is not hard to see that $\{a_{n},b_{n}\colon n\in\omega\}$ is as required. ∎

First, notice that if we prove the statement assuming that $X$ has no endpoints, then the statement holds in general. So, in order to simplify the proof, we assume that $X$ has no endpoints.

Let $\mathcal{B}$ be a countable base of $M$. For each $B\in\mathcal{B}$, let $\mathcal{U}_{B}$ be the set of convex components of $f^{\leftarrow}[B]$. Let $\mathcal{U}=\bigcup\{\mathcal{U}_{B}\colon B\in\mathcal{B}\}$; then $\mathcal{U}$ is a countable set of nonempty open and convex subsets of $X$. For each $U\in\mathcal{U}$, by Lemma 15 there exist $\{a^{U}_{n},b^{U}_{n}\colon n<\omega\}\subset U$ such that $a^{U}_{n}<b^{U}_{n}$ for each $n<\omega$ and $U=\bigcup\{(a^{U}_{n},b^{U}_{n})\colon n<\omega\}$. Define $D=\{a^{U}_{n},b^{U}_{n}\colon U\in\mathcal{U},n<\omega\}$, clearly $D$ is a countable set.

So let $V$ be a convex component of $X\setminus\overline{D}$, we have to prove that $f\restriction V$ is constant. Assume this is not the case and let $p_{0},p_{1}\in V$ such that $f(p)\neq f(q)$. Since $M$ is Hausdorff, there exists $B_{0},B_{1}\in\mathcal{B}$ such that $f(p)\in B_{0}$, $f(q)\in B_{1}$ and $B_{0}\cap B_{1}=\emptyset$. For $i\in\{0,1\}$, let $U_{i}$ be the element of $\mathcal{U}_{B_{i}}$ such that $p_{i}\in U_{i}$. Clearly, $U_{0}\neq U_{1}$ and since both $U_{0}$ and $U_{1}$ are convex, we may assume without loss of generality that $U_{0}<_{s}U_{1}$. For $i\in\{0,1\}$ there exists $n(i)<\omega$ be such that $p_{i}\in(a^{U_{i}}_{n(i)},b^{U_{i}}_{n(i)})\subset U_{i}$. Then,

and this implies that $(p_{0},p_{1})\cap D\neq\emptyset$. But since $V$ is convex, $(p_{0},p_{1})\subset V$, so we obtain a contradiction. Thus, we conclude that $f\restriction V$ is constant and the result is proved. ∎

Let $X$ be a Souslin line. By [5, Lemma III.5.31], we may assume that $X$ is densely ordered and contains no separable interval. By the observations we made before the statement, there is a base $\{U_{\alpha}\colon\alpha<\omega_{1}\}$ of $X$. Recursively we may choose $x_{\alpha}\in U_{\alpha}\setminus\overline{\{x_{\beta}\colon\beta<\alpha\}}$ for all $\alpha<\omega_{1}$. Then $L=\{x_{\alpha}\colon\alpha<\omega_{1}\}$ is a left-separated subset of $X$. Since $L$ is dense in $X$, $L$ is also a Souslin line.

We claim that $L$ is functionally countable. Let $f\colon L\to\mathbb{R}$ a be continuous function. By Proposition 16, there is a countable set $D\subset L$ such that $f[L\setminus\overline{D}]$ is countable. Since $L$ is left-separated, $\overline{D}\subset\{x_{\alpha}\colon\alpha<\beta\}$ for some $\beta<\omega_{1}$. Then it follows that $f[L]\subset\{f(x_{\alpha})\colon\alpha<\beta\}\cup f[L\setminus\overline{D}]$ so $f[L]$ is countable. ∎