Cardinality of connected LOTS

Question

A LOTS (linearly ordered topological space) is a totally ordered set $(X,<)$ together with the corresponding order topology generated by all the open rays $(\leftarrow,x)$ and $(x,\rightarrow)$ for $x\in X$. Note that every LOTS is a $T_2$ space.

If $X$ is a connected LOTS with at least two points, then $|X|\ge 2^{\aleph_0}$.

This follows for example from the non-trivial fact that every LOTS is normal (see Why are ordered spaces normal? [collecting proofs]). So given two distinct points $a$ and $b$, by Urysohn's lemma there is a continuous map $f:X\to[0,1]$ with $f(a)=0$ and $f(b)=1$. If $X$ is also connected, the range $f(X)$ must be the whole interval $[0,1]$ and hence the cardinality of $X$ is at least $|[0,1]|=2^{\aleph_0}$.

Is there a more elementary way to see the result without using the fact that a LOTS is normal?

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I am fine assuming this characterization:

The ordered space $(X,<)$ with its order topology is connected iff

• (1) $X$ is Dedekind-complete (= least upper bound property): Every nonempty subset with an upper bound in $X$ has a least upper bound.
• (2) $X$ is order-dense: Given any two distinct points, there is a third point in between.

This is not very hard to prove (it's Exercise 26G in Willard).

Answer 1

I'm not sure if this counts as elementary enough, but it's probably easier than showing LOTS is normal (the proof that any LOTS is $T_4$ is quite hard, although if you know that, the proof that they are $T_5$ is quite easy).

Lemma 1. A connected LOTS is Dedekind-complete.

Proof: Let $A\subseteq X$ be a non-empty set bounded from above. If $\sup A$ doesn't exist, let $U = \{x\in X : \exists_{a\in A}\ x < a\}$ and $V = \{x\in X : \exists_{y\notin U}\ y < x\}$. Note that $y\notin U$ is equivalent to $y$ being an upper bound of $A$. Since there is no smallest upper bound of $A$, this implies that $V = X\setminus U$, so that $U, V$ are open non-empty sets which partition $X$. $\square$

Remark. A LOTS is connected iff its Dedekind-complete and has a dense ordering.

Lemma 2. A bounded Dedekind-complete LOTS is compact.

Proof: The open rays $(\leftarrow, a) = \{x\in X : x < a\}$ and $(a, \rightarrow) = \{x\in X : a < x\}$ form a subbasis for $X$. Let $\mathcal{U} = \{(\leftarrow, a) : a\in A\}\cup \{(b, \rightarrow): b\in B\}$ be an open cover of $X$. Let $b_0 = \inf B$ then $b_0\in (\leftarrow, a)$ for some $a\in A$ so that there exists $b\in B$ with $b\in (\leftarrow, a)$. Then $X = (\leftarrow, a)\cup (b, \rightarrow)$ so $\{(\leftarrow, a), (b, \rightarrow)\}$ is a finite refinement of $\mathcal{U}$. From Alexander subbasis theorem $X$ is compact. $\square$

Remark. A LOTS is compact iff its bounded and Dedekind-complete.

Lemma 3. Any LOTS is Hausdorff.

Proof: If $x, y\in X$ are distinct, say $x < y$, then either there is $x < z < y$ so that $(\leftarrow, z)$ and $(z, \rightarrow)$ are disjoint neighbourhoods to which $x, y$ belong, or such $z$ doesn't exist so that $(\leftarrow, y)$ and $(x, \rightarrow)$ are disjoint neighbourhood to which $x, y$ belong. $\square$

Theorem. Any connected LOTS is Tychonoff.

Proof: If $X$ is a connected LOTS, let $Y$ be obtained from $X$ by adding biggest element $\infty$ if $X$ has no biggest element, and smallest element $-\infty$ if $X$ has no smallest element. Then $Y$ is bounded and Dedekind-complete LOTS, so that $Y$ is a compact Hausdorff space, hence Tychonoff. Since $X$ is a subspace of Tychonoff space, it is itself Tychonoff. $\square$

Remark. Any LOTS is hereditarily collectionwise normal.

Theorem. If $X$ is a connected LOTS with at least two points then $|X|\geq \mathfrak{c}$.

Proof: Let $a, b\in X$ be distinct, then since $X$ is Tychonoff there exists $f:X\to [0, 1]$ such that $f(a) = 0$ and $f(b) = 1$. Then $f(X)\subseteq [0, 1]$ is a connected set with $0, 1\in f(X)$ so $f(X) = [0, 1]$, so that $f$ is surjective. This shows $|X|\geq \mathfrak{c}$. $\square$

Answer 2

Here is another proof of the result that does not use the fact that $X$ is normal.

Assume Jakobian's Lemma 2 (which is Problem 17E in Willard), namely:

The LOTS $X$ is compact iff the ordered set $(X,<)$ is Dedekind-complete and has a minimum element and a maximum element.

In particular, this implies that if $X$ is connected, every closed interval $[a,b]\subseteq X$ is compact.

(Side note: Corollary: Every connected LOTS is locally compact.)

So assume $X$ is connected with at least two distinct points $a<b$. By order density, choose $c$ and $d$ such that $a<c<d<b$. Starting with the closed interval $[a,b]$ we have constructed two non-degenerate disjoint closed intervals $[a,c]$ and $[d,b]$, one on the left and one on the right. For the next step, repeat the procedure with each of the intervals obtained so far to get 4 pairwise disjoint closed intervals. Repeat countably many times.

Each sequence $\alpha\in\{0,1\}^\mathbb N$ can be used to encode a choice of going left ($0$) or going right ($1$) countably many times, corresponding to a decreasing sequence of nested nonempty closed intervals in $X$. Each of these intervals is compact and the sequence of intervals satisfies the finite intersection property. By compactness the intersection of the sequence of intervals is nonempty and we can pick an element $f(\alpha)$ in that intersection (for example the minimum element in there). And different $\alpha\in 2^{\aleph_0}$ give different $f(\alpha)\in X$, because the intervals at each step of the construction in the previous paragraph were pairwise disjoint. Thus $|X|\ge 2^{\aleph_0}$.

Note: This is similar to the construction of a Cantor set, but the subspace $\{f(\alpha):\alpha\in 2^{\aleph_0}\}\subseteq X$ will in general not be homeomorphic to the Cantor set.