# Dense in a total order and in an order topology

I'm trying to connect the definition of dense in a total ordered set

Def. 1 Given total order $(X,\le)$, a set $S\subseteq X$ is dense in $X$ if for any $x,x'\in S$ such that $x<x'$, there is $y\in S$ such that $x<y<x'$.

with the definition of dense-in-itself in a order topology (induced by the same total order)

Def. 2 Given the order topology $(X,\tau)$ with total order $\le$, a set $S\subseteq X$ without isolated points is called dense-in-itself.

The comparison is taken from the answer: dense in terms of order and in terms of the order topology, and is formulated as

Lemma 1. Given a totally ordered set $(X,\le)$ then $S\subseteq X$ in the induced order topology is dense-in-itself according to Definition 2 if and only if $S$ is dense according to Definition 1.

Proof. If $S$ is dense-in-itself then it has no isolated points and consists only of limit points. Given any two $x,x'\in S$ such that $x<x'$, then $(x,x')$ is not empty because (*)

On the other side, if $S$ is dense according to the total order definition, suppose that $S$ has an isolated point, that is, there is an open set containing only $x$, which contradicts Lemma ??? (the order topology consists of unions of open intervals and rays).

Is the Lemma true? How do I proceed with the proof at point (*)? Intuitively it is clear: To have a hole with $x$ and $x'$ on its ends and $x,y\in S$ then $S$ cannot be open. Should I add the condition that $S\in\tau$ to make the Lemma valid? From the answer in the cited question it seems to be the case, but I need anyway to understand how to prove it because the argument "Since X is dense in terms of topology, there is an x in (a,b)" is not enough for me to understand.

The starting point of the reasoning was to connect the definition of dense according to a total order and this third definition:

Def. 3 Given a topological space $(X,\tau)$, subset $S\subseteq X$ is said to be dense if $\bar{S}=X$. Generally given a subset $D\subseteq S$, $D$ is said to be dense in $S$ if $S\subseteq \bar{D}$.

If I take literally "dense-in-itself" and apply it to Def. 3 I would get the following lemma

Lemma 2

$S\subseteq X$, $S$ is dense-in-itself if and only if $S\subseteq\bar{S}$.

which is wrong since the r.h.s. is always true. It cannot be that all (let restrict it to) open set are dense in order topology?

How are connected Definitions 2 and 3?

Definition: two points $x < x^+$ in a linearly ordered space $(X,<)$ form a jump when $(x,x^+) =\emptyset$, i.e. there are no $z \in X$ with $x < z < x^+$.

In a linear space $(X,<)$ if $S$ is order-dense (def. 1) then $S$ is topologically dense ($\overline{S} =X$ in the order topology $\mathcal{T}_{<}$), and the reverse holds when $(X,<)$ has no jumps.

Proof: If $S$ is order dense, let $U$ be a basic open set of $(X,\mathcal{T}_{<})$; this means that $U = [a,b)$ where $a = \min(X)$ (if it exists) and $b \in X$, or $U = (a,b)$ where $a < b, a,b \in X$ or $U = (a,b]$ where $b=\max(X)$ (if it exists) and $a \in X$. $S$ being order dense implies that there is some $z \in X$ with $a < z < b$, and in all cases this shows that $U \cap S \neq \emptyset$, so that $S$ is topologically dense in $(X,\mathcal{T}_{<})$.

If $X$ has no jumps and $S$ is topologically dense in $(X, \mathcal{T}_{<})$ and $a < b, a,b \in X$, then we know that $U = (a,b)$ is non-empty and open in $\mathcal{T}_{<}$ so that $S \cap U \neq \emptyset$, which says that $S$ is order dense.

A space like $X = [0,1] \cup [2,3]$ in the inherited order is dense in itself in the order topology but $S:= X \cap \mathbb{Q}$ is dense topologically, but not order dense as no point of $S$ lies between $1$ and $2$.

The same holds for $\mathbb{Q} \times \{0,1\}$ in the Double Arrow space $[0,1]\times \{0,1\}$ (in the lexicographic order, and its induced topology), or the endpoints of the Cantor set $C$ (as the usual subset of $[0,1]$), which is countable dense (topologically) but not in the order sense, $C$ having infinitely many jumps.

• In the first sentence you means "form" instead of "for" I think. I cannot edit it. – PeptideChain Jul 17 '18 at 8:48
• @PeptideChain so edited. ta. – Henno Brandsma Jul 17 '18 at 8:49
• You have connected "my" two definitions 1 and 3. One point/argument I miss (and is also given in my first question) is a justification (intermediate step) of the implication $(a,b)\in\tau\land \overline{S}=X\to(a,b)\cap S\neq\varnothing$, corresponding to your paragraph "If $X$ has no jumps....is order dense". – PeptideChain Jul 17 '18 at 12:42
• It can/must be some triviality but I do not get it, yet. – PeptideChain Jul 17 '18 at 12:44
• @PeptideChain see this question for that fact. – Henno Brandsma Jul 17 '18 at 12:54

Without further assumptions, none of these three definitions implies either of the other two, even when considering just subsets of $\mathbb{R}$.

You are entirely correct that "dense-in-itself" when interpreted in the sense of definition $3$ is trivial: every topological space is closed in itself, so every set is dense in its subspace topology. Since there are subsets of $\mathbb{R}$ that do not satisfy definitions $1$ and $2$, $3$ does not imply $1$ or $2$ or vice-versa.

Specific examples can be used to show that neither $1$ or $2$ imply each other as well. We begin by showing that $1$ fails to imply $2$. To see this, consider the singleton set $\{0\}$ as a subspace of $\mathbb{R}$. This set has an isolated point, but is trivially dense-in-itself in the sense of definition $1$. This set also demonstrates that connectivity is not enough to guarantee that $1$ implies $2$.

To see that $2$ does not imply $1$, take $X = [0,1] \cup [2,3]$ as a subspace of $\mathbb{R}$. Here, $X$ is totally ordered by the standard order on $\mathbb{R}$ and has no isolated points, hence satisfies definition $2$. However, $1 < 2$ but there are no points between $1$ and $2$. Thus $X$ does not satisfy definition $1$. Definition $1$ can fail quite dramatically when definition $2$ is satisfied, in fact: the Cantor set offers an example of a subset of $[0,1]$ with no isolated points but with infinitely many points $x, y$ with $x < y$ and no points in between.

• The space $X=(0,1)\cup \{5\},$ as a subpace of $\Bbb R,$ is not even a linear space. No linear order on $X$ can generate the subspace topology. – DanielWainfleet Jul 17 '18 at 6:17
• Good catch. I fixed it by replacing with the singleton example. Are there better examples of this you know of? – Alex Nolte Jul 17 '18 at 10:22
• Thank you for the counterexamples. The spirit of my question is more to connect the definitions, as noted in my first sentence: exactly the further assumptions you have mentioned in your first sentence. – PeptideChain Jul 17 '18 at 12:38
• My comment about $X=(0,1)\cup \{5\}$ was not meant as a criticism of your A. It is obvious that its subspace topology is strictly stronger than its order topology, with the usual order, but not obvious that NO linear order on $X$ will induce the subspace topology. In contrast to the subspace $Y=[0,1)\cup \{5\},$ whose order topology is strictly weaker, but as a subspace. $Y$ is homeomorphic to $\{-5\}\cup [0,1),$ whose subspace & order topologies co-incide. – DanielWainfleet Jul 17 '18 at 16:06