Verification of basic properties of order topology

Question

I have considered the order topology on general ordered sets and came to the following two 'conclusions/conjectures', which I wish to verify:

Let $X$ be an ordered set under a relation $>$, and let $\tau_X$ be it's appropriate order topology.

(1) Then if $>$ is a well order and $X$ uncountable, then $\tau_X$ is not discrete.

(2) If $>$ is a well order, since a well order is isomorphic to some ordinal as an ordered space, then for all $x\in X$ which is not a limit ordinal, we have that $\{ x \}\in \tau_X$.

(3) If $>$ is a dense order, then the singletons are not open.

The question comes from the fact that for a while I thought to myself that the order topology induced by a well order, but I suspect (1) spoils such an answer. The reason I initially thought that it does was (2). (2) seems relatively simple since if $x$ is the successor of $y$ and has a successor $z$, then:

$$ \{x \}= \{ a\in X: a<z \} \cap \{ a\in X: a>y \} \quad \text{and is then open} $$

On the other hand I think that sets of the form:

$$ \tag{*} \underset{i\in [k]}{\bigcap} \{ a\in X: a<z_i \} \cap \underset{j\in [\ell]}{\bigcap} \{ a\in X: a>y_j \} $$

Form a basis of the order topology, and hence (1) and (3) follows.

I would appreciate any corrections, if need be.

Answer 1

As you note, you're basically talking about ordinals $\alpha$ in their order topology.

In these spaces all limit ordinal elements (an ordinal is the set of smaller ordinals) is non-isolated and every non-limit is isolated (such an $\alpha$ is of the form $\beta+1$ for a unique $\beta$, and then $\{\alpha\} = (\beta, \alpha+1)$, which is open in the order topology; this is indeed an intersection of two subbasic elements, as you describe)

If $\alpha$ is uncountable, it contains $\omega$ as an element, and this is not as isolated point (being a limit ordinal). But the only countable infinite ordinal that is discrete is $\omega \simeq \Bbb N$, so in fact all except one infinite ordinals are non-discrete.

In a dense order there indeed are no isolated points: suppose $\{x\}$ were open, then either $x$ is the minimum of $X$ and there is some $x' > x$ such that $\{x\}=\{z \in X: z < x'\}$, but then by being a dense order we have $x''$ with $x < x''< x'$, but this contradicts $\{x\}=\{z \in X: z < x'\}$, as $x''$ is in the right hand side, but not in the left hand side. Similarly when $x$ is the maximum of $X$, and if it's neither, for any $(x_1,x_2)$ that contains $x$ we find elements in between $x_1$ and $x$, showing that the interval never equals $\{x\}$ and the open intervals and half-open segments on the min or max form a base for the order topology.

Answer 2

This is not an answer to the OP, but to the question stated in comments in another answer:

Question. Does all this imply that the discrete topology on uncountable set, cannot be an order topology?

The answer (in the comments, by Henno Brandsma) suggests to use upward Löwenheim-Skolem theorem, to conclude that on every uncountable set one could define an order relation so that the resulting order topology is discrete. Here is a different (I would also say standard) construction to make any uncountable set a discrete ordered set (in ZFC, going through a well-order first).

Let $X$ be any uncountable set, and say $|X|=\kappa$ where $\kappa=\{\alpha:\alpha<\kappa\}$ is an uncountable cardinal (= smallest ordinal with that cardinality).

The main step: Insert a countable increasing sequence in front of each limit ordinal.

That is, say $\gamma<\kappa$ and $\gamma$ is limit. Insert a copy of $\omega=\{0,1,2...\}$ (the first infinite ordinal) in front of $\gamma$. To make it more formal, let the copy be $S(\gamma)=\{(n,\gamma):n\in\omega\}$ (with the order from the first component). If $\beta<\gamma$ define $\beta<(n,\gamma)$ for all $n\in\omega$. Respect the order that was already there, so for example if $\delta$ is another limit ordinal, with $\delta<\gamma$, and if $\delta<\beta<\gamma$ then for all $m,n$ we have $(m,\delta)<\delta<\beta<(n,\gamma)<\gamma$.

Let $Y$ be $\kappa$ together with all increasing sequences inserted, one in front of each limit ordinal. Then the order topology on $Y$ is discrete. On the other hand $X$ and $Y$ have the same cardinality, which shows that for every uncountable set there is an order relation of it that generates the discrete topology.

A variation of this approach would be as follows. Assume that $(X,<)$ is an uncountable linear order (well-order, or not). Let $\Bbb Z$ be the set of all zahlen, integers, with its usual order. Take the product $X\times\Bbb Z$ with the lexicographic (dictionary) order. Then every element $(x,n)$ of $X\times\Bbb Z$ has an immediate predecessor $(x,n-1)$ and an immediate successor $(x,n+1)$, hence the order topology on $X\times\Bbb Z$ is discrete. Note that $X\times\Bbb Z$ has the same infinite (and uncountable) cardinality as $X$, so one could introduce a linear order that generates the discrete topology on any uncountable set. (Take a bijection between $X$ and $X\times\Bbb Z$ if necessary and use the order on $X\times\Bbb Z$ to redefine the order on $X$, though it would be easier to work with $X\times\Bbb Z$ directly.)

There is more (standard constructions along this way) in:
Lutzer, D. J.
On generalized ordered spaces.
Dissertationes Math. Rozprawy Mat. 89 (1971), 32 pp.

And just one more link which I find related and interesting.
Purisch, S.
The orderability and suborderability of metrizable spaces.
Trans. Amer. Math. Soc. 226 (1977), 59–76.