Closure of an Interval in the Order Topology

Question

Here is the problem I am working on:

Let $X$ be an ordered set in the order topology. Show that $\overline{(a,b)} \subseteq [a,b]$. Under what conditions does equality hold?

which is, in my opinion, unsatisfactorily dealt with here. I found most of the answers unhelpful; the only one coming close to helping me is Henno Brandsma's answer. I was able to answer the first part easily enough, but I am having trouble with the second part:

It seems that we have $[a,b] \subseteq \overline{(a,b)}$ if and only if $a,b \in \overline{(a,b)}$, because $[a,b] = (a,b) \cup \{a,b\}$ (i.e., the two sets only differ by the two elements $a$ and $b$). I am not sure what else I could conclude; perhaps that $a \in \overline{(a,b)}$ if and only if $a \in C$ for every closed set $C \supseteq (a,b)$, a similar thing being true of $b$. As I have been solutions to this problem, and evidently there is more to conclude, but I don't see what else.

In Henno Brandsma's answer, he speaks of left and right neighbors. Well, I am unfamiliar with such terminology. Interpreting "right neighbor," it would seem that the claim "$a \in \overline{(a,b)}$ iff $a$ has no right neighbor" is wrong; it seems that it should be "$a$ has no left neighbor in $\overline{(a,b)}$," but perhaps neighbors have a unusual definition in math. I do have the notion of predecessors and successors: If $(a,b) = \emptyset$, then $a$ is the immediate predecessor of $b$, and $b$ the immediate successor of $a$.

EDIT: Okay, I am still having some difficulty. I can't seem to prove the claim "$a \in \overline{(a,b)}$ iff $a$ has no immediate successor." Furthermore, I am having difficulty proving that "$a$ is in the closure of $(a,b)$ iff every open set $O$ containing $a$ intersects $(a,b)$."

Regarding the proof of the main claim, Henno writes that if $a$ is the minimum of $X$, then there is some $y > a$ such that $[a,y) \subseteq O$. I buy that if $a$ is the minimum, then such a $y$ exists (but shouldn't we have $a \ge y$, in case $X$ only contains $a$?); but why is $[a,y)$ contained in $O$? I am also having trouble with the second case.

I think I understand the proof of "If $a$ does have a right neighbor, then $a \notin \overline{(a,b)}$," but I just want make sure I understand why $P = \{x \in X ~|~ x < a^+ \}$. If $X$ has a smallest element $s$, then $P = [s,a^+)$ is just a basis element of the order topology and therefore open; if $X$ has no smallest element, then $P=(- \infty, a^+)$, which is a union of basis elements and therefore open. Is this right?

So, would someone help me prove these two other claims?

Answer 1

Henno’s answer is indeed the only good one, and it is correct. In particular, he is correct when he says that $a\in\operatorname{cl}(a,b)$ iff $a$ has no right neighbor. You need not have any prior familiarity with the term right neighbor: he defines it in his answer, though he does expect that the reader will realize that the right neighbor of $a$ must be to the right of $a$ in the order; I’ve edited his answer to make that explicit. As he defines it, the right neighbor of $a$ is the immediate successor of $a$ (if there is one), and the left neighbor of $a$ is the immediate predecessor of $a$ (again if there is one).

Added in response to OP’s edit: If $X$ is any space and $A\subseteq X$, then a point $x\in X$ is in the closure of $A$ if and only if $U\cap A\ne\varnothing$ for each open nbhd $U$ of $x$. This is one of the two most common (equivalent) definitions of the closure of a set. In particular, if $\langle X,\le\rangle$ is a linearly ordered set with the order topology, $a,b\in X$, and $a<b$, then $a\in\operatorname{cl}(a,b)$ if and only if $U\cap(a,b)\ne\varnothing$ for each open nbhd $U$ of $a$. There is literally nothing to prove here unless you’re familiar only with some other definition of the closure of a set, and in that case you should try to prove that the definition that you know is equivalent to this one.

Regarding the proof of the main claim, Henno writes that if $a$ is the minimum of $X$, then there is some $y > a$ such that $[a,y) \subseteq O$. I buy that if $a$ is the minimum, then such a $y$ exists (but shouldn't we have $a \ge y$, in case $X$ only contains $a$?); but why is $[a,y)$ contained in $O$?

We know that $X\ne\{a\}$: $a,b\in X$, and $a<b$, so $X$ contains at least the two distinct elements $a$ and $b$

The order topology on $X$ has a base consisting of all sets of the following three forms:

• $(x,y)$ for $x,y\in X$ with $x<y$;
• $(\leftarrow,y)$ for $y\in X$; and
• $(x,\to)$ for $x\in X$.

(You are probably more familiar with the notations $(-\infty,y)$ and $(x,\infty)$ for $(\leftarrow,y)$ and $(x,\to)$, respectively.) If $a=\min X$, the only basic open nbhds of $a$ are those of the second type, $(\leftarrow,y)$ for some $y\in X$, since there is no $x\in X$ such that $x<a$. $O$ is an open nbhd of $a$, so it must contain some basic open nbhd of $a$, and there must therefore be a $y\in X$ such that $a\in(\leftarrow,y)\subseteq O$. But if $a=\min X$, then $(\leftarrow,y)=[a,y)$, so there is a $y\in X$ such that $a\in[a,y)\subseteq O$.

I think I understand the proof of "If $a$ does have a right neighbor, then $a \notin \overline{(a,b)}$," but I just want make sure I understand why $P = \{x \in X ~|~ x < a^+ \}$. If $X$ has a smallest element $s$, then $P = [s,a^+)$ is just a basis element of the order topology and therefore open; if $X$ has no smallest element, then $P=(- \infty, a^+)$, which is a union of basis elements and therefore open. Is this right?

It’s correct, but in fact $P$ is a basic open set in both cases, assuming that one uses the most natural base for the order topology. This is the one generated by the subbase consisting of the open rays $(\leftarrow,x)$ and $(x,\to)$ for $x\in X$ and contains precisely the sets of the three types that I mentioned above.