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Here is Prob. 3, Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

Show that every order topology is regular.

First of all, here are some relevant definitions.

Ordered Set:

Let $X$ be a non-empty set, and let $<$ be a relation on $X$ such that

(i) for every element $x \in X$, it is not true that $x < x$, or in other words, $x \not< x$,

(ii) for any two distinct elements $x$ and $y$ of $X$, either $x < y$ or $y < x$, but not both, and

(iii) for any elements $x$, $y$, and $z$ of $X$ such that $x < y$ and $y < z$, we also have $x < z$.

Then $<$ is said to be an order relation on $X$ (more precisely a simple or total order relation), and the set $X$ with an order relation $<$ is said to be an ordered set (more precisely a simply or totally ordered set).

Here is the definition of an immediate predecessor and an immediate successor of an element in a simply ordered set:

Let $x$ is an element of a simply ordered set $X$ with order relation $<$. If there exists an element $w \in X$ such that the open interval $(w, x)$ turns out to be the empty set, then $w$ is said to be an immediate predecessor of $x$ in $X$. And, if there exists an element $y \in X$ such that the open interval $(x, y)$ turns out to be the empty set, then $y$ is said to be an immediate successor of $x$ in $X$. [It can also be shown that $x$ can have at most one immediate predecessor and at most one immediate successor in $X$, althugh we won't have much use for either of these facts.]

Please refer to Sec. 3 in Munkres.

Here is the definition of order topology.

Let $X$ be a simply ordered set having more than one element and with an order relation $<$. Then the order topology on $X$ is the one having as a basis the collection consisting of the following three types of sets:

(i) all the open intervals $(a, b) \colon= \{ \ x \in X \, \colon \, a < x < b \ \}$ for any elements $a$ and $b$ of $X$ such that $a < b$,

(ii) if $X$ has a smallest element $a_0$, then all the closed-open intervals $\left[a_0, b\right) \colon= \left\{ \ x \in X \, \colon \, a_0 \leqq x < b \ \right\}$ for any element $b \in X$ such that $a_0 < b$, and

(iii) if $X$ has a largest element $b_0$, then all the open-closed intervals $\left(a, b_0 \right] \colon= \left\{ \ x \in X \, \colon \, a < x \leqq b_0 \ \right\}$ for any element $a \in X$ such that $a < b_0$.

If $X$ has no smallest element, then there will be no sets of type (ii) in the basis; if $X$ has no largest element, then there will be no sets of type (iii).

Please refer to Sec. 14 in Munkres.

Finally, here is the definition of regular space.

A topological space $X$ is said to be regular if

(i) singleton subsets (and hence finite subsets) of $X$ are closed in $X$, and

(ii) for every element $x \in X$ and every closed set $B$ of $X$ such that $x \not\in B$, there exist disjoint open sets $U$ and $V$ in $X$ such that $x \in U$ and $B \subset V$.

Please refer to Sec. 31 in Munkres.

In what follows, we will be repeatedly using the following result:

Let $X$ be a non-empty set with an order relation $<$, and let $u$ and $v$ be any two elements of $X$ such that $u < v$. Then the closures of the four intervals determined by $u$ and $v$, namely the closures $\overline{(u, v)}$, $\overline{[u, v)}$, $\overline{(u, v]}$, and $\overline{[u, v]}$, are all contained in the closed interval $[u, v]$.

Please refer to Prob. 5, Sec. 17, in Munkres. Here is my Math SE post on this problem.

My Attempt:

Let $X$ be a simply ordered set with order relation $<$, and let $X$ have the order topology.

First we show that any singleton set of $X$ is closed. Let $a \in X$. Then we note that the set $X \setminus \{ a \}$ equals $$ \begin{cases} \left( \bigcup_{x \in X \mbox{ and } x < a} (x, a) \right) \cup \left( \bigcup_{y \in X \mbox{ and } a < y} (a, y) \right) \ & \mbox{ if $a$ is neither a smallest nor a largest element of $X$}, \\ \bigcup_{y \in X \mbox{ and } a < y} (a, y) \ & \mbox{ if $a$ is a smallest element of $X$}, \\ \bigcup_{x \in X \mbox{ and } x < a} (x, a) \ & \mbox{ if $a$ is a largest element of $X$}. \end{cases} $$ Thus $X \setminus \{ a \}$, being a union of basic open sets, is open in each one of these three cases.

Alternatively, we note that $X$ in the order topology is a Hausdorff space (by Theorem 17.11 in Munkres), and we note further that one-point sets in the Hausdorff space $X$ are closed (by Theorem 17.8 in Munkres).

Let $x$ be a point of $X$ and $U$ be an open set of $X$ such that $x \in U$. We need to find an open set $V$ in $X$ such that $ x \in V$ and $\overline{V} \subset U$, by virtue of Lemma 31.1 (a) in Munkres.

Now the following three cases arise according to whether $x$ is (1) neither a smallest element nor a largest element of $X$, (2) a smallest element of $X$, or (3) a largest element of $X$.

Case 1: Suppose that $x$ is neither a smallest nor a largest element of $X$. Since $x \in U$ and since $U$ is open in $X$, there exists an open interval $(a, b)$ of $X$ such that $$ x \in (a, b) \subset U. \tag{0}$$

[ PS: Some correction made after reading one of the first one of the comments below. I had initially considered only three sub-cases under Case 1, whereas there are four sub-cases that need to be consiedred.]

Now the following four sub-cases arise according to whether $x$ has (i) neither an immediate predecessor nor an immediate successor, (ii) an immediate predecessor but no immediate successor, (iii) an immediate successor but no immediate predecessor, or (iv) both an immediate predecessor and an immediate successor in $X$.

Case 1 (i): Suppose $x$ has neither an immediate predecessor nor an immediate successor in $X$. Then since $a < x < b$, there are elements $c$ and $d$ in $X$ such that $$ a < c < x < d < b. \tag{1}$$ Then let us put $$V \colon= ( c, d ).$$ Then $V$ is an open set in $X$ such that $x \in V$. Furthermore we also note that $$ \overline{V} \subset [ c, d], \tag{2}$$ and by (1) we also have $$ [c, d] \subset (a, b). \tag{3} $$ Thus from (0), (1), (2), and (3) we can conclude that $ x \in V$ and $\overline{V} \subset U$.

Case 1 (ii): Suppose that $x$ has an immediate predecessor $x-1$ but no immediate successor in $X$. Then since by (0) we have $a < x < b$, therefore we must also have $$ a \leqq x-1 < x < b. \tag{1}$$ And, as $x$ has no immediate successor in $X$ and as $x < b$, so there exists an element $d \in X$ such that $x < d < b$, and therefore from (1) also we have $$ a \leqq x-1 < x < d < b. \tag{2} $$ Now as $x-1$ is an immediate predecessor in $X$ of $x$, so we have the equality $$ (x-1, d) = [x, d). \tag{3} $$

Let us now put $V \colon= (x-1, d) = [x, d)$. Thus from (2) and (3) we have $x \in V$ and also $$ \overline{ V} \subset [x, d] \subset (a, b), $$ which together with (0) aobve implies that $\overline{V} \subset U$.

Case 1 (iii): Suppose that $x$ has an immediate successor $x+1$ but no immediate predecessor in $X$. Then since by (0) above we have $a < x < b$, we must also have $$ a < x < x+1 \leqq b. \tag{1} $$ And as $x$ has no immediate predecessor in $X$ and as $a< x$, so there exists an element $c \in X$ such that $a < c < x$, and then from (1) we also have $$ a < c < x < x+1 \leqq b. \tag{2} $$ As $x+1$ is the immediate successor in $X$ of $x$, so we have the equality $$ (c, x+1) = (c, x]. \tag{3} $$

Now let us put $V \colon= (c, x+1) = (c, x]$. and therefore from (2) and (3) above we get $x \in V$ and also $$ \overline{V} \subset [c, x] \subset (a, b), $$ which together with (0) above again implies that $\overline{V} \subset U$.

[I am inserting Case 1 (iv) only after reading the first comment below.]

Case 1 (iv): Suppose that $x$ has an immediate predecessor $x-1$ and an immediate successor $x+1$ in $X$. Then by (0) above as $a < x < b$, so we must also have $$ a \leqq x-1 < x < x+1 \leqq b. \tag{1}$$ Then $$(x-1, x+1) = \{x\}. \tag{2}$$ So if we put $V \colon= (x-1, x+1) = \{ x \}$, then $V$ is open in $X$ and being singleton $V$ is also closed in $X$, and therefore from (1) we have $$ \overline{V} = \{ x \} \subset (a, b), $$ which together with (0) above implies that $\overline{V} \subset U$.

Case 2: Suppose that $x$ is a smallest element of $X$. Then as $x \in U$ and as $U$ is an open set of $X$, so there exists an element $b \in X$ such that $x < b$ and also $$ [x, b) \subset U. \tag{0}$$

Now the following two sub-cases arise according to whether $x$ has (i) no immediate successor in $X$ or (ii) an immediate successor in $X$.

Case 2 (i): Suppose that $x$ has no immediate successor in $X$. Then as $x < b$, so there exists an element $d$ in $X$ such that $$ x < d < b. \tag{1} $$

Let us put $V \colon= [x, d)$. Then $V$ is an open set in $X$ such that $x \in V$ and also $$ \overline{V} \subset [x, d] \subset [x, b), $$ which together with (0) above implies that $\overline{V} \subset U$.

Case 2 (ii): Suppose that $x$ has an immediate successor $x+1$ in $X$. Then since $x < b$, we also have $$ x < x+1 \leqq b, \tag{1}$$ and then we have the equality $$ [x, x+1) = \{ x \}. \tag{2}$$ Let us put $V \colon= [x, x+1) = \{ x \}$. Then $V$ is an open set of $X$ such that $x \in V$ and from (1) we also have $$ \overline{V} = \{ x \} \subset [x, b), $$ which together with (0) above also implies that $\overline{V} \subset U$. Note that here we have used the fact that since one-point sets in $X$ are closed and since by virtue of (2) above $V$ is a one-point set in $X$, therefore we have $\overline{V} = V$ itself. Please refer to Sec. 17 in Munkres.

Case 3: Suppose that $x$ is a largest element of $X$. Then as $x \in U$ and as $U$ is an open set of $X$, so there exists an element $a \in X$ such that $a < x$ and also $$ (a, x ] \subset U. \tag{0}$$

Now the following two sub-cases arise according to whether $x$ has (i) no immediate predecessor in $X$ or (ii) an immediate predecessor in $X$.

Case 3 (i): Suppose that $x$ has no immediate predecessor in $X$. Then as $a < x$, so there exists an element $c$ in $X$ such that $$ a < c < x. \tag{1} $$

Let us put $V \colon= (c, x]$. Then $V$ is an open set in $X$ such that $x \in V$ and also $$ \overline{V} \subset [c, x] \subset (a, x ], $$ which together with (0) above implies that $\overline{V} \subset U$.

Case 3 (ii): Suppose that $x$ has an immediate predecessor $x-1$ in $X$. Then since $a < x$, we also have $$a \leqq x-1 < x, \tag{1}$$ and then we have the equality $$ (x-1, x ] = \{ x \}. \tag{2}$$ Let us put $V \colon= (x-1, x ] = \{ x \}$. Then $V$ is an open set of $X$ such that $x \in V$ and from (1) we also have $$ \overline{V} = \{ x \} \subset (a, x ], $$ which together with (0) above also implies that $\overline{V} \subset U$. Note that here we have also used the fact that since one-point sets in $X$ are closed, by virtue of (2) above we have $\overline{V} = V$ itself.

Is this rather lengthy proof of mine correct over all? If so, then is also correct and accurate in each and every individual detail? Is my presentation clear and elementary enough? If not, then where is it in the above write-up that there are issues or errors?

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    $\begingroup$ it is correct overall, and it is correct in the details, as far as I read them, but it is definitely very lengthy. And, strictly speaking it is incomplete, for example if $X=\{0,1,2\}$ and $x=1$, I don't think the cases considered above apply to this $x$. $\endgroup$ – Mirko Jun 16 at 21:47
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    $\begingroup$ Saaqib unless it makes it easier for you, I don't think you need to go to the trouble of repeating definitions which are elementary in subjects like point-set topology. Most readers who will likely answer your questions will know what the order topology or a regular space is. Mentioning this both because I think it will make your questions easier to read, and will save you time. $\endgroup$ – Mariah Jun 16 at 23:42
  • $\begingroup$ @Mirko thank you for the correction you suggested. I've just edit my post to remove the error. $\endgroup$ – Saaqib Mahmood Jun 17 at 3:13
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Assume x in open U.
Thus exists a,b with x in (a,b) subset U.

If exists u,v with a < u < x < v < b, then
x in open (u,v) and $\overline{(u,v)} \subseteq [u,v] \subseteq U.$

If (a,x) is empty and exists v with x < v < b, then
x in open (a,v) = [x,v) and $\overline{[x,v)} \subseteq [x,v] \subseteq U.$

If (x,b) is empty and exists u with a < u < x a simular proof ensues.

If (a,x) and (x,b) are empty, then
(a,b) = {x} is open and as linear ordered
topologies are $T_1,$ {x} is closed subset of (a,b)

Since, in every case there is an open set V with x in V and $\bar V$ subset U, X is regular by the dual notion of regularity.

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  • $\begingroup$ How do you know that "Thus exists a,b with x in (a,b) subset U."? This is not true in general, for example when $x$ is an endpoint ($\min$ or $\max$ of $X$). You proof would be correct if you allow $a=-\infty$ and $b=\infty$, but this would be a different approach than what Munkres does, as described by the OP. In short, I think your answer is formally invalid. It could be debated to what extent it might be accepted as informally correct, but certainly without some proper explanation as to what you mean, I personally would not have accepted it. Things like this make OP consider cases. $\endgroup$ – Mirko Jun 17 at 3:35
  • $\begingroup$ (<-,b) or (a,->) as needed. $\endgroup$ – William Elliot Jun 17 at 4:08
  • $\begingroup$ I know, but this is not how the OP says Munkres does it, and if you do it this way you need to (a) explain what is <- and ->, and (b) prove that the order topology defined with the help of <- and -> is the same as the order topology defined by Munkres without the use of <- and ->. I am not saying that the proof would be difficult, but there are some details, and they are missing from your answer, so to me your answer answers a different question. $\endgroup$ – Mirko Jun 17 at 4:15
  • $\begingroup$ @Mirko. They are rays. $\endgroup$ – William Elliot Jun 17 at 8:21

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