Show that every order topology is regular.
My attempt:
Approach(1): Let $X$ be an ordered set equipped with $\mathcal{T}_o$ order topology. Let $\{ x\}$ be a singleton set in $X$. Then $(x,+\infty)\cup (-\infty, x)=X\setminus \{x\} \in \mathcal{T}_X$. So $\{ x\}$ is closed in $X$. Hence $X$ is $T_1$. Let $x\in X$ and $U\in \mathcal{N}_x$. Since $x\in U\in \mathcal{T}_o$, we have $\exists B\in \mathcal{B}_o$ such that $x\in B\subseteq U$. By definition of basis, $B\in \mathcal{B}_0 = \{ (a,b)| a\lt b \text{ , } a,b\in X\}$$ \cup \{ [a_0,b)| a_0\lt b \text{ , $a_0$ is smallest element(if $\exists$) of $X$} \} $$\cup \{ (a,b_0]| a\lt b_0 \text{ , $b_0$ is largest element(if $\exists$) of $X$}\}$. Assume $B=(a,b)$. Case(1): $\exists p,q \in (a,b)$ such that $p\lt x\lt q$. Let $V=(p,q)$. So $V\in \mathcal{N}_x$. By exercise 5 section 17, $\overline{V}=\overline{(p,q)} \subseteq [p,q]\subseteq (a,b)\subseteq U$. Case(2): $\nexists p,q \in (a,b)$ such that $p\lt x\lt q$, i.e. $(a,b)\setminus \{x\} =\emptyset$. So $(a,b)=\{x\} \in \mathcal{T}_o$. Let $V=\{x\}$. Then $V\in \mathcal{N}_x$. Since $X$ is $T_1$, $\overline{V}=\overline{\{x\}}=\{x\}=(a,b)\subseteq U$. Case(3): $\exists p\in (a,b)$ such that $p\lt x$ and $\nexists q\in (a,b)$ such that $x\lt q$. Note $(p,b)=(p,x]$. Let $V=(p,b)$. so $V\in \mathcal{N}_x$. By exercise 5 section 17, $\overline{V}=\overline{(p,b)}= \overline{(p,x]}\subseteq [p,x]\subseteq (a,b)\subseteq U$. Case(4): $\nexists p\in (a,b)$ such that $p\lt x$ and $\exists q\in (a,b)$ such that $x\lt q$. Let $V=(a,q)=[x,q)$. So $V\in \mathcal{N}_x$. By exercise 5 section 17, $\overline{V}=\overline{(a,q)}=\overline{[x,q)} \subseteq [x,q]\subseteq (a,b)\subseteq U$. Thus, in all cases $\exists V\in \mathcal{N}_x$ such that $\overline{V}\subseteq U$. By lemma 31.1, $X$ is $T_3$. Proof is similar if $B\in \{ [a_0,b)| a_0\lt b \text{ , $a_0$ is smallest element(if $\exists$) of $X$} \}$ or $B\in \{ (a,b_0]| a\lt b_0 \text{ , $b_0$ is largest element(if $\exists$) of $X$ } \}$. Is this proof correct?
Approach(2): Let $\{x\}$ be a singleton set. Since $\{x\} =(-\infty, x] \cap [x, +\infty)$, we have $\{x\}$ is closed in $X$. Hence $X$ is $T_1$. Let $x\in X$ and $B$ is closed in $X$ such that $x\notin B$. So $x\in X-B \in \mathcal{T}_o$. $\exists B’\in \mathcal{B}_o$ such that $x\in B’\subseteq X-B$. By definition of basis, $B’\in \mathcal{B}_0 = \{ (a,b)| a\lt b \text{ , } a,b\in X\}$$\cup \{ [a_0,b)| a_0\lt b \text{ , $a_0$ is smallest element(if $\exists$) of $X$} \}$$\cup \{ (a,b_0]| a\lt b_0 \text{ , $b_0$ is largest element(if $\exists$) of $X$ } \}$. Assume $B’=(a,b)$. Since $(a,b) \subseteq X-B$, we have $B \subseteq X-(a,b)=(-\infty ,a]\cup [b,+\infty)$. Case(1): $\exists p,q \in (a,b)$ such that $p\lt x\lt q$. let $U=(p,q)$ and $V=(-\infty, p)\cup (q,+\infty)$. So $U,V\in \mathcal{T}_o$ with $x\in U$, $B\subseteq (-\infty,a] \cup [b,+\infty)\subseteq V$ and $U\cap V=\emptyset$. Case(2): $\nexists p,q \in (a,b)$ such that $p\lt x\lt q$. So $(a,b)=\{x\} \in \mathcal{T}_o$. Let $U=\{x\}$ and $V=(-\infty, x)\cup (x,+\infty)=X-\{x\}$. It’s easy to check $U\in \mathcal{N}_x$, $B\subseteq V\in \mathcal{T}_o$ and $U\cap V=\emptyset$. Case(3): $\exists p\in (a,b)$ such that $p\lt x$ and $\nexists q\in (a,b)$ such that $x\lt q$. Then $B \subseteq (-\infty, p)\cup (x,+\infty)\in \mathcal{T}_o$ and $(p,b)\in \mathcal{N}_x$. Let $U= (p,b)=(p,x]$ and $V= (-\infty, p)\cup (x,+\infty)$. So $U\cap V=\emptyset$. Case(4): $\nexists p\in (a,b)$ such that $p\lt x$ and $\exists q\in (a,b)$ such that $x\lt q$. Then $B \subseteq (-\infty, x)\cup (q,+\infty)\in \mathcal{T}_o$ and $(a,q)\in \mathcal{N}_x$. Let $U= (a,q)=[x,q)$ and $V= (-\infty, x)\cup (q,+\infty)$. So $U\cap V=\emptyset$. Thus, in all cases $\exists U,V \in \mathcal{T}_o$ such that $x\in U$, $B\subseteq V$ and $U\cap V=\emptyset$. Hence $X$ is $T_3$. Proof is similar if $B\in \{ [a_0,b)| a_0\lt b \text{ , $a_0$ is smallest element(if $\exists$) of $X$} \}$ or $B\in \{ (a,b_0]| a\lt b_0 \text{ , $b_0$ is largest element(if $\exists$) of $X$ } \}$. Is this proof correct?
