TITLE: Why is the subspace topology same as the Order topology?
QUESTION [1 upvotes]: Hi i am reading Topology by Munkres and there in the proof of theorem 27.1 it is written that

Given $a<b$, let $\mathcal A$ be a covering of $[a,b]$ by sets open in $[a,b]$ in the subspace topology(which is the same as the order topology).

I can't understand why is the subspace topology same as the order topology in this case? Is this always true? Btw i know how subspace topology and order topologies are defined. If $(X,\tau)$ is a topological space and $A$ be any subset of X then we define $\tau_A$ to be the set with subsets as : $\tau_A=\{U\cap A | U\in \tau\}$. And order topology is defined in a different way.So how can they be same here? For reference i am attaching the screenshot where i have highlighted the part where this is mentioned.

REPLY [1 votes]: In a L.O.T.S. (Linearly Ordered Topological Space) $(X,<),$ if $Y\subset X$ then the subspace topology $T_S$ on $Y$ may be strictly stronger than the $<$-order topology $T_<$ on Y. We always have $T_<\subseteq T_S.$
E.g. if $X=\Bbb R$ with the usual order and $Y=\{0\}\cup (1,2).$ For $Y$ is order-isomorphic to $[1,2)$ and the order-topology on $[1,2)$ has no isolated points, but in the subspace topology, the set $\{0\}=Y\cap (-1,1)$ is open.
A necessary & sufficient condition for $T_S\ne T_<$ is that there exists $y\in Y$ and $x\in X\setminus Y$ such that
$(i).\, x<y$ and the set $B=\{y'\in Y: y'<y\}$ is not empty and has no largest member, and $\forall y'\in B\,(y'<x),$... OR
$(ii).\, y<x$ and the set $C=\{y'\in Y:y'>y\}$ is not empty and has no smallest member, and $\forall y'\in C\,(y'>x).$
Some sufficient conditions for $T_S=T_<$ are
$(i').\, Y$ is order-dense in $X,$ that is, if $x_1,x_2\in X$ with $x_1<x_2$ then $Y\cap (x_1,x_2)\ne \emptyset,$... OR
$(ii').\, Y$ is convex in $X,$ that is, if $y_1,y_2\in Y$ with $y_1<y_2$ and $y_1<x<y_2$ then $x\in Y.$
Observe that $(ii')$ holds if $Y=[a,b]=\{x\in X:a\le x\le b\}.$
In some cases where $T_<\ne T_S,$ the subspace topology may be generated by a different linear order $<'$ on $Y.$ For example, if $X=\Bbb R$ and $Y= [0,1)\cup [2,3)$ then $Y$ with the subspace topology is homeomorphic to  the ordered space $(0,1]\cup [2,3).$ But being a L.O.T.S. is not a hereditary property in general. No linear order on the real subspace $\{0\}\cup (1,2)$ can generate its subspace topology, but I would not call this obvious.
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