Why is the subspace topology same as the Order topology?

Question

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.

Answer 1

Well, in general to check that two topologies are the same, you need to check that an open set in one topology is also open in the other, and viceversa.

Recall that in an ordered set, you define the order topology by declaring your open sets to have a basis of open rays, and open intervals. On the other hand, the subspace topology on the set $[a,b]$ consists precisely (has a basis) of open intervals. Thus, in this case both topologies agree.

Answer 2

I haven't done topology in a while, but verifying this seems straight-forward by comparing basis elements.

Let $O$ be a basis element of the order topology. Then,

• if $O = [a,x), O = (-\infty, x) \cap [a,b]$
• if $O = (x,y), O = (x,y) \cap [a,b]$
• if $O = (x,b], O = (x,\infty) \cap [a,b]$

(Here $a < x, y < b$.) So $O$ is a basis element in the subspace topology in every case. Conversely, let $U$ be a basis element of the subspace topology. Then $U = R \cap [a,b]$, where $R = (x,y)$ is a basis element of the topology of $X$. (Here $x$ may be $-\infty$ and $y$ may be $\infty$.)

• if $x < a$ and $y \notin [a,b]$, then $U = [a,b]$ or $U = \emptyset$
• if $x < a$ and $y \in [a,b]$, then $U = [a,y)$
• if $x \in [a,b]$ and $y > b$, then $U = (x,b]$
• if $x \in [a,b]$ and $y \le b$, then $U = (x,y)$

again, $U$ is a basis element of the order topology in all cases.

Answer 3

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.