Why is the subspace topology same as the Order topology? 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.


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.
