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.


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.
