show that subspace topology and order topology will be same for a non-degenerate set X

Let X be a non-degenerate ordered set in the order topology. Let $$a < b \in X$$. Then the closed interval $$[a,b]$$ has a subspace topology induced by the order topology on X. But $$[a,b]$$ has the order induced by the order on X. So $$[a,b]$$ also has an order topology induced directly by the inherited order. Are these two topologies on $$[a,b]$$ the same?

This is what I am supposed to show or prove, and above this is what I did. Please note, $$S_0$$ and $$S_1$$ are the basis I declared, $$T_0$$ is the topology generated by a basis on X. $$T_1$$ and $$T_2$$ are subspace and order topologies obtained on X. Then, I tried 2 show that they are equal. Do I need to show them in this way:

$$V = U K$$ = $$(-\infty, \infty)$$, and W = intersection of $$(-\infty, \infty)$$ with $$[a,b]$$ = $$(a,b)$$ ??

Any help will be greatly appreciated.

I presented this in class, my prof told me to complete this since this is incomplete. He is saying that you need to prove the other way around showing that $$T_1$$ is a subset of $$T_2$$ and the opposite of that. Then, you can conclude that $$T_1$$ = $$T_2$$.

I can explain what I did, I took $$B_0$$ to be a basis for a topology on X, and $$B_1$$ to be a basis for the order topology so that I get the sub-basis of $$B_0$$ as $$(-\infty, \infty)$$ and sub-basis of $$B_1$$ as $$[a,b]$$. Then, I took $$T_0$$ to be the topology generated by a basis, $$T_1$$ to be the subspace topology and $$T_2$$ to be the order topology. Then, I let $$W$$ which is a subset of $$[a,b]$$ to be in $$T_1$$ or ($$W \in T_1$$), and went on to show that its in basis $$B_1$$ but I got lost after that.

Can someone help me to fix this?? Any help will be greatly appreciated.

I don't follow your proof very well, try to give it more structure and better notation?

Something more general is true (and useful to know):

Suppose $$(X,\le)$$ is a LOTS (linearly ordered space in the order topology) and $$A \subseteq X$$ is order convex, which means:

$$\forall x,y \in A \forall z \in X: (x \le z \le y) \to (z \in A)\tag{1}$$

(Note that $$[a,b]$$ is indeed a standard example of an order convex subset of $$X$$.) then the subspace topology $$\mathcal{T}_A$$ that $$A$$ gets from $$X$$ is the same as the order topology $$\mathcal{T}_{\le_A}$$ that $$A$$ gets from its inherited order $$\le_A = \le \cap (A \times A)$$.

We have two inclusions, one of which uses the order convexity:

Let $$a \in A$$, then $$(\leftarrow, a)_{\le_A}= (\leftarrow,a)\cap A$$ if we spell out the definitions, and this means that all subbasic elements of $$A$$ in the order topology are also open in $$\mathcal{T}_A$$, as intersection of order-open sets with $$A$$. Hence

$$\mathcal{T}_{\le_A} \subseteq \mathcal{T}_A$$

regardless of $$A$$, really.

To see the reverse, let $$O \subseteq A$$ be open in $$\mathcal{T}_A$$ so that $$O = O' \cap A$$ for some (order) open $$O'\subseteq X$$. We must show that $$O$$ is $$\le_A$$-open so let $$a \in O$$. In particular $$a \in O'$$ so by the definition of the standard base for the order topology there is some open interval $$(c,d)$$ containing $$a$$ such that $$(c,d) \subseteq O'$$ (or a set of the form $$[a,d)$$ if $$a = \min(X)$$ or $$(c,a]$$ if $$a=\max(X)$$, but we'll postpone those cases for now).

If now we have some $$a_L \in A \cap [c,x)$$ and some $$a_R \in A \in (x,d]$$ we see that $$(a_L,a_R)_{\le_A} \subseteq A \cap O'=O$$, so $$a$$ is an interior point of $$O$$ in $$\mathcal{T}_{\le_A}$$. This is the easiest case. If $$a_L$$ exists but no $$a_R \in (a,d] \cap A$$, this means there cannot be some $$a' \in A$$ with $$a' > a$$, or else $$(1)$$ would have forced $$d \in A$$, which is not the case, so we conclude $$(a_L, \rightarrow)_{\le_A} \subseteq A \cap O' = O$$ again allowing the conclusion that $$a$$ is an interior point of $$O$$ in $$\mathcal{T}_{\le_A}$$. The subcase of having no $$a_L$$ but some $$a_R$$ yields a similar conclusion using the $$\le_A$$-subbasic set $$(\leftarrow,a_R)_{\le_A}$$. The final case is that no $$a_L$$,$$a_R$$ at all exists, so that also no other elements of $$A$$ at all can exist (using the order convexity again) and $$A$$ is really singleton and so $$O=\{a\}$$ is trivially open in $$\mathcal{T}_{\le_A}$$.

If $$a = \max(X)$$, so $$a=\max(A)$$ too, we have $$(c,a] \subseteq O'$$ for some $$c \in X$$, and can distinguish two subcases: some $$a_L \in [c,a) \cap A$$, and we can use $$(a_L,a]$$ in $$\mathcal{T}_{\le_A}$$ or no such $$a_L$$ and thus $$a=\min(A)$$ by order convexity and $$A=\{a\}$$ again.

The case $$a=\min(X)=\min(A)$$ is similarly boring.

In all cases we have shown $$O \in \mathcal{T}_{\le_A}$$ and so

$$\mathcal{T}_A \subseteq \mathcal{T}_{\le_A}$$

finishing the proof. Such subcase distinctions are quite common in ordered spaces proofs, as the basis elements vary for minimal and maximal points.

As an example where we have a non-order convex set $$A$$ in $$X$$ where we still have this equality of topologies on $$A$$, consider $$A = \mathbb{Z}$$ in $$\Bbb R$$: both topologies yield the discrete topology on $$A$$.

To see that order convexity is important in general, we can consider in $$\Bbb R$$ the subspace $$A=[0,1) \cup \{2\}$$, where $$\{2\}$$ is open in the subspace topology on $$A$$ but not in the order topology on $$A$$ (see the special role of a max again). Or take $$A=[0,1]\times [0,1]$$ as a subset/subspace of $$\mathbb{R}^2$$ in the lexicographic order topology (an example that's treated in Munkres as well) where $$A$$ is disconnected and non-compact in the subspace topology it inherits but is compact and connected in the lexicographic order topology.