Let $(X,<)$ be a linearly ordered set and give $X$ its order topology. Suppose that $Y \subset X$. There are 2 sensible ways to topologise $Y$:
- View $Y$ as a subspace of $X$ and give it the corresponding subspace topology $\tau_\text{ss}$.
- View $Y$ as a suborder of $X$ and give it the corresponding order topology $\tau_\text{ord}$.
In general, $\tau_\text{ss}$ is finer than $\tau_\text{ord}$, or equivalently, the identity map $(Y,\tau_\text{ss}) \to (Y,\tau_\text{ord})$ is continuous. It is sometimes the case that $\tau_\text{ss} = \tau_{ord}$. For example, if $\tau_\text{ss}$ is compact, then the above identity map is a continuous bijection from a compact space onto a Hausdorff space, hence a homeomorphism.
I would like to know what happens if we only assume $Y$ is closed in $X$. For locally compact $X$, the above would seem to show $Y$ closed implies $\tau_\text{ord} = \tau_\text{ss}$. I'd be interested to see a counterexample for general $X$ - if indeed one exists.
No. For example, let $X$ be $\mathbb{Q}$ with its standard ordering and let $Y$ be $([0,\sqrt{2}] \cap \mathbb{Q})\cup \{2\}$. Then $Y$ is closed in $X$ but the two topologies do not coincide: $2$ is isolated in the subspace topology but not in the order topology.