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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 $Y$:

  1. View $Y$ as a subspace of $X$ and give it the corresponding subspace topology $\tau_\text{ss}$.
  2. 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.

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up vote 7 down vote accepted

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.

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That works, thanks! It's funny I swear I tried more or less the same counterexample about an hour ago and decided it didn't work... – Mike F Nov 19 '11 at 1:59
It might be interesting to note that your space $Y$ is homeomorphic to $\{-1\} \cup ([0,\sqrt 2] \cap \mathbb{Q})$ which is an order topology. I wonder whether we could find a closed subspace $Y$ whose topology is not induced by any linear order on $Y$? Something like $Y = ([\sqrt 2, \sqrt 3] \cap \mathbb{Q}) \cup \{5\}$ maybe? – Mike F Nov 19 '11 at 2:07

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