Difference between Subspace topology and Ordered topology... Recently I am started reading Topology by James R.Munkers Second edition.Right now I study The Subspace topology point and Can't understand the differences between subspace topology and Ordered topology especially the example 2 and example 3.
I think with the basis element of topology we can make difference,I am right or wrong ?


The statement Munkres wants to prove is that the subspace topology is not equal to the order topology. This can be done by showing there exists a set open in the subspace topology that is not open in the order topology, or by showing there exists a set open in the order topology that is not open in the subspace topology.
Munkres shows the first of these two.
He first shows--explicitly using the definition of "subspace topology"--that $\{2\}$ is open in the subspace topology.
He then shows that $\{2\}$ is not open in the order topology. This part is slightly more round-about. He intends to do this by showing that any open set in the order topology that contains $2$ must also contain other points, which then establishes that $\{2\}$ cannot be an open set in the order topology. To accomplish this, he makes use of the (unstated here) fact that if you have an open set that contains $2,$ then you must also have a basic open set that contains $2,$ since any open set (this now being true for any topology) is a union of basic open sets. (If, from among the basic open sets whose union is the given open set, you don't have one of them that contains the element $2,$ then the union isn't going to contain the element $2$ either.) OK, so from our assumption that we have an open set (in the order topology) that contains $2,$ it follows that this open set has as a subset a basic open set that contains the element $2.$ Munkres shows that any basic open set that contains the element $2$ must contain other points (this is pretty much all he says, the rest I've said is for the reader to fill in), and from what I said previously this implies that the open set we started with must contain points other than $2,$ which is what we wanted to show in order to show that $\{2\}$ is not open in the order topology.
