Let be a set.
Then the excluded point topology is given by:
A subset of is open iff
The argument given in Counterexamples in Topology by Steen and Seebach (counterexamples 13-15: 5) goes:
Let . Then is open in as .
So by definition can not be a limit point and so is an isolated point.
So any subset is not dense-in-itself, because it contains isolated points.
So by definition is scattered.
There's a flaw in the above, in that is easily shown to be a limit point (the only one possible in , in fact) and so has no isolated points in it, so is dense-in-itself.
So it appears that is not scattered after all, as it contains (exactly) one dense-in-itself subset.
Is Steen and Seebach wrong? Or have I missed something, e.g. " is required to be open to be described as dense-in-itself" or something? But I can't find any such words.