Let $\displaystyle S$ be a set.

Let $\displaystyle p \in S$.

Then the excluded point topology $\displaystyle T = (S, \tau)$ is given by:

A subset of $\displaystyle H \in \tau$ is open iff $\displaystyle p \notin H .$

The argument given in Counterexamples in Topology by Steen and Seebach (counterexamples 13-15: 5) goes:

Let $\displaystyle x \in S$. Then $\displaystyle \{x\} \subseteq S$ is open in $\displaystyle T$ as $\displaystyle \notin \{x\}$ .

So by definition $\displaystyle x$ can not be a limit point and so is an isolated point.

So any subset $\displaystyle H \subseteq S$ is not dense-in-itself, because it contains isolated points.

So by definition $\displaystyle T$ is scattered.

There's a flaw in the above, in that $\displaystyle p$ is easily shown to be a limit point (the only one possible in $\displaystyle T$, in fact) and so $\displaystyle \{p\}$ has no isolated points in it, so is dense-in-itself.

So it appears that $\displaystyle T$ 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. "$\displaystyle H$ is required to be open to be described as dense-in-itself" or something? But I can't find any such words.