# Thread: Is the Excluded Point Topology a scattered space?

1. ## Is the Excluded Point Topology a scattered space?

Let $S$ be a set.

Let $p \in S$.

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

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

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

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

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

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

So by definition $T$ is scattered.

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

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

2. Please use the $$…$$ tags.
$$\{x\}\subseteq S$$ gives $\{x\}\subseteq S$.