(related to Proposition: Perfect Sets vs. Derived Sets)

By hypothesis, $(X,\mathcal O)$ be a topological space. Let $U$ be a subset $U\subseteq X.$

is if and only if it equals its derived set.

- Assume, $U$ is perfect.
- By definition, $U$ is closed and dense-in-itself.
- Since $U$ is closed, the complement $U^C$ is open and contains no points of $U$, especially no limits of $U$.
- Therefore, $U$ contains all of its limits.
- By definition, $U$ equals its derived set.

- Assume, $U$ equals its derived set.
- By definition, $U$ contains all of its limits.
- The complement $U^C$ contains a neighborhood of each of its (limit) points.
- Therefore, $U^C$ is open.
- Thus, $U$ is closed.
- Moreover, $U$ contains no isolated points, because otherwise, they would be in $U$ since it equals its derived set.
- Therefore, $U$ is dense-in-itself.
- Since $U$ is closed and dense-in-itself, it is perfect.∎

**Steen, L.A.;Seebach J.A.Jr.**: "Counterexamples in Topology", Dover Publications, Inc, 1970