(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, $U$ 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.
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- Steen, L.A.;Seebach J.A.Jr.: "Counterexamples in Topology", Dover Publications, Inc, 1970