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Proof

(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.

"$\Rightarrow$"

• 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.

"$\Leftarrow$"

• 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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References

Bibliography

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