Perfect subset of $\mathbb{R}$: closed with no isolated points

A set $P\subseteq \mathbb{R}$ is perfect when

• $P$ is closed (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), and
• $P$ has no isolated points (Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$): no $x\in P$ admits a real $\epsilon >0$ with $N_{\epsilon }(x)\cap P=\{x\}$.

Equivalently, $P$ is closed and $P\subseteq P^{\mathrm{\prime }}$. By Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$, a point of $P$ is isolated in $P$ exactly when it is not a limit point of $P$, so "no point of $P$ is isolated in $P$" says precisely that every point of $P$ is a limit point of $P$, that is, $P\subseteq P^{\mathrm{\prime }}$. Combined with the characterisation of closedness as $P^{\mathrm{\prime }}\subseteq P$ (The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points), a perfect set is exactly a set with $P=P^{\mathrm{\prime }}$, though only the two conditions above are used below.

• Both conditions are needed and neither implies the other. The set $\{0\}\cup [1,2]$ is closed and has the isolated point $0$, so it is not perfect ($\{0\}\cup [1,2]$ is closed, has an isolated point, and is not perfect ↗); the open interval $(0,1)$ has no isolated points and is not closed, so it is not perfect either.

• $\mathrm{\varnothing }$ is perfect, vacuously: it is closed and has no points at all, hence no isolated ones. This is why Every nonempty perfect subset of $\mathbb{R}$ is uncountable carries the hypothesis that $P$ is nonempty: the empty set is perfect and countable.

• A nonempty perfect set is forced to be large. It is uncountable (Every nonempty perfect subset of $\mathbb{R}$ is uncountable), and the simplest examples are the nondegenerate closed intervals (Every nondegenerate closed interval is perfect, giving a second proof that it is uncountable ↗). A perfect set need not contain any interval, the Cantor set being the standard example of that; it is not constructed anywhere in this library, and the statement is recorded here as orientation only, on the references above.