Limit point, isolated point, adherent point, derived set, and dense subset of $\mathbb{R}$

Let $A \subseteq \mathbb{R}$ and $x \in \mathbb{R}$, with neighbourhoods as in The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$ and closure as in Interior, closure, boundary and exterior of a subset of $\mathbb{R}$.

• $x$ is an adherent point of $A$ when $N_\varepsilon(x) \cap A \ne \varnothing$ for every real $\varepsilon > 0$.
• $x$ is a limit point (or accumulation point) of $A$ when $N^{*}_\varepsilon(x) \cap A \ne \varnothing$ for every real $\varepsilon > 0$: every punctured neighbourhood of $x$ meets $A$.
• $x$ is an isolated point of $A$ when $x \in A$ and there is a real $\varepsilon > 0$ with $N_\varepsilon(x) \cap A = \{x\}$.
• The derived set of $A$ is $A' \;:=\; \{\, x \in \mathbb{R} : x \text{ is a limit point of } A \,\}.$
• $A$ is dense in $\mathbb{R}$ when $\overline{A} = \mathbb{R}$.

A limit point is an adherent point, since $N^{*}_\varepsilon(x) \subseteq N_\varepsilon(x)$; and an element of $A$ is an adherent point of $A$, since $x \in N_\varepsilon(x) \cap A$ (The $\varepsilon$-neighbourhood and the punctured $\varepsilon$-neighbourhood of a point of $\mathbb{R}$). So the adherent points of $A$ are exactly the points of $A \cup A'$, a statement proved as part of 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.

Limit point and isolated point are exact opposites inside $A$. For $x \in A$: $x$ is an isolated point of $A$ exactly when it is not a limit point of $A$. Indeed $N_\varepsilon(x) \cap A = \{x\}$ says precisely that $N^{*}_\varepsilon(x) \cap A = \varnothing$, because $x$ itself always lies in $N_\varepsilon(x) \cap A$ when $x \in A$; so the existence of an $\varepsilon$ witnessing isolation is the negation of the condition defining a limit point. A point of $A$ is therefore either isolated in $A$ or a limit point of $A$, and never both.

A limit point need not belong to the set, and a point of the set need not be a limit point. Both possibilities occur, and the two examples that matter later are $0$, which is a limit point of $\{\, 1/k : k \ge 1 \,\}$ without belonging to it, and $0$ again, which belongs to $\{0\} \cup [1,2]$ as an isolated point.

• Terminology: limit point here is about a set, never about a sequence. This library reserves subsequential limit for the sequential notion (Subsequential limit of a real sequence, and the subsequential limit set), and the two are genuinely different: the constant sequence $x_k = 0$ has $0$ as a subsequential limit, while its set of values $\{0\}$ has no limit point at all. The distinction is the one Subsequential limit of a real sequence, and the subsequential limit set records under "Terminology", and it is respected throughout this page.

• Density is defined through the closure, not through intervals. Saying $\overline{A} = \mathbb{R}$ is equivalent to saying that every nonempty open subset of $\mathbb{R}$ meets $A$, and also to saying that every neighbourhood of every real meets $A$; the equivalences follow from 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 and are used in that form in Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable.

• The derived set need not be comparable with the set. It can be strictly larger, as for $\mathbb{Q}$: every punctured neighbourhood of any real contains a rational, since density supplies one strictly between $x$ and $x + \varepsilon$ (Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable), so the derived set of $\mathbb{Q}$ is all of $\mathbb{R}$. It can be strictly smaller, as for $\{0\}$, whose derived set is empty; and it can be neither, as for $\{0\} \cup (1,2)$, whose derived set is $[1,2]$, a set containing points outside the original and omitting the point $0$ of it. A closed set satisfying $A \subseteq A'$ is called perfect (Perfect subset of $\mathbb{R}$: closed with no isolated points).