Interior, closure, boundary, limit point, isolated point and dense subset of a metric space

Let $(X,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric), let $A \subseteq X$ and let $x \in X$. Balls are as in Open ball, closed ball and sphere in a metric space and open sets as in The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement; recall that a real $r$ written as a radius is always $> 0$.

• $x$ is an interior point of $A$ if $B(x,r) \subseteq A$ for some $r$. The set of interior points is the interior $\operatorname{int}(A)$.
• $x$ is an adherent point of $A$ if $B(x,r) \cap A \ne \emptyset$ for every $r$. The set of adherent points is the closure $\overline{A}$.
• $x$ is a limit point (accumulation point) of $A$ if $B(x,r) \cap (A \setminus \{x\}) \ne \emptyset$ for every $r$. The set of limit points is the derived set $A'$.
• $x$ is an isolated point of $A$ if $x \in A$ and $B(x,r) \cap A = \{x\}$ for some $r$.
• The boundary of $A$ is $\partial A := \overline{A} \setminus \operatorname{int}(A)$.
• $A$ is dense in $X$ if $\overline{A} = X$.

The interior is open, and it is the largest open subset of $A$. If $x \in \operatorname{int}(A)$, fix $r$ with $B(x,r) \subseteq A$; the ball $B(x,r)$ is itself open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed), so every $y \in B(x,r)$ has some $s$ with $B(y,s) \subseteq B(x,r) \subseteq A$, which puts $y$ in $\operatorname{int}(A)$. Hence $B(x,r) \subseteq \operatorname{int}(A)$ and $\operatorname{int}(A)$ is open. It is contained in $A$, since $x \in B(x,r) \subseteq A$ for an interior point $x$; and if $V \subseteq A$ is open then every $v \in V$ has a ball inside $V \subseteq A$, so $V \subseteq \operatorname{int}(A)$.

Two descriptions of the boundary agree. $x \in \partial A$ says that every ball around $x$ meets $A$ and that no ball around $x$ is contained in $A$; the second half says exactly that every ball around $x$ meets $X \setminus A$. So

$\partial A = \{\, x \in X : B(x,r) \cap A \ne \emptyset \text{ and } B(x,r) \cap (X \setminus A) \ne \emptyset \text{ for every } r \,\},$

from which $\partial A = \partial(X \setminus A)$ is immediate.

Elementary containments, straight from the definitions. $A \subseteq \overline{A}$, because $x \in A$ lies in every $B(x,r) \cap A$; $A' \subseteq \overline{A}$, because a ball meeting $A \setminus \{x\}$ meets $A$; and $\operatorname{int}(A) \subseteq A \subseteq \overline{A}$. A point of $A$ is either isolated in $A$ or a limit point of $A$, and not both, according to whether some ball meets $A$ only in $x$.

• The closure is defined here by adherent points and by nothing else. That it is closed, that it is the smallest closed set containing $A$, that for nonempty $A$ it is $\{x : d(x,A) = 0\}$, and that it consists of the limits of sequences from $A$, are theorems (The closure of a nonempty $A$ is $\{x : d(x,A) = 0\}$, equals $A$ together with its limit points, and is the smallest closed superset, A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed) and are proved from this definition.
• Limit point of a set is not the same notion as subsequential limit of a sequence (Subsequential limit of a real sequence, and the subsequential limit set), which this library deliberately keeps under a different name: the constant sequence $x_k = 0$ has $0$ as a subsequential limit, while its set of values $\{0\}$ has no limit point at all.
• Dense is relative to the ambient space, and the ambient space is part of the data: $A$ is dense in $X$ when $\overline{A} = X$, with $\overline{A}$ computed in $(X,d)$. The same $A$ inside a larger space is a different question.