Interior, closure, boundary, exterior, derived set and isolated point in a topological space

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison), let $A\subseteq X$ and let $x\in X$. Neighbourhoods are as in Neighbourhood of a point and neighbourhood base, with this library's convention that a neighbourhood need not be open, so a neighbourhood need not be open.

• The interior of $A$ is $\mathrm{int}(A):=\bigcup \{U\in \mathcal{T}:U\subseteq A\}$.
• The closure of $A$ is $\bar{A}:=\bigcap \{F\subseteq X:F\text{ closed and }A\subseteq F\}$.
• The exterior of $A$ is $\mathrm{ext}(A):=\mathrm{int}(X\setminus A)$.
• The boundary of $A$ is $\mathrm{\partial }A:=\bar{A}\setminus \mathrm{int}(A)$.
• $x$ is a limit point (accumulation point) of $A$ if every neighbourhood $N$ of $x$ satisfies $N\cap (A\setminus \{x\})\mathrm{\ne }\mathrm{\varnothing }$. The set of limit points of $A$ is the derived set $A^{\mathrm{\prime }}$.
• $x$ is an isolated point of $A$ if $x\in A$ and some neighbourhood $N$ of $x$ satisfies $N\cap A=\{x\}$.

Both operators are well posed, and the two names are justified rather than asserted. The interior is a union of open sets, hence open by (T2), it is contained in $A$, and it contains every open $U\subseteq A$: so $\mathrm{int}(A)$ is the largest open subset of $A$. The family being intersected in the definition of $\bar{A}$ is nonempty, since $X$ is closed and contains $A$, so the intersection is a set; it is closed by (C2) of Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison, it contains $A$, and it is contained in every closed $F\supseteq A$: so $\bar{A}$ is the smallest closed superset of $A$. In particular

$\mathrm{int}(A)\subseteq A\subseteq \bar{A},$

and $A$ is open if and only if $A=\mathrm{int}(A)$, and closed if and only if $A=\bar{A}$, in each case because one inclusion is automatic and the other says exactly that $A$ belongs to the family in question.

Interior and closure are exchanged by complementation. For every $A\subseteq X$,

$X\setminus \mathrm{int}(A)=\overline{X\setminus A},X\setminus \bar{A}=\mathrm{int}(X\setminus A)=\mathrm{ext}(A)\mathrm{.}$

Indeed $U\mapsto X\setminus U$ is a bijection between the open subsets of $A$ and the closed supersets of $X\setminus A$, and it turns unions into intersections (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison); applying complementation to the first identity gives the second. So $\mathrm{ext}(A)=X\setminus \bar{A}$, and every statement about interiors has a mirror statement about closures.

The pointwise description of the interior. $x\in \mathrm{int}(A)$ if and only if $A$ is a neighbourhood of $x$. If $x\in \mathrm{int}(A)$ then $\mathrm{int}(A)$ is an open set with $x\in \mathrm{int}(A)\subseteq A$; conversely a neighbourhood relation $x\in U\subseteq A$ with $U$ open puts $x$ in the union defining $\mathrm{int}(A)$. The corresponding description of the closure is proved as the next item, because it is the statement that does the work in every later proof.

$X$ is the disjoint union of the three regions. Since $\mathrm{int}(A)\subseteq \bar{A}$, the three sets $\mathrm{int}(A)$, $\mathrm{\partial }A=\bar{A}\setminus \mathrm{int}(A)$ and $\mathrm{ext}(A)=X\setminus \bar{A}$ are pairwise disjoint and their union is $X$. This is recorded again, with the identities for interior and closure of unions and intersections, in the lemma two items below.

• The boundary is symmetric in $A$ and its complement: $\mathrm{\partial }A=\bar{A}\cap \overline{X\setminus A}$, because $\bar{A}\setminus \mathrm{int}(A)=\bar{A}\cap (X\setminus \mathrm{int}(A))=\bar{A}\cap \overline{X\setminus A}$ by the complementation identity above. Hence $\mathrm{\partial }A=\mathrm{\partial }(X\setminus A)$, and $\mathrm{\partial }A$ is closed, being an intersection of two closed sets.

• A limit point of $A$ need not lie in $A$, and a point of $A$ need not be a limit point of $A$. The points of $A$ that are not limit points of $A$ are exactly its isolated points, directly from the two definitions. The relation $\bar{A}=A\cup A^{\mathrm{\prime }}$ is a theorem, proved next, not a restatement.

• These are the metric notions when the topology is a metric topology. For a metric space the definitions of Interior, closure, boundary, limit point, isolated point and dense subset of a metric space are stated with balls in place of neighbourhoods, and the balls around $x$ are a neighbourhood base at $x$; the identification is carried out where metrizable spaces are defined, later on this page, and it is what allows metric examples to be quoted here without reproof.