Interior, closure, boundary and exterior of a subset of $\mathbb{R}$

Let $A\subseteq \mathbb{R}$, with open and closed sets as in Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen.

• The interior of $A$ is the union of all open subsets of $A$: $A^{\circ }:=\bigcup \{U\subseteq \mathbb{R}:U\text{ open and }U\subseteq A\}\mathrm{.}$
• The closure of $A$ is the intersection of all closed supersets of $A$: $\bar{A}:=\bigcap \{F\subseteq \mathbb{R}:F\text{ closed and }A\subseteq F\}\mathrm{.}$
• The boundary of $A$ is $\mathrm{\partial }A:=\bar{A}\setminus A^{\circ }$.
• The exterior of $A$ is $\mathrm{ext}A:=(\mathbb{R}\setminus A{)}^{\circ }$.

Both operators are well defined and deliver what their names claim. The family whose union defines $A^{\circ }$ always contains $\mathrm{\varnothing }$, and the family whose intersection defines $\bar{A}$ always contains $\mathbb{R}$, so the second family is nonempty and both expressions denote subsets of $\mathbb{R}$ without appeal to any convention about empty unions or intersections. Moreover:

• $A^{\circ }$ is open, being a union of open sets (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, claim 1), and $A^{\circ }\subseteq A$, since every set in the family is a subset of $A$. It is therefore the largest open subset of $A$: any open $U\subseteq A$ is a member of the family and so $U\subseteq A^{\circ }$.
• $\bar{A}$ is closed, being an intersection of a nonempty family of closed sets (Arbitrary unions and finite intersections of open subsets of $\mathbb{R}$ are open, and dually for closed sets, claim 3), and $A\subseteq \bar{A}$, since every set in the family contains $A$. It is therefore the smallest closed superset of $A$: any closed $F\supseteq A$ is a member of the family and so $\bar{A}\subseteq F$.

Pointwise description of the interior. For $x\in \mathbb{R}$,

$x\in A^{\circ }⟺\text{there is }\epsilon >0\text{ with }{N}_{\epsilon }(x)\subseteq A\mathrm{.}$

If $x\in A^{\circ }$ then, $A^{\circ }$ being open and containing $x$, there is $\epsilon >0$ with $N_{\epsilon }(x)\subseteq A^{\circ }\subseteq A$. Conversely if $N_{\epsilon }(x)\subseteq A$ then $N_{\epsilon }(x)$ is an open subset of $A$ (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen) containing $x$, hence $x\in N_{\epsilon }(x)\subseteq A^{\circ }$ (The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$).

The corresponding pointwise description of the closure is not a definitional matter and is proved separately, as 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.

• The four sets partition nothing by themselves, but three of them do. For every $A$ the three sets $A^{\circ }$, $\mathrm{\partial }A$ and $\mathrm{ext}A$ are pairwise disjoint with union $\mathbb{R}$. This is not proved here and is not used on this page; what is used is only the definitions above and the characterisations 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.

• Interior and closure are dual. Complementation exchanges the two families above, since $U$ is open exactly when $\mathbb{R}\setminus U$ is closed, so $\mathbb{R}\setminus A^{\circ }=\overline{\mathbb{R}\setminus A}$ and $\mathbb{R}\setminus \bar{A}=(\mathbb{R}\setminus A{)}^{\circ }=\mathrm{ext}A$. The second identity is the reason the exterior is usually described as "the complement of the closure".

• $A$ is open exactly when $A=A^{\circ }$, and closed exactly when $A=\bar{A}$. For the first: $A^{\circ }\subseteq A$ always, and $A\subseteq A^{\circ }$ holds exactly when $A$ is one of the open subsets of $A$, that is, exactly when $A$ is open. The closure half is the same argument read the other way, and it is recorded as a claim 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 because the rest of that theorem needs it.

• Boundary points may or may not belong to the set. $\mathrm{\partial }(0,1)=\mathrm{\partial }[0,1]=\{0,1\}$, and the two boundary points lie outside the first set and inside the second; the boundary sees only the way the set meets its complement, not which side the edge is assigned to.