Interior, Closure, and Boundary

In this lesson you’ll define interior, closure and boundary, learn their identities and the ones that fail, meet density and nowhere-density, and work through the examples where intuition breaks.

Interior. $\operatorname{int}A$ is the union of all open sets contained in $A$ - equivalently the largest such set. A point is interior when some neighbourhood of it lies entirely inside $A$.

Closure. $\bar A$ is the intersection of all closed sets containing $A$ - equivalently the smallest such set. Also $\bar A = A\cup A'$, the set together with its limit points.

Boundary. $\partial A = \bar A\setminus\operatorname{int}A$. A point is on the boundary when every neighbourhood meets both $A$ and its complement.

Every point of the space falls into exactly one of three classes relative to $A$: interior, boundary, or exterior. That trichotomy is the cleanest way to think about all three notions at once.

The figure classifies five test points against a set built from an open disc, one isolated point, and a segment. The classification is computed by sampling a small ring about each point and seeing what it meets. Two results are worth dwelling on: the isolated point belongs to $S$ and is a boundary point, and every point of the segment is likewise. So $S$ has a nonempty part that contributes nothing at all to its interior.

Being in $A$ does not make you interior to $A$. That is the single most useful correction in this lesson.

• $\operatorname{int}A\subseteq A\subseteq\bar A$
• $A$ is open $\iff A = \operatorname{int}A$
• $A$ is closed $\iff A = \bar A$
• $A$ is clopen $\iff \partial A = \varnothing$
• $\partial A = \partial(X\setminus A)$ - a set and its complement share a boundary
• $\bar A = A\cup\partial A$ and $X = \operatorname{int}A \sqcup \partial A \sqcup \operatorname{ext}A$
• Both operations are idempotent: $\operatorname{int}(\operatorname{int}A) = \operatorname{int}A$ and $\overline{\bar A} = \bar A$
• Duality: $\operatorname{int}(X\setminus A) = X\setminus\bar A$

This is where care is needed. Closure behaves well with unions and badly with intersections; interior is the reverse.

$$\overline{A\cup B} = \bar A\cup\bar B \qquad\text{always}$$ $$\overline{A\cap B} \subseteq \bar A\cap\bar B \qquad\text{containment only}$$ $$\operatorname{int}(A\cap B) = \operatorname{int}A\cap\operatorname{int}B \qquad\text{always}$$ $$\operatorname{int}(A\cup B) \supseteq \operatorname{int}A\cup\operatorname{int}B \qquad\text{containment only}$$

The counterexamples are short. For closure of an intersection, take $A=(0,1)$ and $B=(1,2)$ in $\mathbb{R}$: the intersection is empty so its closure is empty, while $\bar A\cap\bar B = \{1\}$. For interior of a union, take $A=[0,1]$ and $B=[1,2]$: the union has interior $(0,2)$, but neither part’s interior contains 1.

Also note the union rule for closure holds only for finitely many sets. For infinitely many it degrades to a containment, as $\bigcup\{q\}$ over the rationals shows: each singleton is closed, the union is $\mathbb{Q}$, and its closure is all of $\mathbb{R}$.

$A$ is dense in $X$ if $\bar A = X$, equivalently if every nonempty open set meets $A$.

$\mathbb{Q}$ is dense in $\mathbb{R}$, and so are the irrationals. A space with a countable dense subset is called separable, which is why $\mathbb{R}$ is separable and why so many analysis arguments can be run over a countable set.

The opposite extreme:

$A$ is nowhere dense if $\operatorname{int}\bar A = \varnothing$.

$\mathbb{Z}$ is nowhere dense in $\mathbb{R}$; so is the Cantor set, despite being uncountable. Nowhere-dense sets are the “small” sets of topology, and the Baire category theorem - that a countable union of nowhere dense sets cannot fill a complete metric space - is the engine behind many existence proofs.

$\mathbb{Q}$ in $\mathbb{R}$. $\operatorname{int}\mathbb{Q} = \varnothing$, $\bar{\mathbb{Q}} = \mathbb{R}$, so $\partial\mathbb{Q} = \mathbb{R}$. A set whose boundary is the whole space.

A punctured disc. For $A$ the closed unit disc minus its centre, $\partial A$ is the unit circle together with the centre point. Boundaries do not have to look like edges.

In the discrete topology. Every set is clopen, so $\operatorname{int}A = \bar A = A$ and $\partial A=\varnothing$ for every $A$.

In the trivial topology. For any nonempty proper $A$: $\operatorname{int}A = \varnothing$, $\bar A = X$, and $\partial A = X$.

Boundary of a boundary. $\partial(\partial A)\subseteq\partial A$, and equality can fail. For $A=\mathbb{Q}$, $\partial A = \mathbb{R}$ and $\partial(\partial A) = \varnothing$.

Example 1: Find the interior, closure and boundary of $[0,1)$ in $\mathbb{R}$.

Solution. The largest open set inside is $(0,1)$, since no interval around 0 fits. The smallest closed set containing it is $[0,1]$, adding the limit point 1.

$$\operatorname{int} = (0,1), \qquad \bar{} = [0,1], \qquad \partial = \{0,1\}$$

∎

Example 2: Find all three for $\mathbb{Q}$ in $\mathbb{R}$.

Solution. No interval consists only of rationals, so the interior is empty. Every real is a limit of rationals, so the closure is $\mathbb{R}$.

$$\operatorname{int}\mathbb{Q} = \varnothing, \qquad \bar{\mathbb{Q}} = \mathbb{R}, \qquad \partial\mathbb{Q} = \mathbb{R}$$

∎

The boundary is the entire space. Worth sitting with, because it shows “boundary” is not a geometric edge.

Example 3: Find all three for $\mathbb{Z}$ in $\mathbb{R}$.

Solution. No interval fits inside $\mathbb{Z}$, so the interior is empty. $\mathbb{Z}$ has no limit points at all - each integer is isolated - so it is already closed.

$$\operatorname{int} = \varnothing, \qquad \bar{\mathbb{Z}} = \mathbb{Z}, \qquad \partial\mathbb{Z} = \mathbb{Z}$$

Since $\operatorname{int}\bar{\mathbb{Z}} = \varnothing$, the set is nowhere dense. ∎

Example 4: Show $\overline{A\cap B}$ can be strictly smaller than $\bar A\cap\bar B$.

Solution. Take $A=(0,1)$ and $B=(1,2)$ in $\mathbb{R}$. Then $A\cap B=\varnothing$, so

$$\overline{A\cap B} = \varnothing$$

But $\bar A = [0,1]$ and $\bar B=[1,2]$, so $\bar A\cap\bar B = \{1\}$.

$$\varnothing \subsetneq \{1\}$$

∎

Example 5: Show $\operatorname{int}(A\cup B)$ can be strictly larger than $\operatorname{int}A\cup\operatorname{int}B$.

Solution. Take $A=[0,1]$ and $B=[1,2]$. Then $A\cup B=[0,2]$ with interior $(0,2)$.

But $\operatorname{int}A\cup\operatorname{int}B = (0,1)\cup(1,2)$, which misses the point 1.

$$(0,1)\cup(1,2) \subsetneq (0,2)$$

∎

The gap is exactly the shared boundary point, which becomes interior only after the union.

Example 6: Find the boundary of the punctured closed disc.

Solution. Let $A = \{x : |x|\le1, x\ne0\}$ in $\mathbb{R}^2$.

The interior is the punctured open disc, since points of the unit circle have no neighbourhood inside $A$ and the origin is not in $A$.

The closure is the full closed disc: the origin is a limit point of $A$, so it gets added back.

$$\partial A = \{|x|=1\}\cup\{0\}$$

∎

A boundary can include a point in the middle. This is the clearest example that boundary is not the same as visible edge.

Example 7: Show the Cantor set is nowhere dense.

Solution. The Cantor set $C$ is closed, being an intersection of closed sets, so $\bar C = C$.

It contains no interval: after $n$ steps of the construction every remaining piece has length $3^{-n}$, so any interval of positive length is eventually cut. Hence $\operatorname{int}C=\varnothing$.

$$\operatorname{int}\bar C = \varnothing \implies \text{nowhere dense}$$

∎

And yet $C$ is uncountable. Nowhere dense does not mean small in cardinality, only topologically thin.

Image processing. Morphological erosion and dilation are the interior and closure operations applied to pixel sets, and boundary extraction is literally $\bar A\setminus\operatorname{int}A$. Edge detection and noise removal are built from them.

Computational geometry. Robust point-in-polygon and boolean operations on solids must handle boundary cases explicitly, and the regularized set operations used in CAD are defined as the closure of the interior in order to eliminate degenerate slivers.

Feasible regions in optimization. Interior-point methods work strictly inside the feasible set, which is why the interior being nonempty is a stated assumption. When the interior is empty, the problem needs a different method entirely.

Physical boundaries. A boundary condition in a PDE is imposed on $\partial\Omega$, and getting the topological boundary right matters when the domain has cracks or interior holes, where the boundary includes surfaces that are not the outer edge.

Numerical analysis of sets. Approximating a region by a grid amounts to computing an interior approximation and a closure approximation, and the difference between them is the discretization error.

Sampling and density. That a countable set can be dense justifies approximating continuous problems on a countable grid, and separability of the underlying space is the precise hypothesis those approximations need.