What the Interior Laws Tell Us
The interior of a set records the points that lie inside it with some open room around them. Once the interior is defined as the largest open subset, several useful rules follow directly: it stays inside the original set, respects inclusion, does not change when applied again, and interacts especially well with finite intersections. These rules let us simplify expressions without having to find all the open subsets of a set each time.
Let \((X,\mathcal{T})\) be a topological space. Recall that a set is open if it belongs to \(\mathcal{T}\). We use the characterization of interior as the largest open subset: \(\operatorname{int}A\) is open, is contained in \(A\), and contains every open subset of \(A\). Equivalently, it is the union of all open subsets of \(A\).
- Contractive: \(\operatorname{int}A\subseteq A\).
- Monotone: if \(A\subseteq B\), then \(\operatorname{int}A\subseteq\operatorname{int}B\).
- Idempotent: \(\operatorname{int}(\operatorname{int}A)=\operatorname{int}A\).
- Finite intersections: \(\operatorname{int}(A\cap B)=\operatorname{int}A\cap\operatorname{int}B\).
Proof. By the definition of interior, \(\operatorname{int}A\) is an open subset of \(A\). This proves contractiveness. Suppose \(A\subseteq B\). Then \(\operatorname{int}A\) is open and is contained in \(B\), so it is contained in the largest open subset of \(B\), namely \(\operatorname{int}B\). This proves monotonicity. Since \(\operatorname{int}A\) is open, it is itself the largest open subset of \(\operatorname{int}A\): it is an open subset of itself, and every subset of it is contained in it. Therefore \(\operatorname{int}(\operatorname{int}A)=\operatorname{int}A\).
For the intersection law, first note that \(\operatorname{int}A\cap\operatorname{int}B\) is open, because a finite intersection of open sets is open. It is contained in \(A\cap B\), by contractiveness, so the largest-open-subset property gives \(\operatorname{int}A\cap\operatorname{int}B\subseteq\operatorname{int}(A\cap B)\). For the reverse inclusion, \(A\cap B\subseteq A\) and \(A\cap B\subseteq B\). Monotonicity gives \(\operatorname{int}(A\cap B)\subseteq\operatorname{int}A\) and \(\operatorname{int}(A\cap B)\subseteq\operatorname{int}B\). Together these inclusions prove the equality. Applying the two-set equality repeatedly proves the statement for any nonempty finite family. For the empty family, the intersection is \(X\), and \(X\) is open in every topology, so \(\operatorname{int}X=X\). \(\square\)
Using the Laws in Set Expressions
The finite-intersection law gives a direct simplification rule: when an interior is applied to a finite intersection, it can be applied to each factor instead. Conversely, if a factor is already an interior, idempotence and contractiveness help remove redundant operations. For example, the theorem implies \(\operatorname{int}(A\cap\operatorname{int}B)=\operatorname{int}A\cap\operatorname{int}B\). Indeed, the intersection law gives \(\operatorname{int}A\cap\operatorname{int}(\operatorname{int}B)\), and idempotence replaces \(\operatorname{int}(\operatorname{int}B)\) with \(\operatorname{int}B\).
These are rules about intersections and repeated interior operations. Each rule has a specific hypothesis: monotonicity requires an inclusion of sets, and the intersection rule concerns a finite intersection. To use them accurately, first identify the set being placed inside the interior, then check whether the relevant inclusion or finite-intersection structure is present.
Worked Example: Intersections of Intervals in \(\mathbb{R}\)
Use the standard topology on \(\mathbb{R}\). Let \(A=[0,2]\cup\{4\}\) and \(B=(1,5]\). Compute \(\operatorname{int}A\), \(\operatorname{int}B\), and \(\operatorname{int}(A\cap B)\), then compare the result with the finite-intersection law.
The interval \((0,2)\) is open and contained in \(A\). Neither endpoint \(0\) nor \(2\) is an interior point: every open interval around \(0\) contains negative numbers, and every open interval around \(2\) contains numbers greater than \(2\), which are not in \(A\). The point \(4\) is not interior to \(A\), because every open interval around \(4\) contains points other than \(4\), such as \(4+\varepsilon/2\) for any \(\varepsilon>0\), and those points are not in \(A\). Thus \(\operatorname{int}A=(0,2)\).
The set \(B=(1,5]\) has interior \((1,5)\): this is open and contained in \(B\), while every neighborhood of \(5\) contains points greater than \(5\), which are not in \(B\). Hence \(\operatorname{int}B=(1,5)\).
For the intersection, \(A\cap B=(1,2]\cup\{4\}\). The interval \((1,2)\) is open and contained in this intersection. The point \(2\) is not interior because every interval around it contains points greater than \(2\) and less than \(4\), which are not in \(A\cap B\). The isolated piece \(\{4\}\) contributes no interior points: every interval around \(4\) contains points other than \(4\) that are not in the intersection. Therefore \(\operatorname{int}(A\cap B)=(1,2)\).
Check with the law: \(\operatorname{int}A\cap\operatorname{int}B=(0,2)\cap(1,5)=(1,2)\), agreeing with the direct calculation. Also \(A\cap B\subseteq A\), and the computed interiors satisfy \((1,2)\subseteq(0,2)\), as monotonicity requires.
How the Topology Changes the Calculation
The interior laws hold in every topology, but the actual interior depends on which sets are declared open. The next examples use small spaces where the open sets can be listed or tested directly.
Worked Example: Interior in the Sierpinski Topology
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). Find the interior of every subset of \(X\), and check the laws for a particular inclusion and intersection.
The open subsets of \(\{a\}\) are \(\varnothing\) and \(\{a\}\), so \(\operatorname{int}\{a\}=\{a\}\). The only open subset of \(\{b\}\) is \(\varnothing\), so \(\operatorname{int}\{b\}=\varnothing\). Since \(X\) and \(\varnothing\) are open, their interiors are themselves. Thus the complete list is \(\operatorname{int}\varnothing=\varnothing\), \(\operatorname{int}\{a\}=\{a\}\), \(\operatorname{int}\{b\}=\varnothing\), and \(\operatorname{int}X=X\).
Take \(A=\{b\}\) and \(B=X\). Then \(A\subseteq B\), while \(\operatorname{int}A=\varnothing\) and \(\operatorname{int}B=X\); hence \(\operatorname{int}A\subseteq\operatorname{int}B\). Their intersection is \(A\), so \(\operatorname{int}(A\cap B)=\operatorname{int}\{b\}=\varnothing\), and \(\operatorname{int}A\cap\operatorname{int}B=\varnothing\cap X=\varnothing\). This checks monotonicity and the intersection law in this topology.
Check: the set \(\{b\}\) is not open in this topology, and its interior is empty. Thus an interior need not equal the set whose interior is being taken; it is the largest open subset, not a claim that the set itself is open.
Worked Example: Interior in the Finite-Complement Topology
Give \(\mathbb{R}\) the finite-complement topology: the open sets are \(\varnothing\) and all subsets with finite complement. Let \(A=\mathbb{N}\) and \(B=\mathbb{R}\setminus\{1,2\}\). Find \(\operatorname{int}A\), \(\operatorname{int}B\), and \(\operatorname{int}(A\cap B)\).
The set \(\mathbb{N}\) is not open, because \(\mathbb{R}\setminus\mathbb{N}\) is infinite. In fact, no nonempty open set is contained in \(\mathbb{N}\): every nonempty open set has finite complement in \(\mathbb{R}\), so it contains all but finitely many real numbers, including infinitely many numbers outside \(\mathbb{N}\). Hence the only open subset of \(A\) is \(\varnothing\), and \(\operatorname{int}A=\varnothing\).
The complement of \(B\) is the finite set \(\{1,2\}\), so \(B\) is open and \(\operatorname{int}B=B\). Since \(A\cap B=\mathbb{N}\setminus\{1,2\}\), this intersection is still a subset of \(\mathbb{N}\). The same argument as for \(A\) shows it contains no nonempty open set, so \(\operatorname{int}(A\cap B)=\varnothing\).
Check with the law: \(\operatorname{int}A\cap\operatorname{int}B=\varnothing\cap B=\varnothing\), which agrees with the direct result. Also \(A\cap B\subseteq A\), and both interiors are empty, so this instance of monotonicity holds. The calculation illustrates that knowing a set’s elements alone does not determine its interior; the topology matters.
A Common Misuse of Idempotence
Idempotence says that applying interior twice has the same result as applying it once. It does not say that applying interior once leaves every set unchanged. The latter conclusion holds for open sets, but a general set may contain boundary points that are not interior points.
Worked Example: Idempotence Does Not Mean Every Set Is Open
Use the standard topology on \(\mathbb{R}\), and take \(A=[0,1]\). Its interior is \((0,1)\). Applying interior again gives \(\operatorname{int}(\operatorname{int}A)=\operatorname{int}(0,1)=(0,1)\), as idempotence states. But \(\operatorname{int}A=(0,1)\ne[0,1]=A\), because \(0\) and \(1\) are not interior points of \(A\).
Common wrong calculation: from idempotence, one might incorrectly write \(\operatorname{int}A=A\) for \(A=[0,1]\). That would give \((0,1)=[0,1]\), which fails because \(0\in[0,1]\) but \(0\notin(0,1)\). The correct substitution into the law is \(\operatorname{int}(\operatorname{int}[0,1])=\operatorname{int}[0,1]\), or equivalently \(\operatorname{int}(0,1)=(0,1)\).
Check: \((0,1)\subseteq[0,1]\), so contractiveness holds. The interior is open, so taking its interior again leaves it unchanged. These two checks confirm the correct laws without asserting that the original closed interval is open.
Applying the Laws Reliably
For a finite intersection, the theorem gives an exact identity, not just an inclusion. For nested interiors, idempotence removes repeated operations. Monotonicity lets us transfer known set inclusions to their interiors, while contractiveness provides a quick check that a proposed answer cannot extend outside the original set.
When simplifying an expression, keep the operations in their proper roles. For example, from \(A\subseteq B\) one may conclude \(\operatorname{int}A\subseteq\operatorname{int}B\); from a finite intersection one may separate the interior across its factors; and from an interior already inside another interior one may use idempotence where the expression matches its hypothesis. Each conclusion follows from the definitions and the proved laws, so no additional assumptions about the space are needed.
Check Your Understanding
For each question, state the topology when it is not the standard topology on \(\mathbb{R}\), and justify your answer using the definition or the interior laws.
- In any topological space, prove that if \(A\subseteq B\), then \(\operatorname{int}A\subseteq\operatorname{int}B\).
- In the standard topology on \(\mathbb{R}\), compute \(\operatorname{int}([0,3]\cap(1,4))\) directly and using the finite-intersection law.
- In the topology \(\{\varnothing,\{a\},\{a,b\}\}\) on \(X=\{a,b\}\), compute the interior of each subset of \(X\).
- Give an example of a set \(A\subseteq\mathbb{R}\) for which \(\operatorname{int}A\ne A\), and verify idempotence for that set.
- For three subsets \(A,B,C\) of a topological space, write the finite-intersection law for \(\operatorname{int}(A\cap B\cap C)\).