Interior of Finite Intersections

Why Finiteness Matters for Interior

The interior of a set records the points that have some open neighborhood lying entirely inside that set. When several conditions must hold at once, their intersection imposes all of them simultaneously. For a finite list of sets, a point interior to each set has a single open neighborhood that satisfies every condition: intersect the finitely many neighborhoods. This is why interior preserves finite intersections. The finiteness is essential, because a topology guarantees that finite intersections of open sets are open, not that arbitrary intersections of open sets are open.

Throughout, \((X,\mathcal{T})\) is a topological space. A set \(U\in\mathcal{T}\) is open, and the interior of \(A\subseteq X\), written \(\operatorname{int}A\), is the union of all open subsets of \(X\) contained in \(A\). Thus \(\operatorname{int}A\) is open and contained in \(A\); moreover, every open set contained in \(A\) is contained in \(\operatorname{int}A\). These properties follow directly from the definition and the topology axioms.

Proof. First suppose \(n\geq1\). If \(x\in\operatorname{int}(\bigcap_{i=1}^{n}A_i)\), then \(\operatorname{int}(\bigcap_{i=1}^{n}A_i)\) is an open set containing \(x\) and contained in every \(A_i\). It is therefore an open subset of each \(A_i\), so it is contained in each \(\operatorname{int}A_i\). Hence \(x\in\bigcap_{i=1}^{n}\operatorname{int}A_i\).

Conversely, suppose \(x\in\bigcap_{i=1}^{n}\operatorname{int}A_i\). For each \(i\), the definition of interior gives an open set \(U_i\) such that \(x\in U_i\subseteq A_i\). The set \(U=\bigcap_{i=1}^{n}U_i\) is open because a topology is closed under finite intersections. It contains \(x\), and \(U\subseteq\bigcap_{i=1}^{n}A_i\). Thus \(x\in\operatorname{int}(\bigcap_{i=1}^{n}A_i)\). This proves both inclusions. If \(n=0\), the intersection on the left is \(X\), whose interior is \(X\), and the stipulated empty intersection on the right is also \(X\). \(\square\)

The two-set case is often the most useful form: \(\operatorname{int}(A\cap B)=\operatorname{int}A\cap\operatorname{int}B\). The proof shows exactly where the hypothesis enters: finitely many neighborhoods can be intersected and still give an open neighborhood. The theorem does not say that the sets \(A_i\) themselves must be open or closed.

Applying the Identity to Real Intervals

In the examples below, \(\mathbb{R}\) has its standard topology, whose open neighborhoods are built from open intervals. For interval calculations, first determine the intersection, including whether each endpoint belongs to it; then determine its interior. Alternatively, compute the interiors of the individual sets and intersect them. The theorem guarantees the same answer.

Intersections of Complements

Complements often turn several exclusions into a single intersection. For \(A,B\subseteq X\), De Morgan’s law gives \((X\setminus A)\cap(X\setminus B)=X\setminus(A\cup B)\). The finite-intersection theorem can then be applied to the complement sets. In the real line, it is especially important to keep track of endpoints: removing an open interval leaves its endpoints in the complement, but those endpoints may not remain in the interior.

The relation between interiors and complements gives another useful way to read such calculations. For any \(A\subseteq X\), \(\operatorname{int}(X\setminus A)=X\setminus\overline{A}\). Indeed, \(x\in\operatorname{int}(X\setminus A)\) means that some open neighborhood of \(x\) misses \(A\); by the open-neighborhood criterion for closure, this is equivalent to \(x\notin\overline{A}\). This equivalence is valid in every topological space. It can help identify interior points of a complement, but the finite-intersection theorem itself needs no closure calculation.

Proof. If \(x\in\operatorname{int}(\bigcap_{i\in J}A_i)\), then there is an open set \(U\) with \(x\in U\subseteq\bigcap_{i\in J}A_i\). For every \(i\in J\), this same open set satisfies \(U\subseteq A_i\), so \(x\in\operatorname{int}A_i\). Thus \(x\in\bigcap_{i\in J}\operatorname{int}A_i\). This argument proves the stated inclusion, including for an empty family under the convention that its intersection is \(X\). \(\square\)

A Common Pitfall: Treating an Infinite Intersection as Finite

The finite theorem is sometimes applied to a list that is actually infinite. The following calculation shows why that extension is invalid. For each \(n\in\mathbb{N}\), let \(A_n=(-1/n,1/n)\subseteq\mathbb{R}\). Each \(A_n\) is open, so \(\operatorname{int}A_n=A_n\). Also, \(0\) lies in every \(A_n\). If \(x\ne0\), choose \(n\in\mathbb{N}\) large enough that \(n>1/|x|\); then \(1/n<|x|\), so \(x\notin(-1/n,1/n)\). Consequently, \(\bigcap_{n\in\mathbb{N}}A_n=\{0\}\), whose interior is empty. On the other hand, \(\bigcap_{n\in\mathbb{N}}\operatorname{int}A_n=\bigcap_{n\in\mathbb{N}}A_n=\{0\}\). Thus the left side is \(\varnothing\), while the right side is \(\{0\}\).

Incorrect calculation: “Every \(A_n\) is open, so \(\operatorname{int}(\bigcap_n A_n)=\bigcap_n\operatorname{int}A_n=\{0\}\).” The step asserting that the intersection has interior \(\{0\}\) is wrong: a singleton is not open in the standard topology on \(\mathbb{R}\). The correction is to compute both sides separately. The intersection is \(\{0\}\), so its interior is \(\varnothing\); the intersection of the individual interiors is \(\{0\}\). The topology axiom permits finite intersections of open sets to be open, and this example shows that it does not permit that conclusion for arbitrary intersections.

The practical lesson is to count the sets before using the identity. For finitely many interval constraints, intersect their interiors or take the interior of their intersection—the answers agree. For an infinite family, the inclusion from the proposition remains valid, but equality needs a separate argument and may fail.