Boundaries of Unions and Intersections

Boundary Inclusions for Unions and Intersections

A boundary point is one whose every open neighborhood meets both a set and its complement. For a union or intersection, a point can stop being a boundary point even when it is on the boundary of one of the sets being combined: the other set may make the union locally full, or make the intersection locally empty. The resulting general statements are inclusions, not equalities.

Let \((X,\mathcal{T})\) be a topological space, and let \(A,B\subseteq X\). We use the boundary \(\partial A=\overline{A}\cap\overline{X\setminus A}\). The open-neighborhood criterion for closure gives the following local description, which we will use throughout.

For the equivalence, \(x\in\overline{A}\) means every open neighborhood of \(x\) intersects \(A\); applying the same criterion to \(X\setminus A\) gives the condition for \(x\in\overline{X\setminus A}\). Thus membership in their intersection means every neighborhood meets both sets. Negating that condition gives a neighborhood missing at least one of them. A neighborhood missing \(X\setminus A\) is contained in \(A\); one missing \(A\) is disjoint from \(A\).

The Two Inclusions

Proof. We prove each inclusion by showing that a point outside \(\partial A\cup\partial B\) is outside the boundary on the left. Let \(x\notin\partial A\cup\partial B\). By the boundary criterion, there is an open neighborhood \(U\) of \(x\) that is either contained in \(A\) or disjoint from \(A\), and there is an open neighborhood \(V\) of \(x\) that is either contained in \(B\) or disjoint from \(B\). The intersection \(W=U\cap V\) is open and contains \(x\).

For the union, if \(U\subseteq A\), then \(W\subseteq A\subseteq A\cup B\). If instead \(V\subseteq B\), then \(W\subseteq B\subseteq A\cup B\). In either case, \(W\) is a neighborhood contained in the union, so the boundary criterion gives \(x\notin\partial(A\cup B)\). The only remaining possibility is that \(U\) is disjoint from \(A\) and \(V\) is disjoint from \(B\). Then \(W\) is disjoint from \(A\cup B\), again giving \(x\notin\partial(A\cup B)\). These possibilities cover every choice of \(U\) and \(V\).

For the intersection, if \(U\) is disjoint from \(A\), then \(W\) is disjoint from \(A\cap B\). If \(V\) is disjoint from \(B\), then \(W\) is disjoint from \(A\cap B\). If neither neighborhood is disjoint from its respective set, the alternatives provided by the criterion imply \(U\subseteq A\) and \(V\subseteq B\). In that case \(W\subseteq A\cap B\). So in every case there is a neighborhood of \(x\) either disjoint from the intersection or contained in it. Hence \(x\notin\partial(A\cap B)\). We have proved that the complement of \(\partial A\cup\partial B\) is contained in the complement of each boundary on the left, which proves both inclusions. \(\square\)

The proof uses only open neighborhoods and the boundary criterion. It does not require a metric, a separation axiom, or any condition on whether the sets are open or closed. It also covers empty sets and the whole space: the neighborhood argument applies without change.

Calculations in the Real Line

We now work in \(\mathbb{R}\) with its standard topology. For a real interval, its boundary consists of its finite endpoints, and a singleton \(\{c\}\) has boundary \(\{c\}\). These facts follow from the neighborhood criterion: every open interval around an endpoint meets the interval and its complement, while points strictly inside or strictly outside have a neighborhood contained in the interval or disjoint from it. For a singleton, every neighborhood of its point meets the singleton and its complement, while a point different from it has a neighborhood missing it.

Why Equality Can Fail

The first example gives a concrete warning against replacing either inclusion by an equality. In the union, the shared endpoint \(1\) is a boundary point of each operand, but it lies inside the union \([0,2]\); a sufficiently small open interval around \(1\), such as \((0.9,1.1)\), is contained in the union. In the intersection, the points \(0\) and \(2\) are boundary points of the operands, but neither is in the singleton intersection \(\{1\}\); for example, \((0.1,0.2)\) is a neighborhood of \(0.15\), not of those endpoints, so instead take \((-0.1,0.1)\) around \(0\), which misses \(\{1\}\), and \((1.9,2.1)\) around \(2\), which also misses it. Thus these operand-boundary points do not become boundary points of the intersection.

The general proof explains the mechanism without relying on interval geometry. A point outside both operand boundaries has, for each operand, a neighborhood either contained in that set or disjoint from it. Intersecting those neighborhoods produces a neighborhood where the union is entirely present or entirely absent, and where the intersection is likewise entirely present or absent. Such a point cannot be a boundary point of either combined set.

A common mistake is to infer that every boundary point of an operand must remain a boundary point after combining the sets. For \(A=[0,1]\) and \(B=[1,2]\), this would incorrectly place \(1\) in \(\partial(A\cup B)\), or place \(0\) and \(2\) in \(\partial(A\cap B)\). The corrected calculations are \(\partial(A\cup B)=\{0,2\}\) and \(\partial(A\cap B)=\{1\}\). The theorem guarantees containment in the union of operand boundaries; it does not guarantee that every point of that union survives.