Why Interior Does Not Preserve Unions

Why the Two Operations Can Differ

For a set \(A\) in a topological space \(X\), its interior \(\operatorname{int}A\) consists of the points that have an open neighborhood contained in \(A\). When we take a union first, an open neighborhood may be contained in the union even though it is not contained in any one member of the family. That difference is why interior does not, in general, preserve unions.

We use \(\mathbb{R}\) with its standard topology: its open sets are unions of open intervals. A set \(U\) is an open neighborhood of \(x\) if \(U\) is open and \(x\in U\). The interior is the largest open subset of a set, or equivalently the set of points that have an open neighborhood contained in it.

Proposition: For any topological space \(X\) and any family of subsets \((A_i)_{i\in J}\) of \(X\), \[ \bigcup_{i\in J}\operatorname{int}A_i\subseteq\operatorname{int}\left(\bigcup_{i\in J}A_i\right). \] Equality need not hold.

Proof. If \(J\) is empty, both unions are empty, and the inclusion holds. Otherwise, take \(x\in\bigcup_{i\in J}\operatorname{int}A_i\). Then \(x\in\operatorname{int}A_i\) for some \(i\in J\). By the definition of interior, there is an open neighborhood \(U\) of \(x\) with \(U\subseteq A_i\). Since \(A_i\subseteq\bigcup_{i\in J}A_i\), we also have \(U\subseteq\bigcup_{i\in J}A_i\). Thus \(x\in\operatorname{int}(\bigcup_{i\in J}A_i)\), proving the inclusion. \(\square\)

The inclusion says that any point with a neighborhood lying inside one member certainly has a neighborhood lying inside the whole union. The reverse direction would require more: a neighborhood inside the union would have to lie inside a single member. The next example shows why that may fail, even for just two sets.

Adjacent Closed Intervals

Worked Example: Two Closed Intervals Meeting at One Endpoint

Let \(A=[-1,2]\) and \(B=[2,5]\) in the standard real line. Compute \(\operatorname{int}(A\cup B)\) and \(\operatorname{int}A\cup\operatorname{int}B\).

First, the intervals meet at \(2\), and together they contain every real number from \(-1\) to \(5\), including both endpoints. Hence \(A\cup B=[-1,5]\). Its interior is \((-1,5)\): each \(x\) with \(-1<x<5\) has a sufficiently small open interval around it contained in \([-1,5]\), while no open neighborhood of either endpoint \(-1\) or \(5\) is contained in \([-1,5]\). Therefore \[ \operatorname{int}(A\cup B)=(-1,5). \]

Next, the interior of \(A=[-1,2]\) is \((-1,2)\), since its endpoints have no open neighborhoods contained in \(A\), and every point strictly between them does. The same reasoning gives \(\operatorname{int}B=(2,5)\). Thus \[ \operatorname{int}A\cup\operatorname{int}B=(-1,2)\cup(2,5). \]

The point \(2\) belongs to \(\operatorname{int}(A\cup B)\), but it belongs to neither \(\operatorname{int}A\) nor \(\operatorname{int}B\). For example, the neighborhood \((2-\tfrac12,2+\tfrac12)=(\tfrac32,\tfrac52)\) of \(2\) is contained in \(A\cup B=[-1,5]\). But every open interval around \(2\) contains points greater than \(2\), so it cannot be contained in \(A\); it also contains points less than \(2\), so it cannot be contained in \(B\). Consequently, \[ \operatorname{int}A\cup\operatorname{int}B =(-1,2)\cup(2,5) \subsetneq (-1,5) =\operatorname{int}(A\cup B). \]

Check: the only point in \((-1,5)\) missing from \((-1,2)\cup(2,5)\) is \(2\). This verifies the strictness of the inclusion, not just the inclusion itself.

The point \(2\) illustrates the mechanism. Its neighborhood is covered by two pieces: the part to the left lies in \(A\), and the part to the right lies in \(B\). But no neighborhood of \(2\) lies entirely in either piece. Interior asks whether a whole neighborhood fits inside the set being tested; it does not require that the same member of a union contain all the neighborhood unless we are testing membership in that member’s interior.

Theorem (Neighborhood criterion for equality): For a family \((A_i)_{i\in J}\) of subsets of \(X\), \[ \operatorname{int}\left(\bigcup_{i\in J}A_i\right)=\bigcup_{i\in J}\operatorname{int}A_i \] if and only if every point \(x\) having an open neighborhood contained in \(\bigcup_{i\in J}A_i\) also has an open neighborhood contained in some single \(A_i\).

Proof. Suppose first that the two sets are equal. Let \(x\) have an open neighborhood contained in \(\bigcup_{i\in J}A_i\). Then \(x\in\operatorname{int}(\bigcup_{i\in J}A_i)\), so the assumed equality gives \(x\in\bigcup_{i\in J}\operatorname{int}A_i\). Thus \(x\in\operatorname{int}A_i\) for some \(i\in J\), and the definition of interior supplies an open neighborhood of \(x\) contained in that \(A_i\).

Conversely, suppose the stated neighborhood condition holds. The general inclusion already proved gives \(\bigcup_{i\in J}\operatorname{int}A_i\subseteq\operatorname{int}(\bigcup_{i\in J}A_i)\). If \(x\in\operatorname{int}(\bigcup_{i\in J}A_i)\), then \(x\) has an open neighborhood contained in the union. By the condition, it has an open neighborhood contained in some \(A_i\), so \(x\in\operatorname{int}A_i\subseteq\bigcup_{i\in J}\operatorname{int}A_i\). This proves the reverse inclusion. If \(J\) is empty, there is no such \(x\), and both sides are empty, so the criterion also applies. \(\square\)

This criterion isolates the issue: equality holds precisely when every point interior to the union is interior to at least one member. In the adjacent-interval example, the point \(2\) violates that condition. A neighborhood of \(2\) fits in the union, but none fits in either interval individually.

When Equality Does Hold

Worked Example: Overlapping Closed Intervals

Let \(C=[0,3]\) and \(D=[2,5]\) in the standard real line. Find both interiors and compare them.

The intervals overlap on \([2,3]\), so their union is \([0,5]\). As in the preceding computation, its interior is \((0,5)\). The interiors of the individual intervals are \(\operatorname{int}C=(0,3)\) and \(\operatorname{int}D=(2,5)\). Their union is \[ (0,3)\cup(2,5)=(0,5), \] because every point strictly between \(0\) and \(5\) lies in at least one of these two open intervals. Therefore \[ \operatorname{int}(C\cup D)=(0,5) =\operatorname{int}C\cup\operatorname{int}D. \]

Check: points in \((0,2]\) lie in \((0,3)\), points in \((2,3)\) lie in both intervals, and points in \([3,5)\) lie in \((2,5)\). The endpoints \(0\) and \(5\) are excluded from both sides. So there is no point of the interior of the union missing from the union of the interiors.

Here the overlap means there is no uncovered interior point at the join. Unlike the adjacent-interval case, the intervals do not only touch at a single endpoint. The neighborhood criterion, rather than overlap as a general rule, is the exact test: overlapping sets can still fail to have equality in other situations, so the criterion should be applied to the particular family under consideration.

Worked Example: An Infinite Union of Singletons

For each \(t\in(0,1)\), let \(E_t=\{t\}\subseteq\mathbb{R}\). Compare the interior of the union with the union of the interiors.

Each singleton \(E_t\) has empty interior. Indeed, any open neighborhood of \(t\) in \(\mathbb{R}\) contains an open interval around \(t\), and that interval contains points other than \(t\). Hence no open neighborhood of \(t\) is contained in \(\{t\}\), so \(\operatorname{int}E_t=\varnothing\). It follows that \[ \bigcup_{t\in(0,1)}\operatorname{int}E_t=\varnothing. \]

On the other hand, \(\bigcup_{t\in(0,1)}E_t=(0,1)\): every point \(t\) of \((0,1)\) belongs to its singleton \(E_t\), and no singleton in the family contains a point outside \((0,1)\). Since \((0,1)\) is open, its interior is \((0,1)\). Thus \[ \operatorname{int}\left(\bigcup_{t\in(0,1)}E_t\right) =(0,1)\ne\varnothing =\bigcup_{t\in(0,1)}\operatorname{int}E_t. \]

Check: the left-hand side contains every real number strictly between \(0\) and \(1\), while the right-hand side contains no points. The discrepancy is therefore the whole interval \((0,1)\).

A Common Mistake

Common pitfall. A tempting but incorrect calculation is to claim that interior distributes over unions: \[ \operatorname{int}(A\cup B)=\operatorname{int}A\cup\operatorname{int}B. \] For \(A=[-1,2]\) and \(B=[2,5]\), this would give \((-1,2)\cup(2,5)\) for the left-hand side. But direct computation gives \(\operatorname{int}(A\cup B)=\operatorname{int}[-1,5]=(-1,5)\), which includes \(2\). The corrected statement is the one-sided inclusion \[ \operatorname{int}A\cup\operatorname{int}B\subseteq\operatorname{int}(A\cup B); \] equality requires the neighborhood condition in the theorem above.

The distinction matters whenever a union is assembled from pieces. A point may be interior to the combined set because its neighborhood is covered collectively, even though no single piece contains a neighborhood of that point. For unions of open sets, equality does hold: each \(A_i\) is already its own interior, so both sides equal \(\bigcup_{i\in J}A_i\). But without a condition of this kind, the inclusion need not reverse.

Check Your Understanding

For each question, distinguish a neighborhood contained in the union from one contained in a single member.

In the standard real line, compute \(\operatorname{int}([1,4]\cup[4,7])\) and \(\operatorname{int}[1,4]\cup\operatorname{int}[4,7]\). Which point accounts for any difference?
Let \(A=(-2,1]\) and \(B=[1,4)\). Compute both \(\operatorname{int}(A\cup B)\) and \(\operatorname{int}A\cup\operatorname{int}B\). Are they equal?
State the neighborhood criterion that is equivalent to equality between the interior of a union and the union of the interiors.
For \(E_t=\{t\}\) with \(t\in(2,3)\), compute \(\bigcup_{t\in(2,3)}\operatorname{int}E_t\) and \(\operatorname{int}(\bigcup_{t\in(2,3)}E_t)\).
Why does the inclusion \(\bigcup_i\operatorname{int}A_i\subseteq\operatorname{int}(\bigcup_i A_i)\) hold for every family, even when equality fails?