What Happens to Interior Under a Union?
The interior of a set records which of its points have an open neighborhood lying entirely in the set. When sets are combined by a union, an open neighborhood may lie in the union even though it does not lie in any one member of the family. This distinction determines the general rule: the union of the individual interiors is always contained in the interior of the union, but equality requires an additional condition.
Let \((X,\mathcal{T})\) be a topological space, let \(I\) be an index set, and let \(A_i\subseteq X\) for each \(i\in I\). The family may be finite or infinite. We use the convention that the union of an empty family is \(\varnothing\).
The largest-open-subset description gives the useful test in the definition: to establish that \(x\in\operatorname{int}A\), exhibit an open neighborhood \(U\) of \(x\) contained in \(A\). The neighborhood may depend on \(x\). For a union, it need only be contained in the union as a whole; it need not be contained in a single member.
The General Inclusion
Proof. If \(I=\varnothing\), both sides are empty by convention, so the inclusion holds. Now suppose \(I\ne\varnothing\), and take \(x\in\bigcup_{i\in I}\operatorname{int}A_i\). By the definition of union, there is an index \(j\in I\) such that \(x\in\operatorname{int}A_j\). Since \(\operatorname{int}A_j\) is open and contained in \(A_j\), it is an open neighborhood of \(x\) contained in \(A_j\), and \(A_j\subseteq\bigcup_{i\in I}A_i\). Therefore \(x\) has an open neighborhood contained in the full union, so \(x\in\operatorname{int}(\bigcup_{i\in I}A_i)\). This proves the inclusion. \(\square\)
The theorem does not assert equality for every family. It says that each point already interior to one member remains interior to the union. There can also be points whose neighborhoods fit inside the union only by using portions of several members.
Proof. The theorem gives \(\bigcup_{i\in I}\operatorname{int}A_i\subseteq\operatorname{int}(\bigcup_{i\in I}A_i)\). If every point of the latter set belongs to some \(\operatorname{int}A_i\), then the reverse inclusion also holds, so the two sets are equal. Conversely, if the two sets are equal, each point of \(\operatorname{int}(\bigcup_{i\in I}A_i)\) belongs to \(\bigcup_{i\in I}\operatorname{int}A_i\), which means it belongs to \(\operatorname{int}A_i\) for some \(i\in I\). \(\square\)
Unions of Open Sets
A topology is closed under arbitrary unions: if every \(A_i\) is open, then \(\bigcup_{i\in I}A_i\) is open. In this case each member is its own interior, and the whole union is also its own interior. Thus an exact formula holds for families of open sets, including infinite families.
Proof. If \(I=\varnothing\), the union is \(\varnothing\), which is open and has empty interior. Otherwise, the union is open by the union axiom for a topology. The interior of an open set is itself: it is an open subset of itself, and no subset of it can be larger than the set. Also, \(\operatorname{int}A_i=A_i\) for every \(i\), since each \(A_i\) is open. Substituting these equalities gives the formula. \(\square\)
Worked Example: An Increasing Family of Open Intervals
In \(\mathbb{R}\) with its standard topology, let \[ U_n=\left(-1+\frac{1}{n+1},\,1-\frac{1}{n+1}\right), \qquad n\in\mathbb{N}. \] Find \(\bigcup_{n\in\mathbb{N}}U_n\) and the interior of this union.
Each \(U_n\) is an open interval, so \(U_n\) is open and \(\operatorname{int}U_n=U_n\). The intervals are nested: as \(n\) increases, \(1/(n+1)\) decreases, so the left endpoint moves left and the right endpoint moves right. In particular, every interval is contained in \((-1,1)\).
To show that every \(x\in(-1,1)\) belongs to some \(U_n\), note that \(1-|x|>0\). Choose \(n\in\mathbb{N}\) sufficiently large that \(1/(n+1)<1-|x|\). Rearranging gives \(|x|<1-1/(n+1)\), which is exactly the condition that \(x\) lie between the endpoints of \(U_n\). Neither endpoint \(-1\) nor \(1\) belongs to any \(U_n\), since every interval is contained in \((-1,1)\). Hence \(\bigcup_{n\in\mathbb{N}}U_n=(-1,1)\).
Since this union is open, its interior is \((-1,1)\). The individual interiors give the same answer: \[ \bigcup_{n\in\mathbb{N}}\operatorname{int}U_n = \bigcup_{n\in\mathbb{N}}U_n = (-1,1). \]
Check: Every point of the union belongs to at least one open interval \(U_n\), so it has an open neighborhood contained in the union. The endpoints \(-1\) and \(1\) are not in the union at all. Thus the computed interior agrees with the neighborhood test.
Calculating Interiors for General Families
When the members are not all open, the theorem still gives a reliable first calculation: find the interior of each member, take their union, and then check whether the union itself has additional interior points. The first step gives a subset of the answer, not automatically the whole answer.
Worked Example: A Family of Closed Intervals
In \(\mathbb{R}\) with its standard topology, let \(A_n=[1/n,1]\) for \(n\in\mathbb{N}\). Find the union of the sets, the union of their interiors, and the interior of the union.
For \(n=1\), \(A_1=[1,1]=\{1\}\), whose interior is empty: no open interval around \(1\) is contained in the singleton. For \(n\ge2\), the interior is \(\operatorname{int}A_n=(1/n,1)\). Therefore \[ \bigcup_{n\in\mathbb{N}}\operatorname{int}A_n = \bigcup_{n\ge2}(1/n,1) = (0,1). \] Indeed, every \(x\) with \(0<x<1\) lies in \((1/n,1)\) for a sufficiently large \(n\), and neither \(0\) nor \(1\) lies in any of these open intervals.
Now consider the union of the sets themselves. Each \(A_n\) is contained in \((0,1]\). If \(0<x\le1\), choose \(n\in\mathbb{N}\) large enough that \(1/n\le x\); then \(x\in[1/n,1]=A_n\). Thus \(\bigcup_{n\in\mathbb{N}}A_n=(0,1]\).
The interior of \((0,1]\) in \(\mathbb{R}\) is \((0,1)\). Every point strictly between \(0\) and \(1\) has a sufficiently small open interval contained in \((0,1]\). The point \(1\) has no such neighborhood, because every open interval around \(1\) contains points greater than \(1\), which are not in \((0,1]\). Therefore \[ \operatorname{int}\left(\bigcup_{n\in\mathbb{N}}A_n\right) =(0,1) = \bigcup_{n\in\mathbb{N}}\operatorname{int}A_n. \]
Check: The union is \((0,1]\), so its interior cannot contain \(0\) or any point outside \((0,1]\). The point \(1\) is in the union but is not interior, while every point in \((0,1)\) has an open interval inside the union. This confirms the calculation.
A Common Mistake: Reversing the Inclusion
The general inclusion does not give permission to replace it by equality. Here is a concrete calculation showing why the distinction matters. In \(\mathbb{R}\) with the standard topology, take \(A=[0,1]\) and \(B=[1,2]\). The interiors are \(\operatorname{int}A=(0,1)\) and \(\operatorname{int}B=(1,2)\), so \[ \operatorname{int}A\cup\operatorname{int}B=(0,1)\cup(1,2). \] But \(A\cup B=[0,2]\), whose interior is \((0,2)\). In particular, \(1\in\operatorname{int}(A\cup B)\), because the open interval \((0,2)\) contains \(1\) and is contained in \(A\cup B\). Yet \(1\notin\operatorname{int}A\cup\operatorname{int}B\).
The incorrect calculation would be \[ \operatorname{int}(A\cup B) = \operatorname{int}A\cup\operatorname{int}B = (0,1)\cup(1,2). \] It misses the point \(1\). The correction is to use the always-valid inclusion and then test whether the union has additional interior points. Here, the neighborhood \((0,2)\) uses points from both sets, so it is contained in their union but not in either set individually.
This example also explains why calculating the interiors separately is not always enough. For a general family, first compute \(\bigcup_i\operatorname{int}A_i\); then examine points of the full union that are not in this set. Such a point belongs to the interior of the full union precisely when it has an open neighborhood contained in that full union.
A Practical Test
For an arbitrary family, keep the two membership tests separate. A point \(x\) belongs to \(\bigcup_i\operatorname{int}A_i\) if it lies in the interior of at least one member. A point \(x\) belongs to \(\operatorname{int}(\bigcup_i A_i)\) if there is an open neighborhood of \(x\) contained in the union. The second test allows that neighborhood to meet several members, as long as every point of the neighborhood belongs to at least one of them.
The general inclusion is useful even when equality fails: it supplies a guaranteed part of the interior of a union. When the sets are open, the topology’s union axiom gives equality immediately. For other families, the equality criterion pinpoints what remains to check: does every interior point of the full union already lie in the interior of one member?
Check Your Understanding
For each question, distinguish the interior of the full union from the union of the individual interiors.
- Suppose every \(A_i\) is open. Prove directly that \(\operatorname{int}(\bigcup_i A_i)=\bigcup_i\operatorname{int}A_i\), including the empty-family case.
- In \(\mathbb{R}\) with the standard topology, compute \(\operatorname{int}([0,1]\cup[2,3])\) and \(\operatorname{int}[0,1]\cup\operatorname{int}[2,3]\). Are they equal?
- In \(\mathbb{R}\), let \(C_n=[-1/n,1/n]\) for \(n\in\mathbb{N}\). Find \(\bigcup_n C_n\), \(\bigcup_n\operatorname{int}C_n\), and the interior of the union.
- State the pointwise condition that is equivalent to equality between the interior of a union and the union of the individual interiors.
- Give an open-neighborhood argument showing why a point in \(\operatorname{int}(\bigcup_i A_i)\) need not belong to any single \(\operatorname{int}A_i\).