Arbitrary Unions of Open Sets

One Point at a Time in an Arbitrary Union

In the previous tutorial, complements of closed sets were described as open sets, including their maximal open intervals. Here we focus on how open sets behave when they are combined: an arbitrary union of open subsets of $\mathbb{R}$ is open. The key feature is local. If a point belongs to the union, it belongs to at least one particular set in the family, and that set supplies an open interval around the point.

The Arbitrary Unions of Open Sets Theorem was established earlier in this course. We will use it, rather than re-prove it, to build a useful construction: the interior of an arbitrary subset of $\mathbb{R}$. Even when a set is not open, it has a largest open subset contained in it. This construction explains why arbitrary unions matter beyond the theorem itself.

Indexed Families and Their Unions

An indexed family of sets is written $\{U_\alpha:\alpha\in A\}$, where $A$ is an index set. The index set may be finite, countably infinite, or uncountable. Its union consists of the points that belong to at least one member:

\bigcup_{\alpha\in A}U_\alpha = \{x\in\mathbb{R}:\text{there exists }\alpha\in A\text{ such that }x\in U_\alpha\}.

The index is a label, not necessarily a point in any of the sets. Different indices may label the same set, and a point in the union may belong to many members of the family. To establish that a point lies in a union, it is enough to identify one index whose set contains it. That particular set then provides the local information needed around the point.

Theorem (Arbitrary Unions of Open Sets, recalled): If $\{U_\alpha:\alpha\in A\}$ is any family of open subsets of $\mathbb{R}$, then $\bigcup_{\alpha\in A}U_\alpha$ is open.

The theorem includes the empty family, whose union is $\varnothing$, an open set. In the nonempty case, the theorem does not require a bound on the number of members, nor a common interval size that works for every point and every member. Openness is checked separately at each point of the union.

Worked Example: A Continuum of Intervals Fills One Interval

For each $t\in(0,2)$, let $U_t=(t-1,t+1)$. Every $U_t$ is open. We claim that their union is exactly $(-1,3)$:

\bigcup_{0<t<2}(t-1,t+1)=(-1,3).

First, if $y\in U_t$ for some $t\in(0,2)$, then $t-1<y<t+1$. Since $t>0$, we have $t-1>-1$, so $y>-1$. Since $t<2$, we have $t+1<3$, so $y<3$. Thus the union is contained in $(-1,3)$.

For the reverse inclusion, take any $x\in(-1,3)$. Choose $t=(x+1)/2$. If $-1<x\leq1$, then $0<t\leq1<2$. Also $x-1<t$, because $x<3$, and $t<x+1$, because $x>-1$. Therefore $x\in(t-1,t+1)$. If $1<x<3$, then $1<t<2$; the same inequalities $x-1<t<x+1$ follow from $x<3$ and $x>-1$. This case also gives $x\in(t-1,t+1)$. The two cases cover every $x\in(-1,3)$, proving the claimed equality.

The union is open, as the Arbitrary Unions of Open Sets Theorem guarantees. The calculation also illustrates why an uncountable family can be convenient: the intervals slide continuously, and their union has endpoints $-1$ and $3$, neither of which is included.

The Interior of a Set

For an arbitrary set $E\subseteq\mathbb{R}$, consider all the open sets that fit inside $E$. There is at least one such set, since $\varnothing$ is open and $\varnothing\subseteq E$. Taking the union of all of them collects every point that can be included in an open subset of $E$.

Definition (Interior): For $E\subseteq\mathbb{R}$, the interior of $E$, denoted $\operatorname{int}(E)$, is the union of all open sets contained in $E$: $$ \operatorname{int}(E)=\bigcup\{V\subseteq E: V\text{ is open in }\mathbb{R}\}. $$

The sets in this union form a family indexed by the collection of open subsets of $E$. This collection may be large, but arbitrary unions are permitted. In particular, $\operatorname{int}(E)$ is open by the recalled theorem.

Theorem (The Interior Is the Largest Open Subset): For every $E\subseteq\mathbb{R}$, the set $\operatorname{int}(E)$ is open, is contained in $E$, and contains every open set contained in $E$. Equivalently, it is the largest open subset of $E$ with respect to inclusion.

Proof. By definition, $\operatorname{int}(E)$ is a union of open sets, so it is open by the Arbitrary Unions of Open Sets Theorem. Each set in the union is contained in $E$; therefore their union is contained in $E$. Finally, if $V$ is any open set with $V\subseteq E$, then $V$ is one of the sets included in the defining union. Hence $V\subseteq\operatorname{int}(E)$. These three facts prove the claim. $\square$

This maximality property also characterizes the interior uniquely. If an open set $G$ is contained in $E$ and contains every open subset of $E$, then $G=\operatorname{int}(E)$: each contains the other. Thus the construction does not depend on a choice of intervals or on a choice of representation of the open sets in the union.

Worked Example: The Interior of a Set with Two Pieces

Let $E=[-2,0]\cup(1,3]$. We claim that

\operatorname{int}(E)=(-2,0)\cup(1,3).

The proposed interior is open and is contained in $E$, so the largest-open-subset theorem gives $(-2,0)\cup(1,3)\subseteq\operatorname{int}(E)$. For the reverse inclusion, consider a point of $E$ not in the proposed set. The only such points are $-2$, $0$, and $3$. No open interval around $-2$ is contained in $E$, since every interval around $-2$ contains points less than $-2$. No open interval around $0$ is contained in $E$, since every such interval contains points strictly between $0$ and $1$. No open interval around $3$ is contained in $E$, since every such interval contains points greater than $3$.

If an open set $V\subseteq E$ contained any of these three points, openness would give an interval around that point contained in $V$, and hence contained in $E$, contradicting the preceding checks. Thus no open subset of $E$ contains any of the three points. Every point of $\operatorname{int}(E)$ must therefore lie in $(-2,0)\cup(1,3)$, proving the equality.

How the Interior Changes When the Set Changes

The largest-open-subset description immediately yields two structural properties. Enlarging a set cannot remove any open subset that was already inside it. Also, taking the interior twice has no further effect, because the first interior is already open.

Theorem (Monotonicity and Idempotence of the Interior): If $A\subseteq B\subseteq\mathbb{R}$, then $\operatorname{int}(A)\subseteq\operatorname{int}(B)$. For every $E\subseteq\mathbb{R}$, $\operatorname{int}(\operatorname{int}(E))=\operatorname{int}(E)$.

Proof. The set $\operatorname{int}(A)$ is open and is contained in $A$. Since $A\subseteq B$, it is also an open subset of $B$. The largest-open-subset property for $B$ gives $\operatorname{int}(A)\subseteq\operatorname{int}(B)$, proving monotonicity.

For idempotence, $\operatorname{int}(E)$ is open. It is therefore an open subset of itself, so the largest-open-subset property gives $\operatorname{int}(E)\subseteq\operatorname{int}(\operatorname{int}(E))$. On the other hand, the interior of any set is contained in that set, so $\operatorname{int}(\operatorname{int}(E))\subseteq\operatorname{int}(E)$. The two inclusions prove equality. $\square$

Worked Example: A Dense Union Can Be the Whole Line

For each $x\in\mathbb{R}$, let $V_x=(x-1,x+1)$. Each $V_x$ is open. Every real number $y$ belongs to $V_y$, since $y-1<y<y+1$. Thus every real number belongs to the union, and, because each $V_x\subseteq\mathbb{R}$, we have

\bigcup_{x\in\mathbb{R}}(x-1,x+1)=\mathbb{R}.

The union is open, as is $\mathbb{R}$ itself. Notice that no single interval $V_x$ is the whole line. The union reaches every real number because the center varies over all real numbers. This is a basic use of an arbitrary family: its union can be much larger than any one member.

What Arbitrary Unions Do—and Do Not—Guarantee

The theorem about arbitrary unions is an openness guarantee, not a claim that unions preserve every other feature of their members. For instance, the union of open intervals may be unbounded, as in the preceding example, or have several separate pieces. Its shape depends on how the members overlap. In the definition of the interior, that flexibility is useful: all available open subsets can be combined without losing openness.

A common pitfall is to look for one radius that works for the whole union. Openness requires that for each point $x$ in the union, there be some positive radius $r$ such that $(x-r,x+r)$ lies in the union. The radius may depend on $x$. When using an indexed family, the practical first step is to identify a particular member containing $x$; one should not assume that every member contains $x$, or that all members offer the same radius.

The distinction between union and intersection is also important. The arbitrary-union theorem applies to families of any size. It does not say that arbitrary intersections of open sets are open. For example, the open intervals $(-1/n,1/n)$, for positive integers $n$, have intersection $\{0\}$, which is not open. The next tutorial examines why finite intersections behave differently from arbitrary intersections.

For any set $E$, the interior construction gives a systematic answer to the question, “Which points of $E$ can be kept while remaining inside an open set?” A point lies in $\operatorname{int}(E)$ precisely when it belongs to some open subset of $E$. The defining union gathers all such points, while its maximality ensures that no larger open subset of $E$ has been missed.

Check Your Understanding

Use the arbitrary-union theorem and the largest-open-subset property to answer these questions.

In an indexed union, what must be shown to establish that a point belongs to the union?
Why is $\operatorname{int}(E)$ open even if $E$ itself is not open?
If $A\subseteq B$, why must $\operatorname{int}(A)\subseteq\operatorname{int}(B)$?
For $E=[1,4)\cup\{7\}$, determine $\operatorname{int}(E)$ and explain why neither $1$ nor $7$ belongs to it.
Why does the arbitrary-union theorem not imply that an arbitrary intersection of open sets is open?

What You'll Learn