Arbitrary Intersections of Closed Sets

From Finite Open Conditions to Arbitrary Closed Constraints

In the previous tutorial, finitely many open conditions could be combined by taking a common positive neighborhood radius. For an infinite family, that minimum-radius argument may fail: positive radii can have infimum zero. Closed sets behave differently under intersections. There is no requirement to find one neighborhood that lies inside every set; instead, a point belongs to the intersection precisely when it belongs to every member of the family.

The Arbitrary Intersections of Closed Sets Theorem was established earlier in this course. We will use it rather than re-prove it. The aim here is to understand how it supports constructions: an intersection can impose infinitely many closed constraints at once, and the intersection of all closed sets containing a given set produces its smallest closed enlargement.

Arbitrary Intersections as Simultaneous Constraints

Let $\{F_\alpha:\alpha\in A\}$ be a family of subsets of $\mathbb{R}$, indexed by a set $A$. Its intersection is

\bigcap_{\alpha\in A}F_\alpha = \{x\in\mathbb{R}:x\in F_\alpha\text{ for every }\alpha\in A\}.

The index set $A$ may be finite, countably infinite, or uncountable. A point fails to belong to this intersection as soon as one set in the family excludes it. The intersection of an empty family is taken to be $\mathbb{R}$, which is closed.

Theorem (Arbitrary Intersections of Closed Sets, recalled): If $\{F_\alpha:\alpha\in A\}$ is any family of closed subsets of $\mathbb{R}$, then $\bigcap_{\alpha\in A}F_\alpha$ is closed. The empty-family intersection is $\mathbb{R}$.

This theorem applies whether or not the intersection is empty. Closedness does not imply nonemptiness: a family can impose incompatible constraints. Nor does the theorem require that the sets be nested or that there be only countably many of them. Its basic use is direct: identify each condition as a closed set, then take their intersection to describe the points satisfying all conditions.

Worked Example: Infinitely Many Bounds Determine a Closed Interval

For each positive integer $n$, define

F_n=\left[-\frac{1}{n},\,2+\frac{1}{n}\right].

Each $F_n$ is a closed interval. We claim that their intersection is $[0,2]$. First, if $x\in[0,2]$, then $-1/n\leq 0\leq x$ and $x\leq 2\leq 2+1/n$ for every positive integer $n$. Thus $x\in F_n$ for every $n$, so $x\in\bigcap_{n=1}^{\infty}F_n$.

Conversely, suppose $x\in\bigcap_{n=1}^{\infty}F_n$. If $x<0$, then $-x>0$. Choose a positive integer $n>1/(-x)$, using the Archimedean property of the real numbers. Then $1/n<-x$, so $-1/n>x$. This contradicts $x\in F_n$, which requires $-1/n\leq x$. Therefore $x\geq0$. If $x>2$, choose a positive integer $n>1/(x-2)$. Then $1/n<x-2$, so $2+1/n<x$, again contradicting $x\in F_n$. Hence $x\leq2$, proving

\bigcap_{n=1}^{\infty}\left[-\frac{1}{n},\,2+\frac{1}{n}\right]=[0,2].

The endpoints are retained: $0$ and $2$ belong to every interval in the family. The interval theorem ensures the intersection is closed, while the inequalities identify exactly which points it contains.

Worked Example: A Closed Intersection Can Be Empty

For each positive integer $n$, let $G_n=[n,\infty)$. Each $G_n$ is closed. If a real number $x$ belonged to every $G_n$, it would satisfy $x\geq n$ for every positive integer $n$. But the Archimedean property gives a positive integer $n>x$, contradicting $x\geq n$. Thus

\bigcap_{n=1}^{\infty}[n,\infty)=\varnothing.

The empty set is closed, so this conclusion is consistent with the arbitrary-intersection theorem. The example also distinguishes two questions: the theorem guarantees that the intersection is closed, but it does not guarantee that any point satisfies all the constraints.

The Smallest Closed Set Containing a Given Set

Given a set $A\subseteq\mathbb{R}$, there may be many closed sets containing it. For example, $\mathbb{R}$ always contains $A$, but it may be much larger than necessary. Intersecting all closed supersets of $A$ gives a natural candidate for the smallest one. This construction works because the family being intersected is nonempty: $\mathbb{R}$ is one of its members.

Definition (Closure): For $A\subseteq\mathbb{R}$, the closure of $A$, denoted $\overline{A}$, is the intersection of all closed subsets of $\mathbb{R}$ that contain $A$: $$ \overline{A} = \bigcap\{F\subseteq\mathbb{R}:F\text{ is closed and }A\subseteq F\}. $$

Theorem (The Closure Is the Smallest Closed Superset): For every $A\subseteq\mathbb{R}$, the set $\overline{A}$ is closed, contains $A$, and is contained in every closed set that contains $A$.

Proof. Consider the family of all closed sets containing $A$. It is nonempty because it includes $\mathbb{R}$. By the Arbitrary Intersections of Closed Sets Theorem, its intersection $\overline{A}$ is closed. Every member $F$ of this family contains $A$; consequently, every point of $A$ belongs to every such $F$, and hence $A\subseteq\overline{A}$. Finally, if $F$ is any closed set with $A\subseteq F$, then $F$ is one of the sets being intersected. An intersection is contained in each of its members, so $\overline{A}\subseteq F$. These statements prove all three claims. $\square$

The phrase “smallest closed superset” refers to inclusion, not to length or numerical size. The theorem says that every closed set containing $A$ must also contain $\overline{A}$. It therefore gives a way to prove an inclusion involving a closure: it is enough to find a closed set that contains the original set.

Worked Example: The Closure of an Open Interval

Let $A=(2,5)$. The closed interval $[2,5]$ contains $A$, so the smallest-closed-superset theorem gives $\overline{A}\subseteq[2,5]$. To see that no smaller closed set can contain $A$, let $F$ be any closed set with $(2,5)\subseteq F$. For each positive integer $n\geq2$, the number $2+1/n$ lies in $(2,5)$, so $2+1/n\in F$, and $2+1/n\to2$. Also, $5-1/n\in(2,5)$, so $5-1/n\in F$, and $5-1/n\to5$.

By the sequential characterization of closed sets, a closed set contains the limit of every convergent sequence of its points. Thus $2,5\in F$, and therefore $[2,5]\subseteq F$. This holds for every closed $F$ containing $A$, so $[2,5]\subseteq\overline{A}$. Combining the inclusions yields

\overline{(2,5)}=[2,5].

This example shows how the closure construction adds precisely the endpoints required for a closed superset.

A Neighborhood Test for Membership in the Closure

The intersection definition describes closure globally, by referring to every closed set containing $A$. There is also a local test: a point belongs to the closure exactly when every open ball centered at that point meets $A$. In $\mathbb{R}$, the open ball of radius $r>0$ centered at $x$ is $B_r(x)=(x-r,x+r)$.

Theorem (Neighborhood Characterization of Closure): For $A\subseteq\mathbb{R}$ and $x\in\mathbb{R}$, $$ x\in\overline{A} \quad\Longleftrightarrow\quad B_r(x)\cap A\neq\varnothing\text{ for every }r>0. $$

Proof. Suppose first that $x\in\overline{A}$. If some $r>0$ had $B_r(x)\cap A=\varnothing$, then every point of $A$ would lie outside $B_r(x)$. The complement $\mathbb{R}\setminus B_r(x)$ is closed and contains $A$, so the smallest-closed-superset theorem gives $\overline{A}\subseteq\mathbb{R}\setminus B_r(x)$. But $x\in B_r(x)$, contradicting $x\in\overline{A}$. Hence every such ball meets $A$.

For the reverse implication, suppose every ball $B_r(x)$ meets $A$, but $x\notin\overline{A}$. The closure is closed, so its complement is open. Since $x\in\mathbb{R}\setminus\overline{A}$, openness gives some $r>0$ such that $B_r(x)\subseteq\mathbb{R}\setminus\overline{A}$. Since $A\subseteq\overline{A}$, this ball is disjoint from $A$, contradicting the assumption that every ball meets $A$. Therefore $x\in\overline{A}$, completing the proof. $\square$

Worked Example: The Closure of a Set with a Missing Endpoint

Let $A=[-1,0)\cup(0,1]$. Every point of $A$ belongs to $\overline{A}$, since $A\subseteq\overline{A}$. The missing point $0$ also belongs to the closure: for every $r>0$, choose $y=\min(r/2,1/2)$. Then $0<y<1$, so $y\in A$, and $|y-0|=y<r$. Thus every ball around $0$ meets $A$.

If $x>1$, take $r=(x-1)/2>0$. Every $z\in B_r(x)$ satisfies $z>x-r=(x+1)/2>1$, so $B_r(x)\cap A=\varnothing$. If $x<-1$, take $r=(-1-x)/2>0$. Every $z\in B_r(x)$ satisfies $z<x+r=(x-1)/2<-1$, again making the ball disjoint from $A$. These cases account for all points outside $[-1,1]$. The neighborhood characterization therefore gives

\overline{[-1,0)\cup(0,1]}=[-1,1].

The test works at the included endpoints as well: they already lie in $A$, hence in its closure. The only new point needed to fill the closed interval is $0$.

Closure Properties and a Useful Caution

The smallest-closed-superset description immediately yields three basic properties. First, $A\subseteq\overline{A}$. Second, if $A\subseteq B$, then $\overline{A}\subseteq\overline{B}$: the set $\overline{B}$ is closed and contains $A$, so the minimality of $\overline{A}$ applies. Third, $\overline{\overline{A}}=\overline{A}$, because $\overline{A}$ is already closed. Its own smallest closed superset is itself: it is one closed superset of itself, and every set contains itself.

A common pitfall is to infer that every infinite intersection of closed sets is nonempty. The family $[n,\infty)$ disproves that inference. Another is to confuse the closure of a set with the set itself: a set need not be closed, and the closure may add points, as the open-interval example shows. The neighborhood test identifies exactly which points are added: every neighborhood of such a point must meet the original set.

Express the constraints as sets.
Write each required condition as membership in a set, and check that each set is closed when closedness is needed.

Intersect the family.
The intersection consists of the points satisfying every condition; it is closed by the arbitrary-intersection theorem.

Use closure when starting from an arbitrary set.
Intersect all closed supersets to obtain the smallest closed set containing it, or use the neighborhood test to check whether a particular point belongs.

Check Your Understanding

Use the arbitrary-intersection theorem and the closure results to answer the following questions.

What condition must a point satisfy to belong to $\bigcap_{\alpha\in A}F_\alpha$?
Why does the arbitrary-intersection theorem not imply that an intersection of closed sets is nonempty?
State the definition of $\overline{A}$ in terms of closed supersets of $A$.
Why is $\overline{A}$ contained in every closed set that contains $A$?
Use the neighborhood characterization to describe what it means for $x$ to belong to $\overline{A}$.
For $F_n=[-1/n,2+1/n]$, why must a point in every $F_n$ lie between $0$ and $2$?