Closure as the Smallest Closed Superset

Closure as a Least Closed Set

In the previous tutorial, closed sets were connected to the limit points they contain. That criterion helps decide whether a particular set is closed. Closure answers a related but different question: if a set is not closed, what is the least amount that must be added to make it closed? The answer can be used not only to identify missing limit points, but also to compare sets and organize unions.

Recall the Theorem (The Closure Is the Smallest Closed Superset), established earlier in this course. For every $A\subseteq\mathbb{R}$, its closure $\overline A$ is closed, contains $A$, and is contained in every closed set that contains $A$. Equivalently, it is the intersection of all closed supersets of $A$. We will use this property, rather than re-prove it, to establish further laws of closure.

Recall (Smallest Closed Superset): The closure $\overline A$ is a closed set containing $A$, and if $F$ is closed and $A\subseteq F$, then $\overline A\subseteq F$.

“Smallest” here means smallest by inclusion, not smallest in length or number of elements. There may be many closed sets containing $A$, but each of them must contain $\overline A$. This gives a practical strategy: when a convenient closed set $F$ contains $A$, the universal property immediately gives $\overline A\subseteq F$. To prove equality, one must also show that $F\subseteq\overline A$.

Two Basic Laws of Closure

The least-superset description implies that inclusion between sets is preserved by closure. It also implies that once a set has been closed, closing it again adds nothing. These are structural consequences of the universal property, rather than separate ways to define closure.

Theorem (Monotonicity and Idempotence of Closure): If $A\subseteq B\subseteq\mathbb{R}$, then $\overline A\subseteq\overline B$. For every $A\subseteq\mathbb{R}$, $\overline{\overline A}=\overline A$.

Proof. Suppose $A\subseteq B$. The set $\overline B$ is closed and contains $B$, so it also contains $A$. By the Smallest Closed Superset property, $\overline A\subseteq\overline B$. This proves monotonicity.

For idempotence, $\overline A$ is closed. It is therefore itself a closed set containing $\overline A$. Applying the smallest-superset property to the set $\overline A$ gives

\overline{\overline A}\subseteq\overline A.

On the other hand, every set is contained in its closure, so applying that fact to $\overline A$ gives $\overline A\subseteq\overline{\overline A}$. The two inclusions prove equality. $\square$

Monotonicity says that enlarging the original set cannot make its closure smaller. Idempotence says that closure is a one-step operation: after all necessary points have been added, repeating the operation has no further effect. Both laws are useful when an expression contains several nested closures or when one set is compared with another.

Worked Example: Bounding a Closure with a Closed Superset

Let $A=(2,5)$ and $F=[1,6]$. The interval $F$ is closed, and $A\subseteq F$. The Smallest Closed Superset property therefore gives

\overline{(2,5)}\subseteq[1,6].

This is a valid bound, but it does not identify the closure exactly. In fact, $\overline{(2,5)}=[2,5]$. To verify the equality, first note that $[2,5]$ is closed and contains $(2,5)$, so $\overline{(2,5)}\subseteq[2,5]$. Next, every point $x\in(2,5)$ belongs to $\overline{(2,5)}$. The endpoints belong as well: for every $r>0$, the point $2+\min\{r/2,1\}$ lies in $(2,5)$ and is less than distance $r$ from $2$; likewise, $5-\min\{r/2,1\}$ lies in $(2,5)$ and is less than distance $r$ from $5$. By the Neighborhood Characterization of Closure, both endpoints belong to the closure. Thus $[2,5]\subseteq\overline{(2,5)}$, proving equality.

The larger closed set $[1,6]$ gives a useful outer bound, but it is not the smallest such set. This illustrates why containment in a closed superset proves an inclusion, not automatically equality.

Worked Example: Comparing Closures by Inclusion

Let $A=(0,1)$ and $B=[-2,3)$. Since $A\subseteq B$, monotonicity gives $\overline A\subseteq\overline B$. We can identify the two closures directly: $\overline A=[0,1]$, while $\overline B=[-2,3]$. Hence

[0,1]\subseteq[-2,3].

The endpoint $3$ belongs to $\overline B$, although it is not in $B$: every ball centered at $3$ meets $B$. The same is true of $-2$, since every ball centered there meets $B$. The set $[-2,3]$ is closed and contains $B$, and points outside it have a neighborhood disjoint from $B$; these facts verify the stated closure. The example shows that monotonicity does not assert that the closures are equal. It asserts only the inclusion forced by $A\subseteq B$.

Closure and Arbitrary Unions

For finitely many sets, the closure of their union is the union of their closures, as established earlier in the course in the Theorem (Closure of a Finite Union). It is tempting to expect the same equality for an arbitrary family. The smallest-closed-superset property gives a related equality, but the order of operations matters: take the union of the individual closures, then close that union.

Theorem (Closure of a Union and the Closures of Its Members): For any family $\{A_\alpha:\alpha\in I\}$ of subsets of $\mathbb{R}$, $$ \overline{\bigcup_{\alpha\in I}A_\alpha} = \overline{\bigcup_{\alpha\in I}\overline{A_\alpha}}. $$

Proof. Write $A=\bigcup_{\alpha\in I}A_\alpha$ and $C=\overline A$. If $I$ is empty, both unions are empty and both sides equal $\overline{\varnothing}=\varnothing$. Now suppose $I$ is nonempty. For each $\alpha\in I$, $A_\alpha\subseteq A$. By monotonicity, $\overline{A_\alpha}\subseteq\overline A=C$. Thus

\bigcup_{\alpha\in I}\overline{A_\alpha}\subseteq C.

Since $C$ is closed, the Smallest Closed Superset property implies

\overline{\bigcup_{\alpha\in I}\overline{A_\alpha}}\subseteq C =\overline{\bigcup_{\alpha\in I}A_\alpha}.

For the reverse inclusion, $A_\alpha\subseteq\overline{A_\alpha}$ for every $\alpha$, so

\bigcup_{\alpha\in I}A_\alpha \subseteq \bigcup_{\alpha\in I}\overline{A_\alpha}.

Monotonicity of closure now gives

\overline{\bigcup_{\alpha\in I}A_\alpha} \subseteq \overline{\bigcup_{\alpha\in I}\overline{A_\alpha}}.

Together, the two inclusions prove the equality. $\square$

This identity says that closing each member of a family before taking the union does not change the final result, provided the union is closed afterward. It does not say that the union of the individual closures is already closed. With infinitely many sets, new limit points can arise as the sets vary, even when each individual closure is just that set.

Worked Example: A New Limit Point from an Infinite Union

For each $n\in\mathbb{N}_0$, let $A_n=\{1/(n+1)\}$. Each $A_n$ is a singleton and is closed, so $\overline{A_n}=A_n$. Consequently,

\bigcup_{n=0}^{\infty}\overline{A_n} = \left\{1,\frac12,\frac13,\frac14,\ldots\right\}.

This union does not contain $0$. However, $0$ is in the closure of the union: given $r>0$, choose a positive integer $n$ such that $n+1>1/r$. Then $1/(n+1)<r$, so the union has a point within distance $r$ of $0$. Thus

0\in\overline{\bigcup_{n=0}^{\infty}A_n} \quad\text{but}\quad 0\notin\bigcup_{n=0}^{\infty}\overline{A_n}.

In particular, the closure of an infinite union need not equal the union of the individual closures. The extra point $0$ is accounted for when the latter union is closed, exactly as in the theorem.

Using the Least-Superset Property Carefully

The smallest-closed-superset viewpoint is most useful when the goal is an inclusion. To show $\overline A\subseteq F$, it is enough to show that $F$ is closed and $A\subseteq F$. To show $F\subseteq\overline A$, one needs a separate argument, often using the Neighborhood Characterization of Closure or the Sequential Characterization of Closure. An outer closed bound alone cannot establish equality.

A second common pitfall is to assume arbitrary unions of closed sets are closed. They need not be: in the example above, every $A_n$ is closed but their union omits its limit point $0$. This is consistent with the fact that arbitrary intersections of closed sets are closed; the behavior of unions is different. Finite unions of closed sets are closed, but an infinite union may acquire limit points not included in the union.

The principal consequences to keep available are:

If $A\subseteq B$, then $\overline A\subseteq\overline B$.
Taking closure twice has the same effect as taking it once.
The closure of an arbitrary union equals the closure of the union of the individual closures.
A closed superset gives an upper bound for the closure, while equality requires proving the reverse inclusion too.

Find a closed superset.
Choose a closed set $F$ that contains the set $A$ under consideration.

Use minimality.
Conclude $\overline A\subseteq F$ from the Smallest Closed Superset property.

Check the other inclusion.
If equality is needed, prove $F\subseteq\overline A$, for example by checking that every neighborhood of each point of $F$ meets $A$.

Check Your Understanding

Use the smallest-closed-superset property and the results proved here to answer the following questions.

If $A\subseteq B$, what inclusion must hold between $\overline A$ and $\overline B$, and which property proves it?
Why does $\overline{\overline A}=\overline A$?
If $A\subseteq F$ and $F$ is closed, what can be concluded about $\overline A$? Does this alone prove equality?
For an arbitrary family $\{A_\alpha\}$, which set must be closed in order to obtain the equality relating the closure of the union to the closures of its members?
In the example $A_n=\{1/(n+1)\}$, why is $0$ in the closure of the union but not in the union of the individual closures?

Closure as the Smallest Closed Superset

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