Open and Closed Sets: Definitions, Properties and Examples

Definition of open set

The notion of open set rests on the concept of neighborhood. Let \(A\subseteq\mathbb R\). We say that \(A\) is an open set if for every point \(x_0\in A\) there exists a real number \(r>0\) such that

\[ (x_0-r,x_0+r)\subseteq A. \]

In other words, a set is open if each of its points has at least one neighborhood entirely contained in the set.

It is important to note that the radius \(r\) may depend on the chosen point. There need not be a single value of \(r\) that works for all points of the set; what matters is that, once an arbitrary point \(x_0\in A\) is fixed, there exists at least one neighborhood centred at \(x_0\) contained in \(A\).

The definition can also be expressed using quantifiers:

\[ A \text{ open} \iff \forall x_0\in A\ \exists r>0 \text{ such that } (x_0-r,x_0+r)\subseteq A. \]

Let us look at two immediate examples.

Consider the open interval

\[ A=(0,1). \]

Let \(x_0\in(0,1)\). Since \(x_0\) lies strictly between \(0\) and \(1\), the quantities

\[ x_0 \qquad\text{and}\qquad 1-x_0 \]

are both positive. We may therefore choose

\[ r=\frac12\min\{x_0,1-x_0\}. \]

With this choice we obtain

\[ (x_0-r,x_0+r)\subseteq(0,1). \]

Since the point \(x_0\) was arbitrary, the interval \((0,1)\) is an open set.

Now consider the set

\[ B=[0,1]. \]

This set is not open. Indeed, the point \(0\) belongs to \(B\), but no neighborhood of \(0\) is contained in \(B\).

In fact, for every \(r>0\), the neighborhood

\[ (-r,r) \]

contains negative numbers, which do not belong to \(B\). Consequently there is no radius \(r>0\) such that

\[ (-r,r)\subseteq[0,1]. \]

Hence the set \([0,1]\) is not open.

Examples of open sets

The definition of open set can be applied to many sets on the real line. In this section we will examine some of the most important examples.

Open intervals

Consider an open interval

\[ A=(a,b), \qquad a<b. \]

We want to show that \(A\) is an open set. Let \(x_0\in(a,b)\). Since \(x_0\) lies strictly between \(a\) and \(b\), the quantities

\[ x_0-a \qquad\text{and}\qquad b-x_0 \]

are both positive. Set

\[ r=\frac12\min\{x_0-a,\; b-x_0\}. \]

Then \(r>0\) and the neighborhood

\[ (x_0-r,x_0+r) \]

remains entirely contained in the interval \((a,b)\). Hence every point of the interval has a neighborhood contained in the set, and so \((a,b)\) is an open set.

Open half-lines

Now consider the half-line

\[ A=(a,+\infty). \]

Let \(x_0\in A\). Then \(x_0>a\), so the distance between \(x_0\) and the point \(a\) is positive. Choosing

\[ r=\frac{x_0-a}{2}, \]

we have \(r>0\) and

\[ x_0-r = \frac{x_0+a}{2} > a. \]

It follows that

\[ (x_0-r,x_0+r)\subseteq(a,+\infty). \]

Hence \((a,+\infty)\) is also an open set.

By an entirely analogous argument, one shows that the half-line

\[ (-\infty,b) \]

is an open set as well.

The set \(\mathbb R\)

The entire real line is itself an open set. Indeed, once an arbitrary point \(x_0\in\mathbb R\) is fixed, every neighborhood of the form

\[ (x_0-r,x_0+r), \qquad r>0, \]

is contained in \(\mathbb R\). Consequently \(\mathbb R\) satisfies the definition of open set.

The empty set

The empty set

\[ \varnothing \]

is also regarded as an open set. Indeed, the definition requires that every point of the set have a neighborhood contained in the set. Since the empty set contains no point, this condition is automatically satisfied.

For this reason \(\varnothing\) and \(\mathbb R\) are always open sets.

Definition of closed set

The notion of closed set is closely connected to that of open set. Let \(A\subseteq\mathbb R\). We say that \(A\) is a closed set if its complement

\[ \mathbb R\setminus A \]

is an open set.

In symbols:

\[ A \text{ closed} \iff \mathbb R\setminus A \text{ open}. \]

To check that a set is closed, it is therefore not necessary to work directly with the set itself; it is often simpler to study its complement and verify that it is open.

Consider, for instance, the interval

\[ [0,1]. \]

Its complement is

\[ \mathbb R\setminus[0,1] = (-\infty,0)\cup(1,+\infty). \]

Both half-lines are open and, as we will see later, the union of two open sets is again an open set. Consequently \(\mathbb R\setminus[0,1]\) is open, and therefore \([0,1]\) is a closed set.

Now consider the interval

\[ (0,1). \]

Its complement is

\[ \mathbb R\setminus(0,1) = (-\infty,0]\cup[1,+\infty). \]

This set is not open, since neither the point \(0\) nor the point \(1\) has a neighborhood entirely contained in the complement.

Hence \((0,1)\) is not a closed set.

In the following sections we will see a particularly important characterization of closed sets based on accumulation points.

Examples of closed sets

As we did for open sets, let us now examine some significant examples of closed sets.

Closed intervals

Consider the interval

\[ A=[a,b], \qquad a<b. \]

To establish that \(A\) is closed, it suffices to study its complement:

\[ \mathbb R\setminus A = (-\infty,a)\cup(b,+\infty). \]

The two half-lines \((-\infty,a)\) and \((b,+\infty)\) are open. Moreover, as we will see later, the union of open sets is again an open set.

Hence \(\mathbb R\setminus A\) is open, and so \([a,b]\) is a closed set.

Closed half-lines

Consider the half-line

\[ [a,+\infty). \]

Its complement is

\[ \mathbb R\setminus[a,+\infty) = (-\infty,a), \]

which is an open set.

Consequently \([a,+\infty)\) is a closed set.

By the same argument, one shows that the half-line

\[ (-\infty,b] \]

is a closed set as well.

Sets consisting of finitely many points

Consider a set consisting of a single point:

\[ A=\{a\}. \]

Its complement is

\[ \mathbb R\setminus\{a\} = (-\infty,a)\cup(a,+\infty). \]

Being the union of two open sets, it is open. Hence \(\{a\}\) is a closed set.

The same argument shows that every set consisting of finitely many points is a closed set.

The set \(\mathbb R\)

The entire real line is a closed set.

Indeed, its complement is the empty set:

\[ \mathbb R\setminus\mathbb R = \varnothing. \]

Since \(\varnothing\) is an open set, it follows that \(\mathbb R\) is closed.

The empty set

The empty set is a closed set as well.

Indeed,

\[ \mathbb R\setminus\varnothing = \mathbb R. \]

Since \(\mathbb R\) is an open set, it follows that \(\varnothing\) is closed.

We have thus obtained an interesting result: both \(\mathbb R\) and \(\varnothing\) are simultaneously open and closed.

Relationship between open and closed sets

Open sets and closed sets are defined in terms of one another: a set is closed if and only if its complement is open. One should not, however, think that the words "open" and "closed" are necessarily opposites as they are in everyday language.

Indeed, an open set may fail to be closed, a closed set may fail to be open, but there also exist sets that are simultaneously open and closed.

Sets that are open but not closed

The interval

\[ (0,1) \]

is an open set, as we have already shown.

However, it is not closed, since its complement

\[ (-\infty,0]\cup[1,+\infty) \]

is not open.

Hence \((0,1)\) is open but not closed.

Sets that are closed but not open

Consider the interval

\[ [0,1]. \]

We have seen that it is closed because its complement

\[ (-\infty,0)\cup(1,+\infty) \]

is open.

On the other hand, \([0,1]\) is not open, since neither the point \(0\) nor the point \(1\) has a neighborhood entirely contained in the set.

Hence \([0,1]\) is closed but not open.

Sets that are simultaneously open and closed

We have already observed that the empty set \(\varnothing\) is open and that its complement \(\mathbb R\) is open. Consequently \(\varnothing\) is also closed.

Likewise, \(\mathbb R\) is open and its complement \(\varnothing\) is open; hence \(\mathbb R\) is also closed.

The sets

\[ \varnothing \qquad\text{and}\qquad \mathbb R \]

are therefore simultaneously open and closed.

Sets that are neither open nor closed

Finally, there exist sets that are neither open nor closed.

One example is the interval

\[ (0,1]. \]

It is not open, since the point \(1\) has no neighborhood entirely contained in the set.

Moreover, it is not closed, since its complement

\[ (-\infty,0]\cup(1,+\infty) \]

is not open.

Hence \((0,1]\) is neither open nor closed.

In conclusion, the properties of being open and being closed are independent: a set may possess just one of the two properties, both, or neither.

Characterization via accumulation points

One of the most important characterizations of closed sets involves the concept of accumulation point. It allows us to recognize whether a set is closed by looking solely at the location of its accumulation points.

Recall that a point \(x_0\in\mathbb R\) is called an accumulation point of a set \(A\subseteq\mathbb R\) if every deleted neighborhood of \(x_0\) contains at least one point of \(A\).

The following fundamental theorem then holds.

Theorem. A set \(A\subseteq\mathbb R\) is closed if and only if it contains all of its accumulation points.

In symbols:

\[ A \text{ closed} \iff A'\subseteq A, \]

where \(A'\) denotes the derived set of \(A\), that is, the set of all accumulation points of \(A\).

Proof. Suppose first that \(A\) is closed and let \(x_0\in A'\). We want to prove that \(x_0\in A\).

We argue by contradiction and suppose that \(x_0\notin A\).

Since \(A\) is closed, the complement \(\mathbb R\setminus A\) is open. As \(x_0\in\mathbb R\setminus A\), there exists a neighborhood

\[ (x_0-r,x_0+r) \subseteq \mathbb R\setminus A. \]

This neighborhood contains no point of \(A\), contradicting the fact that \(x_0\) is an accumulation point of \(A\).

Hence we must have \(x_0\in A\), and therefore

\[ A'\subseteq A. \]

Let us now prove the converse. Suppose that

\[ A'\subseteq A \]

and consider a point

\[ x_0\in\mathbb R\setminus A. \]

Since \(x_0\notin A\) and all accumulation points belong to \(A\), the point \(x_0\) cannot be an accumulation point of \(A\).

By the definition of accumulation point, there then exists a radius \(r>0\) such that the deleted neighborhood

\[ (x_0-r,x_0+r)\setminus\{x_0\} \]

contains no point of \(A\).

Since moreover \(x_0\notin A\), it follows that the whole interval

\[ (x_0-r,x_0+r) \]

is contained in the complement \(\mathbb R\setminus A\).

We have thus shown that every point of \(\mathbb R\setminus A\) has a neighborhood contained in the complement. Consequently \(\mathbb R\setminus A\) is open.

Hence \(A\) is closed.

Geometric interpretation

The theorem states that a set is closed precisely when it does not "leave out" any of its own accumulation points.

For example, the interval

\[ [0,1] \]

contains all of its accumulation points and is therefore closed.

By contrast, the interval

\[ (0,1) \]

does not contain the accumulation points \(0\) and \(1\). Consequently it is not closed.

This characterization is often the simplest method for deciding whether a set is closed.

Properties of open sets

Open sets enjoy important closure properties that allow new open sets to be built from open sets already known.

In particular, the arbitrary union of open sets is again an open set, while the intersection of finitely many open sets is again an open set.

Union of open sets

Let \(\{A_i\}_{i\in I}\) be a family of open sets. Then

\[ \bigcup_{i\in I}A_i \]

is an open set.

Proof. Set

\[ A=\bigcup_{i\in I}A_i \]

and let \(x_0\in A\).

By the definition of union, there exists at least one index \(j\in I\) such that

\[ x_0\in A_j. \]

Since \(A_j\) is open, there exists a radius \(r>0\) such that

\[ (x_0-r,x_0+r)\subseteq A_j. \]

As \(A_j\subseteq A\), it follows that

\[ (x_0-r,x_0+r)\subseteq A. \]

We have thus shown that every point of \(A\) has a neighborhood contained in \(A\). Hence \(A\) is open.

Finite intersection of open sets

Let \(A_1,A_2,\ldots,A_n\) be open sets. Then

\[ A_1\cap A_2\cap\cdots\cap A_n \]

is an open set.

Proof. Set

\[ A=A_1\cap A_2\cap\cdots\cap A_n \]

and let \(x_0\in A\).

Then

\[ x_0\in A_1,\quad x_0\in A_2,\quad \ldots,\quad x_0\in A_n. \]

Since each set is open, there exist positive radii

\[ r_1,r_2,\ldots,r_n \]

such that

\[ (x_0-r_k,x_0+r_k)\subseteq A_k, \qquad k=1,\ldots,n. \]

Set

\[ r=\min\{r_1,r_2,\ldots,r_n\}. \]

Then \(r>0\) and

\[ (x_0-r,x_0+r) \subseteq A_1\cap A_2\cap\cdots\cap A_n. \]

Hence \(A\) is open.

Why can an infinite intersection fail to be open?

The preceding property cannot be extended to infinite intersections.

Consider, in fact, the family of open intervals

\[ A_n= \left( -\frac1n, \frac1n \right), \qquad n\in\mathbb N. \]

Each \(A_n\) is open.

However,

\[ \bigcap_{n=1}^{\infty} \left( -\frac1n, \frac1n \right) = \{0\}. \]

The set \(\{0\}\) is not open, since no neighborhood of \(0\) is contained in \(\{0\}\).

This example shows that the intersection of an infinite family of open sets may fail to be open.

In summary:

\[ \boxed{ \begin{aligned} &\text{The arbitrary union of open sets is open;}\\[4pt] &\text{the finite intersection of open sets is open.} \end{aligned} } \]

Properties of closed sets

The properties of closed sets are dual to those of open sets. In particular, the arbitrary intersection of closed sets is again a closed set, while the union of finitely many closed sets is again a closed set.

Arbitrary intersection of closed sets

Let \(\{A_i\}_{i\in I}\) be a family of closed sets. Then

\[ \bigcap_{i\in I}A_i \]

is a closed set.

Proof. Set

\[ A=\bigcap_{i\in I}A_i. \]

Using De Morgan's laws we obtain

\[ \mathbb R\setminus A = \mathbb R\setminus \left( \bigcap_{i\in I}A_i \right) = \bigcup_{i\in I} \left( \mathbb R\setminus A_i \right). \]

Since each set \(A_i\) is closed, the complement \(\mathbb R\setminus A_i\) is open.

Moreover, the arbitrary union of open sets is open.

Hence \(\mathbb R\setminus A\) is open, and so \(A\) is closed.

Finite union of closed sets

Let \(A_1,A_2,\ldots,A_n\) be closed sets. Then

\[ A_1\cup A_2\cup\cdots\cup A_n \]

is a closed set.

Proof. Set

\[ A=A_1\cup A_2\cup\cdots\cup A_n. \]

Applying De Morgan's laws once more we obtain

\[ \mathbb R\setminus A = \mathbb R\setminus \left( A_1\cup A_2\cup\cdots\cup A_n \right) = (\mathbb R\setminus A_1) \cap (\mathbb R\setminus A_2) \cap \cdots \cap (\mathbb R\setminus A_n). \]

Since each complement \(\mathbb R\setminus A_k\) is open and the finite intersection of open sets is open, it follows that \(\mathbb R\setminus A\) is open.

Hence \(A\) is closed.

Why can an infinite union fail to be closed?

The preceding property cannot be extended to infinite unions.

Consider, in fact, the sets

\[ A_n= \left[ \frac1n, 1 \right], \qquad n\in\mathbb N. \]

Each \(A_n\) is a closed interval.

Their union is

\[ \bigcup_{n=1}^{\infty} \left[ \frac1n, 1 \right] = (0,1]. \]

The set \((0,1]\) is not closed, since the point \(0\) is an accumulation point that does not belong to the set.

This example shows that the union of an infinite family of closed sets may fail to be closed.

In summary:

\[ \boxed{ \begin{aligned} &\text{The arbitrary intersection of closed sets is closed;}\\[4pt] &\text{the finite union of closed sets is closed.} \end{aligned} } \]