2.6: Open Sets, Closed Sets, Compact Sets, and Limit Points

The open ball in \(\mathbb{R}\) with center \(a \in \mathbb{R}\) and radius \(\delta>0\) is the set

\[B(a ; \delta)=(a-\delta, a+\delta).\]

Definition \(\PageIndex{1}\)

A subset \(A\) of \(\mathbb{R}\) is said to be open if for each \(a \in A\), there exists \(\delta>0\) such that

\[B(a ; \delta) \subset A.\]

Example \(\PageIndex{1}\)

1. Any open interval \(A=(c, d)\) is open. Indeed, for each \(a \in A\), one has \(c<a<d\).
2. The sets \(A=(-\infty, c)\) and \(B=(c, \infty)\) are open, but the \(C=[c, \infty)\) is not open.

Solution

1. Let

\[\delta=\min \{a-c, d-a\}.\]

Then

\[B(a ; \delta)=(a-\delta, a+\delta) \subset A.\]

Therefore, \(A\) is open.

2. The reader can easily verify that A and B are open. Let us show that \(C\) is not open. Assume by contradiction that \(C\) is open. Then, for the element \(c \in C\), there exists \(\delta>0\) such that

\[B(c ; \delta)=(c-\delta, c+\delta) \subset C.\]

However, this is a contradiction because \(c-\delta / 2 \in B(c ; \delta)\), but \(c-\delta / 2 \notin C\).

Theorem \(\PageIndex{1}\)

The following hold:

1. The subsets \(\emptyset\) and \(\mathbb{R}\) are open.
2. The union of nay collection of open subsets of \(\mathbb{R}\) is open.
3. The intersection of a finite number of open subsets of \(\mathbb{R}\) is open.

Proof

The proof of (a) is straightforward.

(b) Suppose \(\left\{G_{\alpha}: \alpha \in I\right\}\) is an arbitrary collection of open subsets of \(\mathbb{R}\). That means \(G_{\alpha}\) is open for every \(\alpha \in I\). Let us show that the set

\[G=\bigcup_{\alpha \in I} G_{\alpha}\]

is open. Take any \(a \in G\). Then there exists \(\alpha_{0} \in I\) such that

\[a \in G_{\alpha_{0}}.\]

Since \(G_{\alpha_{0}}\) is open, there exists \(\delta>0\) such that

\[B(a ; \delta) \subset G_{\alpha_{0}}\]

This implies

\[B(a ; \delta) \subset G\]

because \(G_{\alpha_{0}} \subset G\). Thus, \(G\) is open.

(c) SUppose \(G_{i}, i=1, \ldots, n\), are open subsets of \(\mathbb{R}\). Let us show that the set

\[G=\bigcap_{i=1}^{n} G_{i}\]

is also open. Take any \(a \in G\). Then \(a \in G_{i}\) for \(i=1, \ldots, n\). Since each \(G_{i}\) is open, there exists\(\delta_{i}>0\) such that

\[B\left(a ; \delta_{i}\right) \subset G_{i}.\]

Let \(\delta=\min \left\{\delta_{i}: i=1, \ldots, n\right\}\). Then \(\delta>0\) and

\[B(a ; \delta) \subset G.\]

Thus, \(G\) is open. \(\square\)

Definition \(\PageIndex{2}\)

A subset \(S\) of \(\mathbb{R}\) is called closed if its complement, \(S^{c}=\mathbb{R} \backslash S\), is open.

Example \(\PageIndex{2}\)

The sets \([a, b]\), \((-\infty, a]\), and \([a, \infty)\) are closed.

Solution

Indeed, \((-\infty, a]^{c}=(a, \infty)\) and \([a, \infty)^{c}=(-\infty, a)\) which are open by Example 2.6.1. Since \([a, b]^{c}=(-\infty, a) \cup(b, \infty)\), \([a, b]^{c}\) is open by Theorem 2.6.1. Also, single element sets are closed since, say, \(\{b\}^{c}=(-\infty, b) \cup(b, \infty)\).

Theorem \(\PageIndex{2}\)

The following hold:

1. The sets \(\emptyset\) and \(\mathbb{R}\) are closed.
2. The intersection of any collection of closed subsets of \(\mathbb{R}\) is closed.
3. The union of a finite number of closed subsets of \(\mathbb{R}\) is closed.

Proof

The proofs for these are simple using the De Morgan's law. Let us prove, for instance, (b). Let \(\left\{S_{\alpha}: \alpha \in I\right\}\) be a collection of closed sets. We will prove that the set

\[S=\bigcap_{\alpha \in I} S_{\alpha}\]

is also closed. We have

\[S^{c}=\left(\bigcap_{\alpha \in I} S_{\alpha}\right)^{c}=\bigcup_{\alpha \in I} S_{\alpha}^{c}.\]

Thus, \(S^{c}\) is open because it is a union of opens sets in \(\mathbb{R}\) (Theorem 2.6.1(b)). Therefore, \(S\) is closed. \(\square\)

Example \(\PageIndex{3}\)

It follows from part (c) and Example 2.6.2 that any finite set is closed.

Solution

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Theorem \(\PageIndex{3}\)

A subset \(A\) of \(\mathbb{R}\) is closed if and only if for any sequence \(\left\{a_{n}\right\}\) in \(A\) that converges to a point \(a \in \mathbb{R}\), it follows that \(a \in A\).

Proof

Suppose \(A\) is a closed subset of \(\mathbb{R}\) and \(\left\{a_{n}\right\}\) is a sequence in \(A\) that converges to \(a\). Suppose by contradiction that \(a \notin A\). Since \(A\) is closed, there exists \(\varepsilon>0\) such that \(B(a ; \varepsilon)=(a-\varepsilon, a+\varepsilon) \subset A^{c}\). Since \(\left\{a_{n}\right\}\) converges to \(a\), there exists \(N \in \mathbb{N}\) such that

\[a-\varepsilon<a_{N}<a+\varepsilon.\]

This implies \(a_{N} \in A^{c}\), a contradiction.

Let us now prove the converse. Suppose by contradiction that \(A\) is not closed. Then \(A^{c}\) is not open. SInce \(A^{c}\) is not open, there exists \(a \in A^{c}\) such that for any \(\varepsilon>0\), one has \(B(a ; \varepsilon) \cap A \neq \emptyset\). In particular, for such an \(a\) and for each \(n \in \mathbb{N}\), there exists \(a_{n} \in B\left(a ; \frac{1}{n}\right) \cap A\). It is clear that the sequence \(\left\{a_{n}\right\}\) is in \(A\) and it is convergent to \(a\) (because \(\left|a_{n}-a\right|<\frac{1}{n}\), for all \(n \in \mathbb{N}\)). This is a contradiction since \(a \notin A\). Therefore, \(A\) is closed. \(\square\)

Theorem \(\PageIndex{4}\)

If \(A\) is a nonempty subset of \(\mathbb{R}\) that is closed and bounded above, then \(\max A\) exists. Similarly, if \(A\) is a nonempty subset of \(\mathbb{R}\) that is closed and bounded below, then \(\min A\) exists

Proof

Let \(A\) be a nonempty closed set that is bounded above. Then \(\sup A\) exists. Let \(m = \sup A\). To complete the proof, we will show that \(m \in A\). Assume by contradiction that \(m \notin A\). then \(m \in A^{c}\), which is an open set. So there exists \(\delta>0\) such that

\[(m-\delta, m+\delta) \subset A^{c}.\]

This means there exists no \(a \in A\) with

\[m-\delta<a \leq m.\]

This contradicts the fact that \(m\) is the least upper bound of \(A\) (see Proposition 1.5.1). Therefore, \(max A\) exists. \(square\)

Definition \(\PageIndex{3}\)

A subset \(A\) of \(\mathbb{R}\) is called compact if for every sequence \(\left\{a_{n}\right\}\) in \(A\), there exists a subsequence \(\left\{a_{n_{k}}\right\}\) that converges to a point \(a \in A\).¹

Example \(\PageIndex{4}\)

Let \(a, b \in \mathbb{R}\), \(a \leq b\). We show that the set \(A=[a, b]\) is compact. Let \(\left\{a_{n}\right\}\) be a sequence in \(A\). Since \(a \leq a_{n} \leq b\) for all \(n\), then the sequence is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.4.1), we can obtain a convergent subsequence \(\left\{a_{n_{k}}\right\}\). Say, \(\lim _{k \rightarrow \infty} a_{n_{k}}=s\). We now must show that \(s \in A\). Since \(a \leq a_{n_{k}} \leq b\) for all \(k\), it follows from Theorem 2.1.5, that \(a \leq s \leq b\) and, hence, \(s \in A\) as desired. We conclude that \(A\) is compact.

Solution

Let \(\left\{a_{n}\right\}\) be a sequence in \(A\). Since \(a \leq a_{n} \leq b\) for all \(n\), then the sequence is bounded. By the Bolzano-Weierstrass theorem (Theorem 2.4.1), we can obtain a convergent subsequence \(\left\{a_{n_{k}}\right\}\). Say, \(\lim _{k \rightarrow \infty} a_{n_{k}}=s\). We now must show that \(s \in A\). Since \(a \leq a_{n_{k}} \leq b\) for all \(k\), it follows from Theorem 2.1.5, that \(a \leq s \leq b\) and, hence, \(s \in A\) as desired. We conclude that \(A\) is compact.

Theorem \(\PageIndex{5}\)

A subset \(A\) of \(\mathbb{R}\) is compact if and only if it is closed and bounded.

Proof

Suppose \(A\) is a compact subset of \(\mathbb{R}\). Let us first show that \(A\) is bounded. Suppose, by contradiction, that \(A\) is not bounded. Then for every \(n \in \mathbb{N}\), there exists \(a_{n} \in A\) such that

\[\left|a_{n}\right| \geq n.\]

Since \(A\) is compact, there exists a subsequence \(\left\{a_{n_{k}}\right\}\) that converges to some \(a \in A\) that converges to some \(a \in A\). Then

\[\left|a_{n_{k}}\right| \geq n_{k} \geq k \quad \text { for all } k.\]

Therefore, \(\lim _{k \rightarrow \infty}\left|a_{n_{k}}\right|=\infty\). This is a contradiction because \(\left\{\left|a_{n_{k}}\right|\right\}\) converges to \(|a|\). Thus \(A\) is bounded.

Let us now show that \(A\) is closed. Let \(\left\{a_{n}\right\}\) be a sequence in \(A\) that converges to a point \(a \in \mathbb{R}\). By the definition of compactness, \(\left\{a_{n}\right\}\) has a subsequence \(\left\{a_{n_{k}}\right\}\) that converges to \(b \in A\). Then \(a=b \in A\) and, hence, \(A\) is closed by Theorem 2.6.3.

For the converse, suppose \(A\) is closed and bounded and let \(\left\{a_{n}\right\}\) be a sequence in \(A\). Since \(A\) is bounded, the sequence is bounded and, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a convergent subsequence, \(\left\{a_{n_{k}}\right\}\). Say, \(\lim _{k \rightarrow \infty} a_{n_{k}}=a\). It now follows from Theorem 2.6.3 that \(a \in A\). Thhis shows that \(A\) is compact as desired. \(\square\)

Definition \(\PageIndex{4}\)

(cluster/limit/accumulation point). Let \(A\) be a subset of \(\mathbb{R}\). A point \(a \in \mathbb{R}\) (not necessarily in \(A\)) is called a limit point of \(A\) if for any \(\delta>0\), the open ball \(B(a ; \delta)\) contains an infinite number of points of \(A\).

A point \(a \in A\) whihc is not an accumulation point of \(A\) is called an isolated point of \(A\).

Example \(\PageIndex{5}\)

1. Let \(A=[0,1)\).
2. Let \(A = \mathbb{Z}\).
3. Let \(A=\{1 / n: n \in \mathbb{N}\}\). Then \(a=0\) is the only limit point of \(A\). All elements of \(A\) are isolated points.

Solution

1. Then \(a=0\) is a limit point of \(A\) and \(b=1\) is also a limit pooint of \(A\). In fact, any point of the interval \([0,1]\) is a limit point of \(A\). The set \([0,1)\) has no isolated points.
2. Then \(A\) does not have any limit points. Every element of \(\mathbb{Z}\) is an isolated point of \(\matbb{Z}\).
3. Then \(a=0\) is the only limit point of \(A\). All elements of \(A\) are isolated points.

Example \(\PageIndex{6}\)

If \(G\) is an open subset of \(\mathbb{R}\) then every point of \(G\) is a limit point of \(G\).

Solution

In fact, more is true. If \(G\) is open and \(a \in G\), then \(a\) is a limit point of \(G \backslash\{a\}\). Indeed, let \(\delta>0\) be such that \(B(a ; \delta) \subset G\). Then \((G \backslash\{a\}) \cap B(a ; \delta)=(a-\delta, a) \cup(a, a+\delta)\) and, thus \(B(a ; \delta)\) contains an infinite number of points of \(G \backslash\{a\}\).

The following theorem is a variation of the Bolzano-Weierstrass theorem.

Theorem \(\PageIndex{6}\)

Any infinite bounded subset of \(\mathbb{R}\) has at least one limit point.

Proof

Let \(A\) be an infinite subset of \(\mathbb{R}\) and let \(\left\{a_{n}\right\}\) be a sequence of \(A\) such that

\[a_{m} \neq a_{n} \text { for } m \neq n\]

(see Theorem 1.2.7). Since \(\left\{a_{n}\right\}\) is bounded, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a convergent subsequence \(\left\{a_{n_{k}}\right\}\). Set \(b=\lim _{k \rightarrow \infty} a_{n_{k}}\). Given \(\delta>0\), there exists \(K \in \mathbb{N}\) such that \(a_{n_{k}} \in B(b ; \delta)\) for \(k \geq K\). Since the set \(\left\{a_{n_{k}}: k \geq K\right\}\) is infinite, it follows that \(b\) is a limit point of \(A\). \(\square\)

The following definitions and results provide the framework for discussing convergence within subsets of \(\mathbb{R}\).

Definition \(\PageIndex{5}\)

Let \(D\) be a subset of \(\mathbb{R}\). We say that a subset \(V\) of \(D\) is open in \(D\) if for every \(a \in V\), there exists \(\delta >0\) such that

\[B(a ; \delta) \cap D \subset V.\]

Theorem \(\PageIndex{7}\)

Let \(D\) be a subset of \(\mathbb{R}\). A subset \(V\) of \(D\) is open if \(D\) if and only if there exists an open subset \(G\) of \(\mathbb{R}\) such that

\[V=D \cap G.\]

Proof

Suppose \(V\) is open in \(D\). By definition, for every \(a \in V\), there exists \(\delta_{a}>0\) such that

\[B\left(a ; \delta_{a}\right) \cap D \subset V.\]

Define

\[G=\cup_{a \in V} B\left(a ; \delta_{a}\right)\]

Then \(G\) is a union of open subsets of \(\mathbb{R}\), so \(G\) is open. Moreover,

\[V \subset G \cap D=\cup_{a \in V}\left[B\left(a ; \delta_{a}\right) \cap D\right] \subset V.\]

Therefore, \(V=G \cap D\).

Let us now prove the convers. Suppose \(V=G \cap D\), where \(G\) is an open set. For any \(a \in V\), we have \(a \in G\), so there exists \(\delta>0\) such that

\[B(a ; \delta) \subset G.\]

It follows that

\[B(a ; \delta) \cap D \subset G \cap D=V.\]

The proof is now complete. \(\square\)

Example \(\PageIndex{7}\)

Let \(D=[0,1)\) and \(V=\left[0, \frac{1}{2}\right)\).

Solution

We can write \(V=D \cap\left(-1, \frac{1}{2}\right)\). Since \(\left(-1, \frac{1}{2}\right)\) is open in \(\mathbb{R}\), we conclude from Theorem 2.6.7 that \(V\) is open in \(D\). Notice that \(V\) itself is not an open subset of \(\mathbb{R}\).

The following theorem is now a direct consequence of Theorems 2.6.7 and 2.6.1.

Theorem \(\PageIndex{8}\)

Let \(D\) be a subset of \(\mathbb{R}\). The following hold:

1. The subsets \(\emptyset\) and \(D\) are open in \(D\).
2. The union of any collection of open sets in \(D\) is open in \(D\).
3. The intersection of a finite number of open sets in \(D\) is open in \(D\).

Proof

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Definition \(\PageIndex{6}\)

Let \(D\) be a subset of \(\mathbb{R}\). We say that a subset \(A\) of \(D\) is closed in \(D\) if \(D \backslash A\) is open in \(D\).

Theorem \(\PageIndex{9}\)

A subset \(K\) of \(D\) is closed in \(D\) if and only if there exists a closed subset \(F\) of \(mathbb{R}\) such that

\[K=D \cap F.\]

Proof

Suppose \(K\) is a closed set in \(D\). Then \(D \backslash K\) is open in \(D\). By Theorem 2.6.7, there exists an open set \(G\) such that

\[D \backslash K=D \cap G.\]

It follows that

\[K=D \backslash(D \backslash K)=D \backslash(D \cap G)=D \backslash G=D \cap G^{c}.\]

Let \(F=G^{c}\). Then \(F\) is a closed subset of \(\mathbb{R}\) and \(K=D \cap F\).

Conversely, suppose that there exists a closed subset \(F\) of \(\mathbb{R}\) such that \(K=D \cap F\). Then

\[D \backslash K=D \backslash(D \cap F)=D \backslash F=D \cap F^{c}.\]

Since \(F^{c}\) is an open subset of \(\mathbb{R}\), applying Theorem 2.6.7 again, one has the \(D \backslash K\) is open in \(D\).

Therefore, \(K\) is closed in \(D\) by definition. \(\square\)

Example \(\PageIndex{8}\)

Let \(D=[0,1)\) and \(K=\left[\frac{1}{2}, 1\right)\).

Solution

We can write \(K=D \cap\left[\frac{1}{2}, 2\right]\). Since \(\left[\frac{1}{2}, 2\right]\) is closed in \(\mathbb{R}\), we conclude from Theorem 2.6.9 that \(K\) is closed in \(D\). Notice that \(K\) itself is not a closed subset of \(\mathbb{R}\).

Corollary \(\PageIndex{10}\)

Let \(D\) be a subset of \(\mathbb{R}\). A subset \(K\) of \(D\) is closed in \(D\) if and only if for every sequence \(\left\{x_{k}\right\}\) in \(K\) that converges to a point \(\bar{x} \in D\) it follows that \(\bar{x} \in K\).

Proof

Let \(D\) be a subset of \(\mathbb{R}\) Suppose \(K\) is closed in \(D\). By Theorem 2.6.9, there exists a closed subset \(F\) of \(\mathbb{R}\) such that

\[K=D \cap F.\]

Let \(\left\{x_{k}\right\}\) be a sequence in \(K\) that converges to a point \(\bar{x} \in D\). Since \(\left\{x_{k}\right\}\) is also a sequence in \(F\) and \(F\) is a closed subset of \(\mathbb{R}\), \(\bar{x} \in F\). Thus, \(\bar{x} \in D \cap F=K\).

Let us prove the converse. Suppose by contradiction that \(K\) is not closed in \(D\) or \(D \backslash K\) is not open in \(D\). Then there exists \(\bar{x} \in D \backslash K\) such that for every \(\delta>0\), one has

\[B(\bar{x} ; \delta) \cap D \nsubseteq D \backslash K.\]

In particular, for every \(k \in \mathbb{N}\),

\[B\left(\bar{x} ; \frac{1}{k}\right) \cap D \nsubseteq D \backslash K.\]

For each \(k \in \mathbb{N}\), choose \(x_{k} \in B\left(\bar{x} ; \frac{1}{k}\right) \cap D\) such that \(x_{k} \notin D \backslash K\). Then \(\left\{x_{k}\right\}\) is a sequence in \(K\) and, moreover, \(\left\{x_{k}\right\}\) converges to \(\bar{x} \in D\). Then \(\bar{x} \in K\). This is acontradiction. We conclude that \(K\) is closed in \(D\). \(square\)

The following theorem is a direct consequence of Theorems 2.6.9 and 2.6.2.

Theorem \(\PageIndex{11}\)

Let \(D\) be a subset of \(\mathbb{R}\). The following hold:

1. The subsets \(\emptyset\) and \(D\) are closed in \(D\).
2. The intersection of any collection of closed sets in \(D\) is closed in \(D\).
3. The union of a finite number of closed sets in \(D\) is closed in \(D\).

Proof

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Example \(\PageIndex{9}\)

Consider the set \(D=[0,1)\) and the subset \(A=\left[\frac{1}{2}, 1\right)\).

Solution

Clearly, \(A\) is bounded. We showed in Eample 2.6.8 that \(A\) is closed in \(D\). However, \(A\) is not compact. We show this by finding a sequence \(\left\{a_{n}\right\}\) in \(A\) for which no subsequence converges to a point in \(A\).

Indeed, consider the sequence \(a_{n}=1-\frac{1}{2 n}\) for \(n \in \mathbb{N}\). Then a_{n} \in A\) for all \(n\). Moreover, \(\left\{a_{n}\right\}\) converges to \(1\) and, hence, every subsequence also converges to \(1\). Since \(1 \notin A\), it follows that \(A\) is not compact.

Exercise \(\PageIndex{1}\)

Prove that a subset \(A\) of \(\mathbb{R}\) is open if and only if for any \(x \in A\), there exists \(n \in \mathbb{N}\) such that \((x-1 / n, x+1 / n) \subset A\).

Answer

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Exercise \(\PageIndex{2}\)

Prove that the interval \([0,1)\) is neither open nor closed.

Answer

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Exercise \(\PageIndex{3}\)

Prove that if \(A\) and \(B\) are compact subsets of \(\mathbb{R}\), then \(A \cup B\) is a compact set.

Answer

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Exercise \(\PageIndex{4}\)

Prove that the intersection of any collection of compact subsets of \(\mathbb{R}\) is compact.

Answer

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Exercise \(\PageIndex{5}\)

Find all limit points and all isolated points of each of the following sets:

1. \(A=(0,1)\).
2. \(B=[0,1)\).
3. \(C=\mathbb{Q}\).
4. \(D=\{m+1 / n: m, n \in \mathbb{N}\}\).

Answer

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Exercise \(\PageIndex{6}\)

Let \(D=[0, \infty)\). Classify each subset of \(D\) below as open in \(D\), closed in \(D\), neither or both. Justify your answers.

1. \(A=(0,1)\).
2. \(B=\mathbb{N}\).
3. \(C=\mathbb{Q} \cap D\).
4. \(D=(-1,1]\).
5. \(E=(-2, \infty)\).

Answer

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¹ This definition of compactness is more commonly reffered to as sequential compactness.