Two Different Conditions
The proof of Heine–Borel showed how closedness and boundedness work together: boundedness places a set inside a finite interval, while closedness lets us use its open complement. This tutorial examines the two conditions themselves and their distinct roles. In particular, we will connect them to a sequential test: in a closed bounded set, every sequence has a convergent subsequence whose limit still belongs to the set.
The conditions should not be confused. Boundedness restricts how far points of a set can lie from the origin; it does not say that the set contains its limit points. Closedness concerns limit points, but does not prevent a set from extending arbitrarily far along the real line. The two properties are independent, and the examples below will separate them before we consider their combination.
The number \(M\) is one possible bound; it need not be the smallest possible one. The interval formulation follows directly from the absolute-value formulation: if \(|x|\leq M\), then \(x\in[-M,M]\). Conversely, if \(E\subseteq[a,b]\), then \(|x|\leq\max\{|a|,|b|\}\) for every \(x\in E\).
The two descriptions of closedness give different ways to work with it. The complement definition is often useful in open-cover arguments. The sequential definition is useful when a set is described by a sequence or when we need to track limits. We will use the Sequential Criterion for Closedness as an established result rather than prove it again here.
Worked Examples: Checking the Conditions
Worked Example: Solving an Inequality Gives a Closed Bounded Set
Let \(E=\{x\in\mathbb{R}:|2x-1|\leq 5\}\). To identify the set, rewrite the absolute-value inequality as a double inequality:
Thus \(E=[-2,3]\). It is bounded because every \(x\in E\) satisfies \(|x|\leq 3\). It is closed because every closed interval is closed. This example illustrates a useful first step: simplify a set description before deciding which topological properties it has.
Worked Example: Bounded but Not Closed
Consider \(E=\mathbb{Q}\cap[0,1]\), the rational numbers in the unit interval. Every \(x\in E\) satisfies \(0\leq x\leq 1\), so \(E\) is bounded. But \(\sqrt{2}/2\) is not rational and lies in \([0,1]\): it is positive, and \(\sqrt{2}<2\) gives \(\sqrt{2}/2<1\).
By the density of the rationals, there is a sequence of rational numbers in \([0,1]\) converging to \(\sqrt{2}/2\). More explicitly, for each positive integer \(n\), rational approximation gives \(q_n\in\mathbb{Q}\) with \(|q_n-\sqrt{2}/2|<1/n\). For all sufficiently large \(n\), these \(q_n\) lie in \([0,1]\); discarding the finitely many initial terms gives a sequence in \(E\) with the same limit. Since its limit is not in \(E\), the Sequential Criterion for Closedness shows that \(E\) is not closed.
Worked Example: Closed but Not Bounded
The set \(\mathbb{Z}\) of integers is closed. To check this using the complement, take any \(x\notin\mathbb{Z}\). There is an integer \(m\) with \(m<x<m+1\). Set \(r=\min\{x-m,m+1-x\}/2\), which is positive. Then \((x-r,x+r)\) contains no integer, so it is contained in \(\mathbb{R}\setminus\mathbb{Z}\). Every point of the complement therefore has an open interval around it lying in the complement, proving that the complement is open.
The integers are not bounded: for every \(M\geq0\), the integer \(n=\lfloor M\rfloor+1\) satisfies \(n>M\), so \(|n|>M\). Thus \(\mathbb{Z}\) is closed but not bounded. Closedness alone does not keep a set within a finite interval.
How Boundedness Behaves
Boundedness is preserved when points are removed: a subset cannot extend farther than the set containing it. It is also preserved under a finite union. These elementary facts are useful when a set is assembled from simpler pieces.
Proof. Suppose \(A\subseteq B\) and \(B\) is bounded. Choose \(M\geq0\) such that \(|x|\leq M\) for every \(x\in B\). Since every \(x\in A\) also belongs to \(B\), the same inequality holds for every \(x\in A\). Hence \(A\) is bounded.
Now let \(B_1,\ldots,B_m\) be bounded, where \(m\) is a positive integer. For each \(j\), choose \(M_j\geq0\) such that \(|x|\leq M_j\) for every \(x\in B_j\). The finite set of bounds has a maximum \(M=\max\{M_1,\ldots,M_m\}\). If \(x\in\bigcup_{j=1}^m B_j\), then \(x\in B_j\) for at least one \(j\), so \(|x|\leq M_j\leq M\). Therefore the union is bounded. The empty union is empty and is bounded as well. \(\square\)
For closedness, the corresponding finite-union property was established earlier in the course. Together, these facts let us verify that finite unions of closed bounded sets are closed and bounded. Heine–Borel then tells us that such a union is compact. The two checks remain logically separate: a bound for the union does not itself prove closedness, and closedness does not itself give a bound.
Worked Example: Combining Closed Bounded Pieces
Let \(K=[-3,-1]\cup\{4\}\). The interval \([-3,-1]\) and the singleton \(\{4\}\) are closed, so their finite union is closed. Both pieces are bounded; more directly, every \(x\in K\) satisfies \(|x|\leq4\). Thus \(K\) is closed and bounded, and the Heine–Borel Theorem implies that \(K\) is compact.
The point \(4\) does not make the set unbounded: boundedness requires one common bound, not that all points lie in the same connected piece. The bound \(4\) works simultaneously for both pieces.
The Sequential Meaning of Closed and Bounded
A bounded sequence of real numbers has a convergent subsequence by the Bolzano–Weierstrass Theorem. If all its terms lie in a closed set, the subsequential limit also lies in that set, by the Sequential Criterion for Closedness. The next theorem records this consequence and proves its converse. It supplies a way to recognize closed bounded sets by what they do to every sequence of their points.
Proof. First suppose \(E\) is closed and bounded, and let \((x_n)\) be a sequence in \(E\). If the range \(\{x_n:n\geq1\}\) is finite, at least one value occurs infinitely many times. The corresponding constant subsequence converges to that value, which belongs to \(E\).
If the range is infinite, it is an infinite bounded subset of \(\mathbb{R}\). By the Bolzano–Weierstrass Theorem, it has an accumulation point \(x\). We can choose indices \(n_1<n_2<\cdots\) such that \(|x_{n_k}-x|<1/k\): at each stage, the ball of radius \(1/k\) around \(x\) contains infinitely many distinct points of the range, while only finitely many terms have indices at most \(n_{k-1}\). Hence \((x_{n_k})\) converges to \(x\). Because every \(x_{n_k}\) belongs to the closed set \(E\), the Sequential Criterion for Closedness gives \(x\in E\).
Conversely, suppose every sequence in \(E\) has a subsequence converging to a point of \(E\). If \(E\) were unbounded, we could choose \(x_n\in E\) with \(|x_n|>n\) for each positive integer \(n\). Any subsequence \((x_{n_k})\) satisfies \(|x_{n_k}|>n_k\geq k\), so it cannot converge to a real number. This contradicts the assumed subsequence property. Therefore \(E\) is bounded.
To show that \(E\) is closed, take any \(x\in\overline{E}\). If \(E\) is empty, its closure is empty, so there is no such \(x\); otherwise, the Neighborhood Characterization of Closure lets us choose \(y_n\in E\) with \(|y_n-x|<1/n\) for each \(n\). Then \(y_n\to x\). By the assumed property, some subsequence \((y_{n_k})\) converges to a point \(y\in E\). But every subsequence of a sequence converging to \(x\) also converges to \(x\), so uniqueness of limits gives \(y=x\). Thus \(x\in E\). We have shown \(\overline{E}\subseteq E\); since \(E\subseteq\overline{E}\), it follows that \(E=\overline{E}\), and hence \(E\) is closed. \(\square\)
Worked Example: A Sequence in a Closed Bounded Set
Take \(K=[-3,-1]\cup\{4\}\) and define \(x_n=-2\) when \(n\) is odd and \(x_n=4\) when \(n\) is even. Every term belongs to \(K\). The full sequence does not converge: its odd-indexed terms are always \(-2\), while its even-indexed terms are always \(4\), and these values differ. But the odd-indexed subsequence is constantly \(-2\), so it converges to \(-2\in K\); the even-indexed subsequence is constantly \(4\), so it converges to \(4\in K\).
This illustrates the theorem’s exact claim. It guarantees a convergent subsequence with its limit in the set; it does not claim that the original sequence converges.
Why Both Conditions Matter
Heine–Borel states that a subset of \(\mathbb{R}\) is compact exactly when it is closed and bounded. The sequential characterization explains why both conditions matter from the perspective of sequences. Boundedness prevents sequences from escaping without bound, while closedness keeps subsequential limits from falling outside the set. The rational numbers in \([0,1]\) illustrate the second failure: sequences can converge to points missing from the set. The integers illustrate the first: a sequence of integers can move arbitrarily far away.
A common mistake is to infer closedness from the fact that a set is bounded, or boundedness from the fact that it is closed. Neither inference is valid. Check each property against its own definition, and handle the empty set explicitly when needed: it is both closed and bounded, and the sequential condition holds vacuously because there are no sequences with all terms in the empty set.
Find one number \(M\geq0\) that bounds \(|x|\) for every point in the set.
Use an open complement or test whether limits of convergent sequences in the set remain in it.
If both properties hold, Heine–Borel gives compactness, and every sequence in the set has a subsequence converging within it.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- Why is containment in a bounded interval equivalent to having a uniform bound on \(|x|\)?
- Give an example of a bounded set that is not closed, and identify a sequence that demonstrates the failure of closedness.
- In the proof that finite unions of bounded sets are bounded, why is it important that there are only finitely many sets?
- Why does an unbounded set fail the sequential condition in the Sequential Characterization of Closed Bounded Sets?
- Does the sequential characterization say that every sequence in a closed bounded set converges? Explain the distinction.