Bolzano-Weierstrass and Compactness

Bolzano–Weierstrass as a Compactness Principle

The Bolzano–Weierstrass Theorem connects boundedness and accumulation points: every bounded infinite subset of \(\mathbb{R}\) has an accumulation point in \(\mathbb{R}\). Compact sets in the real line are closed and bounded by the Heine–Borel Theorem, so Bolzano–Weierstrass immediately gives information about infinite subsets of a compact set. There is also a converse: if every infinite subset of a set has an accumulation point belonging to that set, then the set is compact.

The location of the accumulation point matters. Bolzano–Weierstrass alone puts it somewhere in \(\mathbb{R}\); it does not say that the point belongs to the original set. Closedness supplies that missing step. Conversely, if accumulation points of infinite subsets are always required to lie in the set, that requirement rules out both unboundedness and missing limit points.

This property differs slightly from saying that every infinite subset has an accumulation point in \(\mathbb{R}\). The latter point might lie outside \(K\). The accumulation-point property requires a point of \(K\), and that requirement is precisely what makes it characterize compactness here.

Compactness Gives Accumulation Points Inside the Set

Proof. Let \(E\subseteq K\) be infinite. Since \(K\) is compact, it is bounded. Every subset of a bounded set is bounded, so \(E\) is bounded. By the Bolzano–Weierstrass Theorem, \(E\) has an accumulation point \(p\in\mathbb{R}\). In particular, \(p\in\overline{E}\). Because \(E\subseteq K\), monotonicity of closure gives \(\overline{E}\subseteq\overline{K}\). Compact subsets of \(\mathbb{R}\) are closed, so \(\overline{K}=K\). Therefore \(p\in K\), as required. \(\square\)

Each hypothesis has a role. Boundedness allows us to apply Bolzano–Weierstrass. Closedness ensures that the accumulation point cannot fall outside the set. Compactness supplies both of these properties in \(\mathbb{R}\), by the Heine–Borel Theorem and the theorem that compact subsets of \(\mathbb{R}\) are closed.

The Converse: The Accumulation-Point Property Forces Compactness

Proof. The forward direction is the theorem just proved. For the converse, suppose \(K\) has the accumulation-point property. We will show that \(K\) is bounded and closed, then apply the Heine–Borel Theorem.

First, \(K\) is bounded. Suppose instead that \(K\) is unbounded. We can choose points \(y_n\in K\), for \(n\geq1\), so that \[ |y_n|>n. \] These points are distinct: if \(m>n\), then \(|y_m|>m>n\), while \(|y_n|\) need not be at most \(n\), so this inequality by itself does not guarantee distinctness. To ensure distinctness, choose recursively with the stronger requirement \[ |y_n|>\max\bigl(n,|y_1|,\ldots,|y_{n-1}|\bigr). \] Unboundedness makes each choice possible. Let \(D=\{y_n:n\geq1\}\), an infinite subset of \(K\).

We show that \(D\) has no accumulation point in \(\mathbb{R}\). Fix \(q\in\mathbb{R}\). For all sufficiently large \(n\), \(n>|q|+2\), so \[ |y_n-q|\geq |y_n|-|q|>n-|q|>2. \] Thus all but finitely many points of \(D\) lie outside the ball of radius \(2\) centered at \(q\). Among the remaining finitely many points, discard \(q\) if it is one of them. If any remain, their distances from \(q\) have a positive minimum \(d\); choose a radius smaller than both \(2\) and \(d\). If none remain, choose any radius smaller than \(2\). In either case, the resulting punctured ball around \(q\) contains no point of \(D\). Hence \(q\) is not an accumulation point of \(D\). Since \(q\) was arbitrary, \(D\) has no accumulation point, contradicting the accumulation-point property. Therefore \(K\) is bounded.

Next, \(K\) is closed. Suppose, to the contrary, that \(K\) is not closed. Then there is a point \(p\in\overline{K}\setminus K\). We construct distinct points \(x_n\in K\) approaching \(p\). Choose \(x_1\in K\) with \(0<|x_1-p|<1\), which is possible because \(p\in\overline{K}\) and \(p\notin K\). After choosing \(x_1,\ldots,x_{n-1}\), choose \[ 0<r_n<\min\left(\frac1n,\frac12|x_1-p|,\ldots,\frac12|x_{n-1}-p|\right). \] Since \(p\in\overline{K}\setminus K\), the ball of radius \(r_n\) around \(p\) contains some \(x_n\in K\). It satisfies \(0<|x_n-p|<r_n\), so it differs from every previously chosen point: its distance to \(p\) is less than half the distance of each earlier point to \(p\). The set \(D=\{x_n:n\geq1\}\) is therefore infinite. Also, \(|x_n-p|<1/n\), so \(x_n\to p\), and \(p\) is an accumulation point of \(D\).

We verify that \(D\) has no accumulation point other than \(p\). Fix \(q\neq p\), and write \(d=|q-p|>0\). For all sufficiently large \(n\), \(|x_n-p|<d/3\), whence \[ |x_n-q|\geq |q-p|-|x_n-p|>2d/3. \] Only finitely many points of \(D\) remain. Discard \(q\) if it is among them. If any of these remaining points exist, choose a radius smaller than their finitely many positive distances from \(q\) and smaller than \(2d/3\); if none exist, choose any radius smaller than \(2d/3\). The punctured ball of this radius about \(q\) contains no point of \(D\). Thus \(q\) is not an accumulation point. Consequently \(p\) is the only accumulation point of \(D\), and \(p\notin K\). This contradicts the property that every infinite subset of \(K\) has an accumulation point belonging to \(K\). So \(K\) is closed.

We have shown that \(K\) is closed and bounded. The Heine–Borel Theorem now implies that \(K\) is compact. This proves the converse and completes the theorem. \(\square\)

How to Use the Characterization

The characterization is especially useful when a problem concerns infinite subsets rather than open covers. To prove that a set is compact, it is enough to take an arbitrary infinite subset and find an accumulation point inside the set. To show that a set is not compact, it suffices to find one infinite subset whose accumulation points all lie outside the set or which has no accumulation point at all.

Keep the distinction between two statements clear. Bolzano–Weierstrass says that a bounded infinite subset of \(\mathbb{R}\) has an accumulation point in \(\mathbb{R}\). The characterization says that every infinite subset of a compact set has one in that compact set. The first conclusion follows from boundedness; the second also depends on closedness. For the converse, requiring accumulation points to stay inside the set rules out unbounded sets and sets with missing limit points. Together, these observations explain why the characterization is equivalent to compactness on the real line.

The real-line setting matters: the final step uses the Heine–Borel Theorem, which identifies compactness with closedness and boundedness in \(\mathbb{R}\). The theorem therefore provides a convenient bridge from the Bolzano–Weierstrass principle to compactness without replacing the open-cover definition.