The real line has many equivalent completeness principles. Good analysis depends on choosing the form that best matches the problem: suprema for monotone arguments, compactness for subsequences, and Cauchy completeness for approximation.
Monotone convergence on the line
Monotone sequences are the simplest limiting processes controlled by order.
Theorem. If \((x_n)\) is increasing and bounded above in \(\mathbb R\), then \(x_n\to\sup_n x_n\). The decreasing bounded case follows by applying this to \(-x_n\).
Proof. Let \(s=\sup\{x_n:n\ge1\}\). For every \(\varepsilon>0\), \(s-\varepsilon\) is not an upper bound, so some \(x_N>s-\varepsilon\). For \(n\ge N\), monotonicity gives \(s-\varepsilon\lt x_N\le x_n\le s\), hence \(x_n\to s\).
Nested intervals
Interval completeness turns a sequence of shrinking constraints into a single point.
Theorem. If closed intervals \([a_n,b_n]\) are nested and \(b_n-a_n\to0\), then \(\bigcap_n[a_n,b_n]\) consists of exactly one point.
Proof. The left endpoints are increasing and bounded above by every \(b_m\), so \(a_n\to x=\sup a_n\). For each fixed \(m\), \(a_n\le b_m\) for \(n\ge m\), hence \(x\le b_m\), and also \(a_m\le x\). Thus \(x\) belongs to all intervals. If \(y\) also belongs to all intervals, then \(|x-y|\le b_n-a_n\to0\), so \(x=y\).
Bolzano-Weierstrass compactness
Bounded sequences on the line always have convergent subsequences.
Theorem. Every bounded sequence in \(\mathbb R\) has a convergent subsequence.
Proof. Put the sequence in a closed interval. Bisect the interval and choose a half containing infinitely many terms. Repeat. This gives nested intervals with lengths tending to zero and infinitely many sequence terms in each interval. Choosing indices successively inside these intervals produces a subsequence converging to the unique point in their intersection.
Cauchy completeness
The Cauchy criterion lets one prove convergence without first knowing the limit.
Theorem. A real sequence converges if and only if it is Cauchy.
Proof. Convergent sequences are Cauchy by the triangle inequality. Conversely, a Cauchy sequence is bounded, hence has a convergent subsequence by Bolzano-Weierstrass. If \(x_{n_k}\to x\), then for large \(n\) and \(k\) with \(n_k\) also large, \(|x_n-x|\le |x_n-x_{n_k}|+|x_{n_k}-x|\) is arbitrarily small.
Heine-Borel on the real line
Compactness converts open-cover problems into finite arguments.
Theorem. A subset of \(\mathbb R\) is compact if and only if it is closed and bounded.
Proof. Compact sets are bounded because the cover by intervals \((-n,n)\) has a finite subcover, and closed because limits of convergent sequences from a compact set must remain in the set. Conversely, a closed bounded set is contained in a compact interval; every sequence has a convergent subsequence by Bolzano-Weierstrass, and closedness keeps the limit in the set. In metric spaces, sequential compactness is equivalent to open-cover compactness.