Completeness of the Real Line

From Cauchy Estimates to a Real Limit

A Cauchy sequence has terms that become arbitrarily close to one another. That condition compares terms in the sequence; it does not itself name a point to which they are close. To prove that a Cauchy sequence of real numbers converges, we need a property of the real numbers that supplies a candidate limit. Here that property is the least-upper-bound property, together with an explicit estimate showing that the sequence approaches the candidate.

We use \(\mathbb{R}\) with its standard topology and metric \(d(x,y)=|x-y|\). A sequence \((x_n)\) in \(\mathbb{R}\) is Cauchy if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(|x_m-x_n|<\varepsilon\) whenever \(m,n\ge N\). It converges to \(L\in\mathbb{R}\) if, for every \(\varepsilon>0\), there is an \(N\in\mathbb{N}\) such that \(|x_n-L|<\varepsilon\) whenever \(n\ge N\). A metric space is complete if every Cauchy sequence in it converges to a point of that space.

The equivalence in the definition follows by changing signs: if \(S\) is nonempty and bounded below, then \(-S=\{-s:s\in S\}\) is nonempty and bounded above. Its supremum exists, and \(\inf S=-\sup(-S)\). We will use this property to construct a limit from the lower bounds of the sequence’s tails.

A Cauchy Sequence Is Bounded

Proof. Let \((x_n)\) be Cauchy. Apply the Cauchy condition with \(\varepsilon=1\). There is \(N_0\in\mathbb{N}\) such that \(|x_n-x_{N_0}|<1\) for every \(n\ge N_0\). The triangle inequality gives \(|x_n|\le |x_{N_0}|+|x_n-x_{N_0}|<|x_{N_0}|+1\) for these indices. The terms \(x_1,\ldots,x_{N_0-1}\) are a finite list. Therefore the number \[ M=\max\{|x_1|,\ldots,|x_{N_0-1}|,\ |x_{N_0}|+1\} \] is a finite bound for all terms; if \(N_0=1\), the finite list before \(N_0\) is empty and we take \(M=|x_1|+1\). Thus \(|x_n|\le M\) for every \(n\in\mathbb{N}\). \(\square\)

The boundedness lemma ensures that every tail of the sequence has a finite infimum. To get a single candidate limit, we take the supremum of those tail infima. The estimates that follow show why this particular real number works.

Completeness of the Real Line

Proof. Let \((x_n)\) be a Cauchy sequence in \(\mathbb{R}\). For each \(n\in\mathbb{N}\), define its \(n\)-th tail by \(T_n=\{x_k:k\ge n\}\), and let \(a_n=\inf T_n\). The set \(T_n\) is nonempty because it contains \(x_n\). It is bounded below because the whole sequence is bounded, by the lemma. Hence \(a_n\in\mathbb{R}\) exists by the least-upper-bound property.

If \(m\ge n\), then \(T_m\subseteq T_n\), so every lower bound for \(T_n\) is also a lower bound for \(T_m\). In particular, \(a_n\le a_m\). Thus \((a_n)\) is nondecreasing. Also, \(a_n\le x_n\le M\), where \(M\) is a bound for the sequence, so the set \(\{a_n:n\in\mathbb{N}\}\) is nonempty and bounded above. Define \[ L=\sup\{a_n:n\in\mathbb{N}\}\in\mathbb{R}. \] We now prove that \(x_n\to L\).

Let \(\varepsilon>0\) be given. Since \((x_n)\) is Cauchy, there is \(N\in\mathbb{N}\) such that \(|x_j-x_k|<\varepsilon/3\) whenever \(j,k\ge N\). In particular, for every \(k\ge N\), \[ x_N-\varepsilon/3 < x_k < x_N+\varepsilon/3. \] Fix any \(n\ge N\). Every element of \(T_n\) is greater than \(x_N-\varepsilon/3\), so \(x_N-\varepsilon/3\) is a lower bound for \(T_n\). Since \(a_n\) is its greatest lower bound, \(a_n\ge x_N-\varepsilon/3\). As \(L\ge a_n\), this gives \(L\ge x_N-\varepsilon/3\).

For every \(m\ge N\), we have \(a_m\le x_m<x_N+\varepsilon/3\). For \(m<N\), monotonicity gives \(a_m\le a_N\le x_N\). Therefore \(x_N+\varepsilon/3\) is an upper bound for all the \(a_m\), and the definition of \(L\) gives \(L\le x_N+\varepsilon/3\). Also, \(x_n\) lies between \(x_N-\varepsilon/3\) and \(x_N+\varepsilon/3\). Both \(x_n\) and \(L\) therefore lie in an interval of length \(2\varepsilon/3\), so \[ |x_n-L|\le 2\varepsilon/3<\varepsilon. \] This holds for every \(n\ge N\). By the definition of convergence, \(x_n\to L\). The limit is a real number because it was obtained as a supremum in \(\mathbb{R}\). \(\square\)

The proof has two distinct ingredients. The Cauchy condition makes all sufficiently late terms close to one another. The least-upper-bound property supplies a real number \(L\) against which those terms can be compared. The final epsilon estimate turns these ingredients into convergence; closeness between terms alone is not the definition of convergence.

Worked Examples of the Estimates

A Common Mistake and Why Completeness Matters

Common pitfall. A small difference between late terms does not mean the late terms are close to zero. For the constant sequence \(x_n=5\), every pair of terms satisfies \(|x_m-x_n|=|5-5|=0\), so the sequence is Cauchy. But it does not converge to \(0\): for every \(n\), \(|x_n-0|=5\), which is not less than \(\varepsilon=1\). Its limit is \(5\), since \(|x_n-5|=0<\varepsilon\) for every \(\varepsilon>0\). The Cauchy condition controls differences between terms; to prove convergence, one must identify a candidate limit and estimate distance to that point.

The theorem says more than that the three example sequences have limits. It says that the same conclusion holds for every Cauchy sequence of real numbers, whether or not a formula for its limit is obvious. The proof relies on the least-upper-bound property: it ensures the tail infima determine a real number. The epsilon argument then verifies that this number is the limit. Both parts are needed for the stated proof.