Convergence Controls Distances Within the Tail
Convergence describes how far the terms of a sequence lie from one fixed point. The Cauchy condition instead compares terms with one another, without first specifying a limit. In a metric space, the triangle inequality connects these two kinds of distance: if two sufficiently late terms are each close to a limit, they must be close to each other. We will prove this implication and use its contrapositive to detect sequences that cannot converge.
Throughout, \((X,d)\) is a metric space with its metric topology, and \((x_n)\) is a sequence in \(X\). The distance \(d(x,y)\) is nonnegative, is zero exactly when \(x=y\), is symmetric, and satisfies the triangle inequality \(d(x,z)\le d(x,y)+d(y,z)\). Convergence and the Cauchy condition below refer to this metric.
The quantifiers have a similar shape, but the distances differ. Convergence measures each late term from the same point \(x\); the Cauchy condition measures every pair of late terms from each other. The index \(N\) may depend on \(\varepsilon\), but once \(N\) is chosen, the required bound must hold for all the indicated indices.
The Main Implication
Proof. Suppose \((x_n)\) converges to \(x\in X\). We must show that for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(m,n\ge N\) gives \(d(x_m,x_n)<\varepsilon\). Fix \(\varepsilon>0\). By convergence, applied with \(\varepsilon/2>0\), there is \(N\in\mathbb{N}\) such that \(k\ge N\) implies \(d(x_k,x)<\varepsilon/2\). Thus, whenever \(m,n\ge N\), the triangle inequality gives \(d(x_m,x_n)\le d(x_m,x)+d(x,x_n)=d(x_m,x)+d(x_n,x)<\varepsilon/2+\varepsilon/2=\varepsilon\). This is exactly the Cauchy condition. \(\square\)
The choice \(\varepsilon/2\) is what makes the two estimates add to the requested \(\varepsilon\). The proof uses only the metric axioms, especially the triangle inequality; it does not require \(X\) to be a subset of \(\mathbb{R}\), or require a special formula for its metric.
Proof. The theorem states that convergence implies the Cauchy condition. Its contrapositive says that failure of the Cauchy condition implies failure of convergence. More explicitly, failure to be Cauchy means that there exists \(\varepsilon_0>0\) such that for every \(N\in\mathbb{N}\), there are \(m,n\ge N\) with \(d(x_m,x_n)\ge\varepsilon_0\). If the sequence converged, the theorem would make it Cauchy, contradicting this condition. Therefore it does not converge. \(\square\)
Notice the strictness in this negation: the Cauchy requirement is \(d(x_m,x_n)<\varepsilon\), so its failure supplies a pair with distance greater than or equal to some fixed positive \(\varepsilon_0\). To disprove convergence by this corollary, one fixed \(\varepsilon_0\) must work against every proposed tail index \(N\).
Worked Estimates
Worked Example: A Convergent Sequence in the Real Line
Take \(\mathbb{R}\) with its standard topology and usual metric \(d(s,t)=|s-t|\), and let \(x_n=3+2/n\). First verify that \(x_n\to3\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>2/\varepsilon\). For every \(n\ge N\), \(|x_n-3|=|(3+2/n)-3|=2/n\le2/N<\varepsilon\). So the sequence converges to \(3\).
Now check the Cauchy estimate directly. Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>4/\varepsilon\). If \(m,n\ge N\), then \[ |x_m-x_n|=|(3+2/m)-(3+2/n)|=|2/m-2/n| \le 2/m+2/n\le 4/N<\varepsilon. \] The same \(N\) works for every pair \(m,n\) in the tail, so the sequence is Cauchy.
Check: the convergence calculation uses \(2/N<\varepsilon\), while the pairwise estimate uses \(4/N<\varepsilon\). Both follow from the chosen \(N>4/\varepsilon\), since \(N>0\) gives \(2/N<4/N<\varepsilon\). The direct estimate agrees with the theorem.
Worked Example: A Sequence in a Finite Metric Space
Let \(X=\{a,b\}\) have the discrete metric, which gives the discrete topology: \(d(u,v)=0\) if \(u=v\), and \(d(u,v)=1\) if \(u\ne v\). Define \(x_1=b\) and \(x_n=a\) for every \(n\ge2\).
We verify convergence to \(a\). Given any \(\varepsilon>0\), choose \(N=2\). If \(n\ge2\), then \(x_n=a\), so \(d(x_n,a)=d(a,a)=0<\varepsilon\). Thus \(x_n\to a\), including when \(0<\varepsilon\le1\); the exceptional first term does not affect the tail.
For the Cauchy condition, take any \(\varepsilon>0\) and again set \(N=2\). Whenever \(m,n\ge2\), both terms equal \(a\), so \(d(x_m,x_n)=d(a,a)=0<\varepsilon\). Hence the sequence is Cauchy. As a check on the metric computation, the only other possible distance in \(X\) is \(d(a,b)=d(b,a)=1\), but neither point \(b\) occurs in the tail starting at \(N=2\).
Worked Example: A Non-Cauchy Sequence Cannot Converge
In \(\mathbb{R}\) with its standard topology, consider \(x_n=(-1)^n\). We show that it fails the Cauchy condition. Set \(\varepsilon_0=1\). For any \(N\in\mathbb{N}\), choose \(m=N\) and \(n=N+1\). Consecutive powers of \(-1\) have opposite signs, so \(x_m\) and \(x_n\) are \(1\) and \(-1\) in some order. Therefore \[ |x_m-x_n|=|1-(-1)|=2\ge1=\varepsilon_0. \] This pair lies at or beyond every proposed tail index \(N\), so no \(N\) can make all pairwise distances in the tail less than \(1\). The sequence is not Cauchy, and the corollary shows that it cannot converge in \(\mathbb{R}\).
Check: the obstruction is the same for every \(N\): the distance between the selected consecutive terms is exactly \(2\), not merely a distance that stays above \(1\) for some initial portion of the sequence. This verifies the quantifiers needed to rule out convergence.
A Common Epsilon Error
A frequent mistake is to apply convergence with the requested \(\varepsilon\) and then assume that two distances each less than \(\varepsilon\) add to less than \(\varepsilon\). The triangle inequality gives a bound of less than \(2\varepsilon\), which may be too large. The following calculation makes the problem explicit.
In \(\mathbb{R}\), let \(x_n=(-1)^n/n\), which converges to \(0\): for every \(\delta>0\), choosing \(N>1/\delta\) gives \(|x_n-0|=1/n\le1/N<\delta\) for \(n\ge N\). For the requested pairwise tolerance \(\varepsilon=1/10\), the terms with indices \(12\) and \(13\) satisfy \[ |x_{12}-0|=1/12<1/10,\qquad |x_{13}-0|=1/13<1/10, \] but \[ |x_{12}-x_{13}|=|1/12-(-1/13)|=1/12+1/13=25/156>1/10, \] because \(250>156\). Thus the two individual bounds by \(1/10\) do not give a pairwise bound by \(1/10\).
The correction is to begin with the desired pairwise tolerance \(\varepsilon\) and use convergence with the smaller tolerance \(\varepsilon/2\), as in the theorem. In this example, for any \(\varepsilon>0\), choose \(N>2/\varepsilon\). Then for \(m,n\ge N\), \[ |x_m-x_n|\le |x_m|+|x_n|=1/m+1/n\le2/N<\varepsilon. \] This direct calculation confirms that the sequence is Cauchy and illustrates the same half-tolerance adjustment.
What the Implication Does—and Does Not—Say
The theorem gives a necessary condition for convergence: a sequence that converges must be Cauchy. Its contrapositive is useful because failure of the Cauchy condition can be witnessed using distances between terms, without having to guess a possible limit. The quantifier pattern to look for is one fixed \(\varepsilon_0>0\) such that every tail, no matter how far out it begins, contains a pair at distance at least \(\varepsilon_0\).
The theorem does not assert that every Cauchy sequence converges in every metric space. That reverse implication is a separate question about the space and is not needed to apply the result proved here. For the present direction, the complete mechanism is the triangle inequality: two distances to a common limit bound the distance between the terms.
Check Your Understanding
For each question, keep track of which indices must work and whether the distance bound is strict.
- In the proof of the theorem, why is convergence applied with \(\varepsilon/2\) rather than \(\varepsilon\)?
- State the negation of the Cauchy condition using quantifiers and a fixed positive \(\varepsilon_0\).
- In \(\mathbb{R}\), show that \(x_n=5-1/n\) converges to \(5\), and use the theorem to conclude that it is Cauchy.
- For \(x_n=(-1)^n\), explain why choosing \(m=N\) and \(n=N+1\) disproves the Cauchy condition for every \(N\).
- Does the theorem say that every Cauchy sequence converges in every metric space? Explain what implication it proves.