Cauchy sequences and completeness
Separate tail agreement from the existence of a limit and detect the missing points of an incomplete space.
On this page
The question for today
What you’ll understand
- Distinguish the Cauchy condition from convergence without assuming a candidate limit.
- Use completeness plus a closed subset to justify existence of a limit.
- Explain why vanishing consecutive steps alone do not prove the Cauchy property.
Suppose successive numerical approximations seem to settle down, but their exact answer is unknown. A convergence test referring to the answer would be circular. The Cauchy condition instead asks whether all sufficiently late approximations agree with one another to any requested tolerance. Completeness is the promise that this internal agreement leads to an actual point of the space. The distinction matters in iterative algorithms, infinite series, and function spaces, where an apparently missing limit may mean the ambient space was chosen too narrowly.
Compare every pair in the tail
For a Cauchy sequence, after a tolerance is selected there is one cutoff beyond which every pair of terms is close. No candidate limit appears. A convergent sequence is Cauchy: put each late term within half the requested tolerance of its limit and use the triangle inequality between the two terms. The converse is not a consequence of the metric axioms. It is exactly the extra property called completeness. The real line is complete with its usual distance, while the rational numbers with the restricted distance are not.
Checking consecutive terms is much weaker. The harmonic partial sums have consecutive differences 1/(n+1), tending to zero, but terms with indices n and 2n remain separated by at least one half. There are n added fractions, each at least 1/(2n). Thus no late cutoff controls every pair to a tolerance less than one half. Small individual steps can accumulate over a long journey. A useful Cauchy proof must control the whole accumulated tail, often by a convergent geometric bound.
A hole can prevent convergence
Take rational decimal truncations of the positive square root of two. Their real distances from that irrational number are at most powers of one tenth, so the triangle inequality proves that the rational truncations are Cauchy. If they converged to a rational number in the rational metric space, they would converge to the same number in the real line. Uniqueness of real limits would make that rational equal to the square root of two, an impossibility. The sequence agrees internally while its limiting point lies outside the chosen space.
A closed subset of a complete metric space is complete. A Cauchy sequence in the subset first converges somewhere in the ambient complete space; closedness then puts its limit back into the subset. Each assumption has a different job. Conversely, a complete subspace of any metric space is closed: a convergent ambient sequence in it is Cauchy, hence converges inside it, and uniqueness identifies the two limits. Saying merely that a set is closed is insufficient when the ambient space itself is incomplete.
Completeness concerns the metric, not just open sets
Equivalent local neighborhoods do not always preserve completeness. Put d(x,y)=|arctan(x)−arctan(y)| on the real line. This metric gives the usual topology because arctangent is a homeomorphism onto its image. Yet the sequence of positive integers is Cauchy for d: the transformed sequence approaches the endpoint pi/2. No real x has that arctangent, so the d-space is incomplete. The usual real metric is complete. Uniform quantitative control between metrics, rather than topological equivalence alone, is needed to transfer Cauchy behavior.
Completeness is not compactness or boundedness. The real line is complete but unbounded, and a bounded open interval is incomplete with usual distance. Every finite set equipped with a genuine metric is complete, including a finite list of representable real numbers with inherited distance. That fact does not guarantee accurate arithmetic: rounding and overflow are properties of computational operations, not failures of metric completeness. Identify the ideal space and metric first, then separately analyze computational error. An approximation that terminates successfully does not establish a theorem about every possible Cauchy sequence.
Work it through
A geometric iteration has a certified tail
In a complete metric space, suppose d(x_(n+1),x_n) is at most 3·(1/2)^n for n starting at zero. Bound the distance from x_n to the limit and choose n guaranteeing error below 0.01.
For m greater than n, add consecutive distances using the triangle inequality. The geometric tail is at most 3·(1/2)^n/(1−1/2)=6·2^(−n).
This upper bound tends to zero independently of m, proving that the sequence is Cauchy. Completeness produces a limit x, and passing m to infinity keeps the bound d(x_n,x) at most 6·2^(−n).
At n=10 the bound is 6/1024, approximately 0.00586, below 0.01. At n=9 the bound is about 0.01172, so n=10 is the first integer certified by this estimate.
Ten iterations, with the stated index convention, certify error below 0.01. The bound is a guarantee and need not equal the actual error.
The living notebook
Explore the model
Interactive lab · optimization
Inspect a geometric tail
Choose step size 0.5 and nonzero initial x in the lab's quadratic gradient descent. Its exact iterate ratio is one half. Step forward and predict the sum of all future absolute step sizes. The displayed finite history suggests the bound, while the geometric-series argument proves it; the lab does not test completeness of arbitrary spaces.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA bound on the entire future tail is stronger than observing a small latest step.
Pause & reason
Check your understanding
Your turn
Try it yourself
Determine whether [0,1) is complete with usual distance. Then explain why [0,1] is complete without checking every possible sequence separately.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
For the first set, approach the missing endpoint from below. For the second, use a closed subset of the complete real line.
Compare with the solution
The sequence x_n=1−1/n lies in [0,1) for positive n and converges to 1 in the real line. It is Cauchy, but cannot converge within [0,1), since real limits are unique. Thus [0,1) is incomplete. The interval [0,1] is closed in the complete real line, so any Cauchy sequence in it converges in the real line and has its limit in the interval. Hence [0,1] is complete.
Keep the idea
Take it with you
- Cauchy means uniform agreement between every pair of sufficiently late terms.
- Completeness converts this internal agreement into existence of a limit in the space.
- Closedness transfers completeness from an ambient complete metric space; topology alone does not determine completeness.
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