Completions and dense subsets
Construct missing limits from Cauchy sequences and understand density as approximation at every scale.
On this page
The question for today
What you’ll understand
- Distinguish density from equality or large measure.
- Explain how equivalent Cauchy sequences represent the same added point.
- State the uniqueness and extension properties of metric completion with their hypotheses.
An incomplete space can be repaired without changing any distances between its original points. The repair adds ideal limits for Cauchy sequences and identifies different approximation procedures that approach the same ideal point. The rational numbers sitting densely inside the real numbers are the guiding example, but the construction works without decimal expansions. It also clarifies why a dense collection of simple objects can determine a continuous theory, while not every arbitrary rule on that collection extends to its completion.
Density means approximation at every positive scale
A subset D is dense in X if its closure is all of X. Equivalently, every nonempty open ball in X meets D. This is a local approximation statement, not a claim that D contains most points by number or measure. The rationals are countable and still dense in the real line. Between any two distinct reals there is a rational, so every real can be approximated to every positive tolerance using rational numbers. The irrational numbers are also dense: density alone does not distinguish their very different cardinality.
For a metric space, density lets you select a sequence from D converging to any given x: choose its nth point within 1/n of x. No single finite tolerance establishes density, because a gap could remain below the resolution tested. A finite grid in an interval is not dense in that interval, though a nested sequence of successively finer grids can have a dense union. The distinction matters in approximation: fixing one discretization and allowing arbitrarily refined discretizations are different mathematical operations.
Add limits without double counting
Consider all Cauchy sequences in X. Two are declared equivalent when the distance between their nth terms tends to zero. Reflexivity and symmetry follow from the metric, while transitivity follows by adding distances through a third sequence. One equivalence class becomes one point of the completion. The same missing real number can be approached by many different rational sequences, and this identification prevents the construction from creating a separate point for every method of approximation. Original points are represented by constant sequences.
The distance between two new points is defined as the limit of their termwise distances. That limit exists because the reverse triangle inequality bounds the difference between two such distances by the sum of the two original sequences' tail distances. The resulting real sequence is Cauchy. Equivalent representatives produce the same limit, so the definition does not depend on arbitrary choices. The triangle inequality passes to the limit, and zero distance means exactly that the representatives belong to the same equivalence class.
What completion preserves and what extensions require
The constant-sequence embedding preserves every original distance and has dense image. The constructed space is complete: a Cauchy sequence of its ideal points can be approximated by original points with errors tending to zero, producing a single Cauchy sequence in X representing the desired limit. Any two completions of X are related by a unique isometry that agrees with their embeddings of X. The phrase unique here includes the requirement of fixing the original space; unrelated relabelings are not being counted as different completions.
A uniformly continuous map from X into a complete metric space extends uniquely to the completion by sending an approximating Cauchy sequence to the limit of its images. Uniform continuity ensures the images are Cauchy and that equivalent sequences have identical image limits. Ordinary continuity alone need not suffice. The map 1/x on the positive rational numbers is continuous, but cannot extend continuously with a finite real value at the new point zero. A dense domain determines an extension if one exists; existence requires additional control.
Work it through
Two approximations, one added endpoint
Let X=(0,1) with usual distance. Compare a_n=1/(n+2) and b_n=1/(n+2)^2 as candidates for points in its completion.
Both sequences converge to zero in the real line, so each is Cauchy in X. Neither converges to an element of X because zero is excluded.
Their nth-term distance is 1/(n+2)−1/(n+2)^2, a nonnegative quantity bounded by 1/(n+2), which tends to zero. Thus the two sequences are equivalent.
Both therefore represent the same added point. Sequences approaching the other endpoint 1 give another class, and the complete interval [0,1] contains X isometrically and densely.
The completion is isometric to [0,1], and both displayed sequences represent its endpoint zero. Different convergence speeds do not create different completion points.
The living notebook
Explore the model
Interactive lab · optimization
Different speeds can approach the same point
For positive initial x, compare step sizes 0.5 and 0.75. The exact sequences are and , so their termwise difference tends to zero. They illustrate two representatives of the missing endpoint of (0,1) when started inside that interval. The lab does not build equivalence classes or display a completion automatically.
- 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 completion identifies approximation procedures whose mutual distance tends to zero.
Pause & reason
Check your understanding
Your turn
Try it yourself
Let D be the rational numbers in [0,1]. Explain why f(q)=2q+3 extends uniquely to a continuous real-valued function on [0,1], and identify the extension.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
A single bound on output differences proves uniform continuity. Then approximate an arbitrary real input by rationals.
Compare with the solution
For q and r in D, the output difference is 2|q−r|, so f is uniformly continuous. The interval [0,1] is a completion of D with the inherited metric and the target real line is complete. If q_n approaches x, the values 2q_n+3 approach 2x+3, giving the extension. Any continuous extension agrees with this value by density and uniqueness of limits. Thus the extension is F(x)=2x+3 everywhere on [0,1].
Keep the idea
Take it with you
- Density permits arbitrarily accurate approximation without implying equality or large measure.
- Completion points are equivalence classes of mutually convergent Cauchy approximations.
- Uniform continuity and a complete target guarantee extension; density guarantees uniqueness once continuity is imposed.
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