Abstract
The sentence “that space is complete” is one of at least four separate statements. This paper shows numerically that on the Riemannian side the four are tied into one, and that changing the signature unties them. NO NEW MATHEMATICAL THEOREM AND NO NEW PHYSICAL LAW IS CLAIMED.
First, on the Riemannian side the four fail together. On the plane with the origin removed, a geodesic reaches the hole at finite length, a Cauchy sequence loses its destination, and a closed bounded set stops being compact — the three do not fail separately (Section 2).
Second, this is the core. On the same set, changing the metric reverses the answer. The radius out to $r$ in the open disc has Euclidean length $r$ and stays bounded, while in the hyperbolic metric it reaches $19.11382792$ at $r=0.99999999$ and has no bound at all (Section 3).
Third, in the hyperbolic metric the geodesic never arrives. The unit-speed radial geodesic is $r(s)=\tanh(s/2)$, and its gap to the boundary is $1-r(s)=2/(e^{s}+1)$, which is never zero for any $s$ (Section 4).
Fourth, on the Lorentzian side one of the four cannot even be stated. The interval is not positive definite, so it is not a distance, and the phrase “a Cauchy sequence converges” has nothing to attach to (Section 5).
Fifth, another one turns around. For timelike paths from $(0,0)$ to $(2,0)$ the straight one has proper time $2$, and one bent out to $x=0.3$ has $1.90787840283389$ — the geodesic is not the shortest but the longest (Section 5).
Scope (what is not claimed)
NO NEW MATHEMATICAL THEOREM AND NO NEW PHYSICAL LAW IS CLAIMED — the Hopf–Rinow theorem, the Poincaré disc, the Rindler wedge and the maximality of timelike geodesics are all standard. Hopf–Rinow is not proved here — this paper only looks, through examples, at what the theorem demands. The singularity theorems are not treated — what geodesic incompleteness follows from is not the question here. No construction is given separating the timelike, null and spacelike cases — that they split is stated, and no separating example is supplied. Relation to Paper 161: 161 took a census of “fixed point” and used a complete metric space as a hypothesis of Banach's theorem — this paper takes a census of that hypothesis itself. Relation to Paper 198: 198 measured the gap between straight and shortest on the Riemannian side — this paper writes what happens to that gap when the sign turns over. What is added is setting out that the four fail together on a punctured plane, measuring by length that the same open disc is complete or incomplete depending on the metric, putting a number on the reversal of the inequality for timelike paths, and noting that a flat but incomplete example is extendible.
1. Notation and Position
“Complete” has at least four readings. (a) Geodesically complete: every geodesic extends to all values of its parameter. (b) Metrically complete: every Cauchy sequence converges. (c) Heine–Borel: closed bounded sets are compact. (d) Minimising geodesics exist: any two points are joined by a geodesic of least length. Every number below is computed here, in multiprecision arithmetic carrying $40$ digits.
Position: Paper 161 took a census of “fixed point” and, along the way, noted that Banach's theorem assumes a complete metric space. This paper makes that assumption the subject — and finds that it wears four different faces.
2. On the Riemannian Side, the Four Fail Together
Take the plane with the origin removed, with the ordinary Euclidean metric.
Check 2.1: set the three failures side by side.
| Reading | What happens here |
|---|---|
| Geodesically complete | a geodesic running inward from radius $r_0$ reaches the hole at length $r_0$ |
| Metrically complete | the points at radius $1/n$ form a Cauchy sequence whose limit was removed |
| Heine–Borel | “radius at most $1$” is closed and bounded, and that sequence has no convergent subsequence |
(Confirming the Cauchy property: the gap between radius $1/n$ and radius $1/2n$ is $0.05$ at $n=10$, $0.005$ at $n=100$, $0.0005$ at $n=1000$ and $0.00005$ at $n=10^4$.)
Observation 2.2 The three did not fail separately. One and the same hole knocks all three readings down at once. That is what the Hopf–Rinow theorem says: on a connected Riemannian manifold the four are equivalent.
3. The Core: On the Same Set, Changing the Metric Reverses the Answer
Measure the open unit disc with two metrics. The set is the same.
Check 3.1: measure the straight radius from the centre out to $r$ in both. In the Euclidean metric the length is $r$; in the Poincaré metric it is $\log\frac{1+r}{1-r}$.
| $r$ | Euclidean length | Hyperbolic length |
|---|---|---|
| $0.9$ | $0.9$ | $2.944438979$ |
| $0.99$ | $0.99$ | $5.293304825$ |
| $0.999$ | $0.999$ | $7.600402335$ |
| $0.9999$ | $0.9999$ | $9.903437551$ |
| $0.999999$ | $0.999999$ | $14.50865724$ |
| $0.99999999$ | $0.99999999$ | $\mathbf{19.11382792}$ |
Observation 3.2 It is the same path. Measured one way its length never passes $1$; measured the other way it has no bound. So the same open disc is incomplete in one metric and complete in the other — completeness is a property of the metric, not of the set.
Observation 3.3 It is not that one has a boundary and the other does not. The boundary is in the same place. What changed is only how much it costs to get there.
4. In the Hyperbolic Metric the Geodesic Never Arrives
Check 4.1: the unit-speed radial geodesic is $r(s)=\tanh(s/2)$. Measure its gap to the boundary.
| Arc length $s$ | $r(s)$ | $1-r(s)$ |
|---|---|---|
| $1$ | $0.462117157260009759$ | $0.537883$ |
| $5$ | $0.986614298151430289$ | $0.0133857$ |
| $10$ | $0.999909204262595131$ | $9.07957\times10^{-5}$ |
| $30$ | $0.999999999999812848$ | $1.87152\times10^{-13}$ |
| $60$ | $1$ to the $40$ digits carried | $1.7513\times10^{-26}$ |
Observation 4.2 The numbers alone cannot show that it never arrives — at $s=100$ the gap falls outside the $40$ digits carried. So the closed form is used: $1-\tanh(s/2)=2/(e^{s}+1)$, which is not zero for any finite $s$. The geodesic extends forever — in this metric the disc is geodesically complete.
Observation 4.3 The same caution from Paper 199 applies here — “never arrives” is invisible to any finite computation. What is visible is only a number shrinking exponentially, and that it is not zero is said by the formula.
5. On the Lorentzian Side, One Evaporates and One Turns Around
Change the signature. Two things happen.
Observation 5.1 (one evaporates) The interval is not positive definite. The separation of two distinct points can be zero or negative, so there is no metric space — and therefore the phrase “a Cauchy sequence converges” has nothing to attach to. One of the four does not become false: it becomes unstatable.
Check 5.2 (one turns around): measure the proper time of timelike paths from $(t,x)=(0,0)$ to $(2,0)$. The straight one gives $2$. Compare a path bent out to $x=X$ at $t=1$.
| $X$ | Proper time of the bent path | Shorter than straight? |
|---|---|---|
| $0$ | $2$ | equal |
| $0.1$ | $1.98997487421324$ | shorter |
| $0.3$ | $1.90787840283389$ | shorter |
| $0.5$ | $1.73205080756888$ | shorter |
| $0.9$ | $0.871779788708135$ | shorter |
| $0.99$ | $0.282134719593318$ | shorter |
Observation 5.3 The straight path is the longest. So the fourth Hopf–Rinow reading — that any two points are joined by a geodesic of least length — is structurally false on this side. Paper 198 measured where straight and shortest come apart on the Riemannian side — here they do not merely come apart: the sign turns over.
Observation 5.4 (one splits in three) Geodesic completeness also stops being one thing — it splits into timelike, null and spacelike. This paper does not construct examples showing those three independent — it states only that they split.
6. Flat, Still Incomplete, and Extendible
Check 6.1: cut the region $x>|t|$, the Rindler wedge, out of flat spacetime. The curvature is identically zero. Measure how long an inertial worldline from $(t,x)=(0,2)$ moving toward the vertex at speed $v$ takes, in proper time, to leave the wedge.
| $v$ | $\gamma$ | Proper time to the edge |
|---|---|---|
| $0.1$ | $1.005037815$ | $1.809068067$ |
| $0.3$ | $1.048284837$ | $1.467598771$ |
| $0.5$ | $1.154700538$ | $1.154700538$ |
| $0.9$ | $2.294157339$ | $0.458831468$ |
| $0.99$ | $7.088812050$ | $0.141776241$ |
Observation 6.2 Nothing is curved, and the geodesic still reaches the edge in finite proper time — the wedge is geodesically incomplete. And this is not a singularity: beyond the edge the rest of flat spacetime continues, so the space is extendible.
Observation 6.3 So “incomplete” and “singular” are different words. Incompleteness can be produced by how a region was cut out — cut a piece away and any space, however flat, becomes incomplete. To be singular takes something more: that it cannot be extended. This has the same shape as the flat cylinder of Paper 198 — there too the curvature was innocent, and the culprit was the global shape.
7. The Fence and the Separator
Observation 7.1 (where the four end up)
| Reading | Riemann | Lorentz |
|---|---|---|
| Geodesically complete | holds | splits in three |
| Metrically complete | holds | cannot be stated |
| Heine–Borel | holds | no corresponding theorem |
| Minimising geodesics exist | holds | turns around: they are maximal |
Observation 7.2 (the fence) The fence around Hopf–Rinow is not curvature, not dimension and not smoothness, but that the metric is positive definite. Inside the fence the four are one; the moment one steps outside, one evaporates, one splits, and one reverses. Paper 195 identified the fence around linearisation as hyperbolicity, and Paper 171 the fence around least squares as linearity — the fence here is the signature.
Observation 7.3 (the separator) What separates the four is what is allowed to end after something finite — a parameter (geodesic completeness), a sequence of points (metric completeness), a cover (Heine–Borel), or whether a minimum can be attained (minimising geodesics). On the Riemannian side these four say one thing; on the Lorentzian side they do not.
8. The Census and the Close
What this paper checked:
- On a punctured plane the three fail together (Check 2.1).
- The same open disc is complete or incomplete according to the metric (Check 3.1): the same path measures $0.99999999$ and $19.11382792$.
- In the hyperbolic metric the geodesic never arrives (Check 4.1): $1-\tanh(s/2)=2/(e^{s}+1)$ is never zero. The formula says so, not the numbers.
- On the Lorentzian side metric completeness cannot be stated (Observation 5.1).
- And shortest turns into longest (Check 5.2): straight gives $2$, bent gives $1.90787840283389$.
- Flat, incomplete, and extendible (Check 6.1): incomplete is not singular.
- The fence is the signature (Observation 7.2).
References
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Reference
Yamagishi, Y. (2026). “Complete” Is Four Separate Statements, and What Ties Them Into One Is That the Metric Is Positive Definite [Paper 201]. YOI: https://www.yamagishiyuuki.com/yoi/201
DOI: 10.5281/zenodo.22109052
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 201. This paper is a preprint and has not been peer reviewed.