Chapters 1½ and 2½ met sets between one and two dimensions and between two and three. This one asks the question from the other end: can a dimension be a number like π, and what has to be true of the space around it? The answer is yes for a set and no for a vector space, and the construction fits on one line.
A vector space has a whole-number dimension: the number of independent directions. A set has other dimensions that measure how it fills space, and those take any real value from 0 up to the dimension of the space it sits in. The similarity dimension of Chapter 1½, log N / log(1/r), is one of them; the tent curve came out at exactly 4/3.
That upper limit is the whole constraint. A set with dimension 3.14159… is bigger than anything that fits in ordinary three-dimensional space, so it needs at least four dimensions around it. Four is enough, because 3.14159… is less than 4. Chapter 4, the hypercube, is exactly the four-dimensional container this set lives in.
Let s = π − 3 = 0.14159…, and let C be the Cantor set made of two pieces, each a copy of the whole scaled by r, one at each end of [0, 1]. Choose r so the similarity equation holds:
2 · rs = 1, so r = 2−1/s = 0.0074812… (that is 1/r = 133.67…)
Since r is less than 1/2 the two pieces do not touch, so the set is a genuine Cantor dust, and its dimension is log 2 / log(1/r) = s. The step from the similarity equation to the Hausdorff dimension is a standard theorem about self-similar sets with non-overlapping pieces (the open set condition; Moran and Hutchinson). It is quoted here, not derived.
Now take C × [0, 1]³, a Cantor dust times a solid cube, which sits in R⁴. The box-counting argument is elementary and worth seeing, because the number π − 3 does all the work. The level-k picture of C has 2k intervals of length rk. Cover C × [0,1]³ by cubes of side ε = rk: each interval needs one cube across, and the cube [0,1]³ needs (1/ε)³ of them, so the count is
N(ε) = 2k · (r−k)³ = (r−k)s · (r−k)³ = ε−(3+s) = ε−π
using 2 = r−s. So at the scales ε = rk the box count grows exactly like ε−π. That gives the box-counting dimension at those scales. That the Hausdorff dimension equals it is the standard product theorem for Hausdorff dimension (the product of a set with R³, where both Hausdorff and box dimension of the cube are 3, adds the dimensions). That step is the product formula for Hausdorff dimension found in standard texts on fractal geometry (Falconer's, for example); it is quoted, not derived.
A Cantor dust with r this small is almost invisible, so the plate shows what self-similarity says in pictures: each row is the left stripe of the row above, magnified by 1/r. Every row looks the same. Drag the slider to choose the target dimension; r changes and the stripes get wider as the dimension approaches 4. The stripes are drawn full height because they stand for slabs C × R³ seen edge-on in one direction.
pi-verify.py does arithmetic and counts. It does not prove the dimension theorems. Its output for the π case:
s = pi-3 = 0.1415926536 r = 2^(-1/s) = 0.0074812066 (1/r = 133.6683) log2/log(1/r) = 0.141592653590 (s = 0.141592653590) Moran sum 2*r^s = 1.000000000000 pieces disjoint (r < 1/2): True dimension of C x R^3 = 3.141592653590 container R^4 needs dim <= 4: True
| target d | s = d − 3 | r = 2−1/s | pieces disjoint |
|---|---|---|---|
| 3.1 | 0.1 | 0.000976563 | yes |
| 3.25 | 0.25 | 0.0625 | yes |
| 3.5 | 0.5 | 0.25 | yes |
| π | 0.14159 | 0.00748121 | yes |
| 3.9 | 0.9 | 0.462937 | yes |
At the level-aligned scales ε = rk the exact count log N / log(1/ε) equals s at every level checked (1 to 6). Between those scales the count wobbles, which is what a very sparse set does: at ε = 10−1 … 10−10 the estimate ranges from 0.30 down to 0.162, and a straight regression over those decades gave 0.158 against the true 0.142. That is the expected behaviour of box counting on a gappy set over a finite range, and it is the reason to prefer the exact level count to a fitted slope.
The constructions in Chapter 1½ cut a unit square into a grid and keep N of m² cells, so their dimension is log N / log m with whole numbers N and m. In the unit 4-cube the same recipe allows any N up to m⁴. That gives dimensions that are logarithms of whole numbers, and it can get as close to π as you like: the verifier found m = 37 with N = 84,460 pieces giving 3.141593, an error of about 4 × 10−8.
It does not give π exactly. Hitting it would need mπ to be a whole number, and this check does not settle whether that can happen. The construction above sidesteps the question by using a ratio that is not a reciprocal of a whole number (1/r = 133.67…), which a lattice cannot do. The honest statement is: the lattice route approximates π, and the non-lattice ratio realises it exactly.
Proposition (lattice dimensions are dense). Let 0 < d ≤ 4, let m ≥ 2 be a whole number with x = md ≥ 2, and let N = ⌊x⌋. Then 1 ≤ N ≤ m4 and
0 ≤ d − ln N / ln m ≤ 2 / (x ln m), and also ≤ ln 2 / ln m. Consequently the dimensions ln N / ln m of lattice sets in the unit 4-cube are dense in (0, 4], and every interval of dimensions contains infinitely many distinct ones.
Proof. From x − 1 < N ≤ x and x ≥ 2 we get x/2 ≤ N ≤ x, and x = md ≤ m4, so N fits in the m4 cells of the grid. Taking logarithms, ln x − ln 2 ≤ ln N ≤ ln x = d ln m, which gives 0 ≤ d − ln N / ln m ≤ ln 2 / ln m. For the sharper rate use N/x > 1 − 1/x and the inequality ln(1 − t) ≥ −2t for 0 ≤ t ≤ 1/2: then ln N ≥ ln x − 2/x, and dividing by ln m gives the bound 2/(x ln m). Fix d and let m grow: ln 2 / ln m → 0 and md → ∞, so for every ε > 0 some m, N give a lattice dimension within ε of d. That is density. If an open interval contained only finitely many lattice dimensions, choose a point d′ in the interval different from all of them and take ε smaller than its distance to each: the construction then produces a lattice dimension within ε of d′ that is none of the finitely many, a contradiction. So every interval contains infinitely many. ∎
The N pieces are N distinct cells of an m-grid, so they do not overlap, and the dimension of the resulting set is ln N / ln m by the same quoted theorem as in §2. For d = π and m = 37 this construction gives N = 84,459 with error 3.2 × 10−6 against the bound 6.6 × 10−6; lattice-density-check.py tests the bound on about 180,000 random pairs (d, m) with no violation, and the largest error observed was 0.61 of the sharp bound. The cruder bound ln 2 / ln m and the density statement itself are machine-checked in LatticeDensity.lean (see the list below); the sharp bound is proved on paper only. This is an elementary statement and may well be known; no search of the literature has been made.
The lattice route would hit π exactly only if mπ = N for whole numbers m and N. As far as the author knows this is an open question; even "is 2π rational?" is listed as unknown. Below is every route considered, with what it actually gives.
| Route | What it gives | Status |
|---|---|---|
| 1 · Certified bracketing | mπ lies strictly between two consecutive integers for every m from 2 to 1,000,000 (999,999 values, none undecided). By hand: 8 < 2π = 8.82497… < 9 and 31 < 3π = 31.54428… < 32. | Proved for those m, by interval arithmetic (pi-integer-power.py). Nothing beyond the limit. |
| 2 · Six exponentials theorem | For any three distinct primes, at least one of pπ is transcendental, hence not a whole number. So at most two primes p can have pπ algebraic. (Take xi = ln pi and y = (1, π). The theorem is usually stated with two x's and three y's; the set of products xiyj is the same with the roles swapped.) The four exponentials conjecture would sharpen this to at most one prime. | Quoted theorem (see M. Waldschmidt's survey slides, Hyderabad 2003); the consequence is one step. Speaks about primes, not directly about composite m. |
| 3 · Gelfond–Schneider, Baker | Nothing. They need an algebraic exponent, or algebraic coefficients in a linear form of logarithms, and π is neither. | Does not apply. |
| 4 · Schanuel's conjecture | Would, as far as the author knows, settle whether mπ is algebraic. | Conditional on an unproved conjecture; the implication is not derived here. |
| 5 · Heuristic | Fractional parts of mπ behave like random numbers, so an exact whole-number hit has probability zero. The nearest approach for m up to 106 is 2.9 × 10−7 (at m = 973,492), about what a random sample of that size would give. | Evidence only, not proof. |
Each route stalls at a known place. Route 1 is finite and cannot reach every m. Route 2 constrains primes and leaves up to two of them open, and a composite mπ is a product of prime powers of π that could in principle be algebraic even if none of the factors is. Routes 3 and 4 need either a hypothesis that fails or a conjecture. The five nearest misses found by route 1:
| m | distance to nearest whole number |
|---|---|
| 973,492 | 2.858 × 10−7 |
| 751,568 | 5.523 × 10−7 |
| 433,759 | 1.089 × 10−6 |
| 941,693 | 2.817 × 10−6 |
| 473,615 | 3.626 × 10−6 |
The certificate rests on the interval arithmetic of the library used (outward-rounded enclosures of π and of mπ at 80 digits); it is as reliable as that library. None of this affects the dimension-π set of §2, which never needs mπ to be a whole number.
PiDimension.lean proves, for every s > 0, that 2·rs = 1 for r = 2−1/s; that the pieces are disjoint for 0 < s < 1; and the box-count identity 2k·(ε−1)3 = (ε−1)3+s, also with 3 + s = π. It compiles on Lean v4.32.0 with only the standard axioms. It does not prove that the Hausdorff dimension equals π; that step is the quoted theorem. The certificates for mπ in §4 are computed, not Lean-checked.LatticeDensity.lean proves that for 0 < d ≤ 4, m ≥ 2 and md ≥ 2, the integer N = ⌊md⌋ satisfies 1 ≤ N ≤ m4 and 0 ≤ d − ln N / ln m ≤ ln 2 / ln m, and from that, for every d in (0, 4] and every ε > 0, the existence of m and N with 1 ≤ N ≤ m4 and |ln N / ln m − d| < ε. It compiles on Lean v4.32.0 with only the standard axioms. It does not prove the sharp 2 / (md ln m) bound, and it does not prove that the grid construction has Hausdorff dimension ln N / ln m; that is the quoted theorem again.