CONSTRUCTING SPACE-FILLING CURVES OF COMPACT CONNECTED MANIFOLDS

YING-FEN LIN AND NGAI-CHING WONG

Abstract. Let M be a compact connected (topological) manifold of finite or infinite dimension n. Let 0 ≤ r ≤ 1 be arbitrary but fixed. We construct in this paper a spacefilling curve f from [0, 1] onto M, under which M is the image of a compact set A of Hausdorff dimension r. Moreover, the restriction of f to A is one-to-one over the image of a dense subset provided that 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)} . The proof is based on the special case where M is the Hilbert cube [0, 1]^{ω}.

1. Introduction

Following the first example given by Peano in 1890, we know that every n-dimensional cube [0, 1]^{n} has a space-filling curve (see, e.g. [11]). In other words, [0, 1]^{n} is a continuous image of the unit interval [0, 1]. This fact is eventually generalized to give

Theorem 1 (Mazurkiewicz and Hahn, (see e.g. [11, p. 106])). Let X be a metrizable space. Then X is a continuous image of [0, 1] if and only if X is compact, connected, and locally connected.

As a consequence of Theorem 1, in addition to finite dimensional cubes [0, 1]^{n}, n = 1, 2, …, the Hilbert cube ℍ = [0, 1]^{ω}, i.e. the product space of countably infinitely many copies of [0, 1], also has a space-filling curve. It is known that every separable infinite dimensional compact convex set in a Fréchet space is affinely homeomorphic to ℍ (see, e.g. [1, p. 100] or [8, p. 40]). Consequently, there are also space-filling curves of such spaces.

A metric space M is called a Hilbert cube manifold if for each x in M, there is a base of neighborhoods of x in which every member is homeomorphic to an open subset of ℍ (see, e.g. [1, p. 298]). When M is compact, it is equivalent to saying that there exist compact subsets U_{1}, …, U_{k} of M such that M is covered by the interiors of U_{1}, …, U_{k} and each of them is homeomorphic to ℍ. In this paper, compact (topological) manifolds M are either modeled on [0, 1]^{n} if dim M = n < ∞, or modeled on ℍ = [0, 1]^{ω} if dim M = ∞.

The existence of a space-filling curve of any compact connected manifold is ensured by Theorem 1. In this paper, we shall construct a computable space-filling curve f of the Hilbert cube ℍ. Similar results has been obtained for finite dimensional cubes [0, 1]^{n} in [5] for n = 1, 2, . . .. In [3], M. Bestvina and J. J. Walsh showed that for positive integers n > m ≥ 2, there is a surjective continuous function g from ℝ^{n} onto ℝ^{m} that is one-to-one over the image of a dense subset. Moreover, if ℝ^{m} = g(K) for any σ-compact subset K of ℝ^{n} then dim K = n. In our construction, for any pre-assigned r between 0 and 1, we can construct explicitly a space-filling curve f from [0, 1] onto [0, 1]^{n}, n = 1, 2, …, ω, maps a compact set A of dimension r onto [0, 1]^{n}. Moreover, the restriction of f to A is one-to-one over the image of a dense subset provided 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)}. Similar conclusions are carried to compact connected manifolds, which supplement the results in [2, 3, 4].

Date: April 24, 2000; submitted to Archiv der Math.

1991 Mathematics Subject Classification. 54F15, 54F20, 54F45.

Key words and phrases. space-filling curves, Hilbert cube manifolds, Hausdorff dimensions.

Partially supported by National Science Council of the Republic of China under grants: NSC 86-2115-M110-002, 37128F.

A part of this paper was finished when the second author was visiting the Goldsmiths College, University of London. He would like to thank Professor Cho-Ho Chu for the warm hospitality.

There is a variety of applications of space-filling curves. To name a few, we mention [6] for embedding Urysohn space into C[0, 1], [12] for classifying geometric finiteness of Kleinian groups, and [9] for converting integral equations in n variables into one involving one variable. See also [13] for more interesting information.

2. Main results

Recall that the Hilbert cube ℍ can be embedded into the separable Hilbert space l_{2} as the set \{(x_{n}) : 0 ≤ x_{n} ≤ \frac{1}{n}\} in norm topology (see, e.g. [1, p. 100]). For computational ease, we identify ℍ as the norm compact convex set \{(x_{n}) : 0 ≤ x_{n} ≤ \frac{1}{2^{n−1}} \} in l_{2}, and

frequently write ℍ = ∏_{n=1}^{∞} [0, \frac{1}{2^{n−1}} ] in l_{2} if no confusion arises.

Lemma 2. We can construct a space-filling curve f of the Hilbert cube ℍ, under which ℍ is the image of a compact subset A of [0, 1] of Hausdorff dimension zero. Moreover, the restriction of f to A is one-to-one over the image of a dense subset.

Proof. We take a sequence of integers {q_{k}} such that q_{k} ≥ 2^{k} + 2 for k = 1, 2, … and

lim_{k→∞} \frac{k}{log_{2} q_{k}} = 0. Let A_{1}, A_{2}, A_{3} … be compact subsets of the interval [0, 1] defined by

A_{l} = \{∑_{k=1}^{∞} \frac{t_{k}}{q_{1} ⋯ q_{2^{k}}} : t_{k} = 1, 2, 3, …, 2^{k}, k = 1, 2, 3, …, 2^{l}\}

for all l = 1, 2, 3, . . .. Observe that

A = ⋂_{l=1}^{∞} A_{l} = \{∑_{k=1}^{∞} \frac{t_{k}}{q_{1} ⋯ q_{2^{k}}} : t_{k} = 1, 2, 3, …, 2^{k}, k = 1, 2, 3, …\}

is compact. Since A_{l} is a disjoint union of 2 × 2^{2} × 2^{3} × ⋯ × 2^{2^{l}} = 2^{(2^{l}+1)2^{l−1}} intervals each of length \frac{1}{q_{1}···q_{2^{l}}}, the Hausdorff p-dimensional measure of A_{l} for any p > 0 is

H_{p}^{∗}(A_{l}) = 2^{(2^{l}+1)2^{l−1}}(\frac{1}{q_{1} ⋯ q_{2^{l}}})^{p}, l = 1, 2, 3, ….

Thus,

H_{p}^{∗}(A) = lim_{l→∞} H_{p}^{∗}(A_{l}) = lim_{l→∞} \frac{2 · 2^{2} ⋯ 2^{2^{l}}}{q_{1}^{p} · q_{2}^{p} ⋯ q_{2^{l}}^{p}} .

Let ϵ(k) = \frac{k}{log_{2} q_{k}}. Then k = log_{2} q_{k}^{ϵ(k)}, or 2^{k} = q_{k}^{ϵ(k)}. Since ϵ(k) → 0^{+} and q_{k} → ∞ as k → ∞, we have

\frac{2^{k}}{q_{k}^{p}} = \frac{q_{k}^{ϵ(k)}}{q_{k}^{p}} = q_{k}^{ϵ(k)−p} → 0 if p > 0.

Consequently, the Hausdorff dimension of A is

dim A = inf{p > 0 : H_{p}^{∗}(A) = 0} = 0.

Our desired space-filling curve f : [0, 1] → ℍ is given by sending t in [0, 1] to the point

(x_{1}(t), x_{2}(t), x_{3}(t), …) in ℍ = ∏_{n=1}^{∞} [0, \frac{1}{2^{n−1}} ] ⊆ l_{2}. More precisely, we write t in its q-

expansion t = ∑_{k=1}^{∞} \frac{t_{k}}{q_{1} ⋯ q_{k}} where t_{k} belongs to \{0, 1, 2, …, q_{k} − 1\}, and write

x_{1}(t) = 0.x_{11} x_{12} x_{13} ⋯; x_{2}(t) = 0.0 x_{22} x_{23} ⋯; x_{3}(t) = 0.0 0 x_{33} ⋯; ⋮ } in base 2 expansion.

Denote by (a)_{2} the base 2 representation of a. We assign q_{0} = t_{0} = x_{nk} = 0 for k = 0, 1, 2, …, n − 1, where n = 1, 2, …, and

x_{11} = \begin{cases}y_{1} & if 1 ≤ t_{1} ≤ 2^{1}, (t_{1} − 1)_{2} = y_{1}, \\ 1 & if 2^{1} + 1 ≤ t_{1} ≤ q_{1} − 2, \\ 0 & if t_{1} = 0 = t_{0} or q_{1} − t_{1} = 1 = q_{0} − t_{0}, \\ 1 & if t_{1} = 0 ≠ t_{0} or q_{1} − t_{1} = 1 ≠ q_{0} − t_{0}; \end{cases}

(x_{12}, x_{22}) = \begin{cases}(y_{1}, y_{2}) & if 1 ≤ t_{2} ≤ 2^{2}, (t_{2} − 1)_{2} = y_{1}y_{2}, \\ (1, 1) & if 2^{2} + 1 ≤ t_{2} ≤ q_{2} − 2, \\ (x_{11}, 0) & if t_{2} = 0 = t_{1} or q_{2} − t_{2} = 1 = q_{1} − t_{1}, \\ (1 − x_{11}, 1) & if t_{2} = 0 ≠ t_{1} or q_{2} − t_{2} = 1 ≠ q_{1} − t_{1}; \end{cases}

⋮

In general, (x_{1n}, x_{2n}, …, x_{nn})

= \begin{cases}(y_{1}, y_{2}, …, y_{n}) & if 1 ≤ t_{n} ≤ 2^{n}, and \\ & (t_{n} − 1)_{2} = y_{1}y_{2} ⋯ y_{n}, \\ (1, 1, …, 1) & if 2^{n} + 1 ≤ t_{n} ≤ q_{n} − 2, \\ (x_{1 n−1}, x_{2 n−1}, …, x_{n−1 n−1}, 0) & if t_{n} = 0 = t_{n−1} or \\ & q_{n} − t_{n} = 1 = q_{n−1} − t_{n−1}, \\ (1 − x_{1 n−1}, 1 − x_{2 n−1}, …, 1 − x_{n−1 n−1}, 1) & if t_{n} = 0 ≠ t_{n−1} or \\ & q_{n} − t_{n} = 1 ≠ q_{n−1} − t_{n−1}. \end{cases}

A routine verification will show that even for those t having two distinct q-expansions, the values of x_{1}(t), x_{2}(t), x_{3}(t), … are unique. We check that f is (uniformly) continuous on [0, 1]. For ϵ > 0, fix a positive integer n such that

∑_{k=n+1}^{∞} (\frac{1}{2^{k−1}} )^{2} < \frac{ϵ}{2}.

For x in ℍ, write x = (x_{1}, x_{2}, …, x_{n}, …) in l_{2}. Observe that

∥x∥_{2}^{2} = ∑_{k=1}^{∞} x_{k}^{2} < ∑_{k=1}^{n} x_{k}^{2} + \frac{ϵ}{2}.

Let m be a positive integer such that \frac{n}{2^{m}} < \frac{ϵ}{2}. Let δ = \frac{1}{q_{1}q_{2}···q_{m+1}} . Suppose t, t′ ∈ [0, 1] such that |t − t′| < δ. We write t, t′ in their q-expansions with infinitely many nonzero digits t_{k} and t′_{k}. In this way, t_{k} = t′_{k} for k = 1, 2, …, m. Let x = f(t) and x′ = f(t′). The first m digits of x_{k} and x′_{k} agree, and thus |x_{k} − x′_{k}| ≤ \frac{1}{2^{m}}, for k = 1, 2, . . .. Then

∥x − x′∥_{2}^{2} < ∑_{k=1}^{n} |x_{k} − x′_{k}|^{2} + \frac{ϵ}{2} ≤ \frac{n}{2^{m}} + \frac{ϵ}{2} < \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ.

It is plain that the image of A under this curve is the entire of ℍ.

Finally, let ℍ_{0} be the subset of ℍ consisting of points x such that f^{−1}(x) contains more than one points in A. Let A_{0} = f^{−1}(ℍ_{0}) ∩ A. It is not difficult to see that a point x = (x_{1}, x_{2}, …) ∈ ℍ_{0} if and only if at least one coordinate x_{i} has a finite binary expansion. Correspondingly, the q-expansion of any point t in A_{0}, when f(t) = x, will have a special

form t = ∑_{k=1}^{∞} \frac{t_{k}}{q_{1}q_{2} ⋯ q_{k}} in which the ith digits of the binary expansion of t_{k}−1 are eventually

constant as k → ∞. Obviously, A\A_{0} is dense in A, ℍ\ℍ_{0} is dense in ℍ, and f is one-to-one from A\A_{0} onto ℍ\ℍ_{0}.

In the following, we denote by [a] the greatest integer part of a real number a.

Lemma 3. For each real number G ≥ 1, there exists a sequence of positive integers {q_{k}}, chosen from {[G], [G] + 1}, such that

lim_{k→∞} (q_{1}q_{2} ⋯ q_{k})^{\frac{1}{k}} = G.

Proof. Set q_{1} = [G]. We shall choose subsequent q_{k} to satisfy the inequalities

[G] G^{k−1} ≤ q_{1}q_{2} ⋯ q_{k} ≤ ([G] + 1)G^{k−1}.

Suppose q_{1}, q_{2}, …, q_{k−1} are chosen accordingly. In case q_{1}q_{2} ⋯ q_{k−1} ≥ G^{k−1}, we set q_{k} = [G]; otherwise, we set q_{k} = [G] + 1. It is easy to see that q_{k} does not violate the above inequalities. Finally, we observe that

(\frac{[G]}{G})^{\frac{1}{k}} ≤ \frac{(q_{1}q_{2} ⋯ q_{k})^{\frac{1}{k}}}{G} ≤ (\frac{[G] + 1}{G})^{\frac{1}{k}}

for all k = 1, 2, . . .. Hence, lim_{k→∞} (q_{1}q_{2} ⋯ q_{k})^{\frac{1}{k}} = G.

Lemma 4. For 0 < r ≤ 1, we can construct a space-filling curve f of the Hilbert cube ℍ, under which ℍ is the image of a compact subset A of [0, 1] of Hausdorff dimension r. Moreover, the restriction of f to A is one-to-one over the image of a dense subset.

Proof. Let G = 2^{1/r} ≥ 2. Utilizing Lemma 3, we get a sequence {p_{k}} of positive integers chosen from {[G], [G] + 1} such that

lim_{k→∞} (p_{1}p_{2} ⋯ p_{k})^{\frac{1}{k}} = G.

Set

q_{1} =p_{1}p_{2} ≥ 2^{2} = 2^{1} + 2,
q_{2} =p_{3}p_{4}p_{5} ≥ 2^{3} > 2^{2} + 2,
q_{3} =p_{6}p_{7}p_{8}p_{9} ≥ 2^{4} > 2^{3} + 2,
⋮

In general, for n = 1, 2, 3, …, we set

q_{n} = p_{φ(n−1)+1} ⋯ p_{φ(n)} ≥ 2^{n+1} ≥ 2^{n} + 2,

where φ(0) = 0, and

φ(n) = 2 + 3 + ⋯ + (n + 1) = \frac{n(n + 3)}{2}, n = 1, 2, ….

With the sequence {q_{n}} in hand, we can proceed as in the proof of Lemma 2 and obtain a compact subset A of [0, 1] whose Hausdorff p-dimensional measure is

H_{p}(A) = lim_{l→∞} \frac{2 · 2^{2} ⋯ 2^{2^{l}}}{(q_{1}q_{2} ⋯ q_{2^{l}})^{p}} = lim_{l→∞} \frac{2^{2^{l−1}(2^{l}+1)}}{(p_{1}p_{2} ⋯ p_{φ(2^{l})})^{p}}

= lim_{l→∞} (\frac{2^{2^{l}+1/2^{l}+3}}{(p_{1}p_{2} ⋯ p_{φ(2^{l})})^{p/φ(2^{l})}} )^{φ(2^{l})}.

It is plain that H_{p}(A) = ∞ whenever G^{p} < 2, and H_{p}(A) = 0 whenever G^{p} > 2. Hence, dim A = r. The rest of the proof goes exactly as in that of Lemma 2.

The finite dimensional version of Lemmas 2 and 4 has been obtained earlier. It is, however, still open to us that if the upper bound \frac{log 2^{n}}{log(2^{n}+2)} can be removed from the following statement.

Lemma 5 ([5, Theorem 2]; see also [7]). Let n ≥ 2 be any positive integer and 0 ≤ r ≤ 1. There exists a continuous curve f from [0, 1] onto [0, 1]^{n} under which [0, 1]^{n} is the image of a compact set A of Hausdorff dimension r. Moreover, the restriction of f to A is one-to-one over the image of a dense subset provided 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)}.

Here comes the main result of this paper.

Theorem 6. Let 0 ≤ r ≤ 1 and M be a compact connected manifold of dimension n, where n = 1, 2, …, ω. We can construct a space-filling curve f of M under which the entire manifold M is the image of a compact subset A of [0, 1] of Hausdorff dimension r. Moreover, the restriction of f to A is one-to-one over the image of a dense subset provided 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)} (= 1 if dim M = ω).

Proof. Suppose M is a compact, connected manifold of dimension n (1 ≤ n ≤ ω). Then there exists a family of compact subsets {U_{1}, U_{2}, …, U_{m}} of M in which each U_{i} is homeomorphic to [0, 1]^{n}, and M ⊆ ⋃_{i=1}^{m} int U_{i}. Without loss of generality, we can assume by connectedness of M that, (U_{1} ∪ ⋯ ∪ U_{k}) ∩ U_{k+1} ≠ ∅ for k = 1, 2, …, m − 1. There are homeomorphisms h_{1}, h_{2}, …, h_{m} from U_{1}, U_{2}, …, U_{m} onto [0, 1]^{n}, and space-filling curves g_{1}, g_{2}, …, g_{m} from [0, \frac{1}{2m−1}], [\frac{2}{2m−1}, \frac{3}{2m−1}], …, [\frac{2m−2}{2m−1}, 1] onto [0, 1]^{n}, respectively.

Suppose p_{1} is a point in U_{1} ∩ U_{2}. Let

h_{1}^{−1}(α_{1}, α_{2}, …) = p_{1} = h_{2}^{−1}(β_{1}, β_{2}, …),

where (α_{1}, α_{2}, …) and (β_{1}, β_{2}, …) are in [0, 1]^{n}. Note that the surjective maps

f_{1} = h_{1}^{−1} ∘ g_{1} : [0, \frac{1}{2m − 1}] → U_{1} and f_{2} = h_{2}^{−1} ∘ g_{2} : [\frac{2}{2m − 1}, \frac{3}{2m − 1}] → U_{2}

are continuous. Let (α′_{1}, α′_{2}, …) = h_{1}(f_{1}(\frac{1}{2m−1})) = g_{1}(\frac{1}{2m−1}) in [0, 1]^{n}. Extend f_{1} to [0, \frac{3}{2(2m−1)}] by setting

f_{1}(\frac{1}{2m − 1} + λ\frac{1}{2(2m − 1)}) = h_{1}^{−1}(λα_{1} + (1 − λ)α′_{1}, λα_{2} + (1 − λ)α′_{2}, …)

for 0 ≤ λ ≤ 1. In particular,

f_{1}(\frac{3}{2(2m − 1)}) = h_{1}^{−1}(α_{1}, α_{2}, …) = p_{1}.

Similarly, let (β′_{1}, β′_{2}, …) = h_{2}(f_{2}(\frac{2}{2m−1})) = g_{2}(\frac{2}{2m−1}) in [0, 1]^{n}. Extend f_{2} to [\frac{3}{2(2m−1)}, \frac{3}{2m−1}] by setting

f_{2}(\frac{2}{2m − 1} − λ\frac{1}{2(2m − 1)}) = h_{2}^{−1}(λβ_{1} + (1 − λ)β′_{1}, λβ_{2} + (1 − λ)β′_{2}, …)

for 0 ≤ λ ≤ 1. In particular,

f_{2}(\frac{3}{2(2m − 1)}) = h_{2}^{−1}(β_{1}, β_{2}, …) = p_{1}.

Therefore, f_{1} and f_{2} agree at the point of the intersection of their domains. As a result, f_{1} ∪ f_{2} is continuous from [0, \frac{3}{2m−1}] onto U_{1} ∪ U_{2}.

In a similar manner, we can construct a continuous function f = ⋃_{k=1}^{m} f_{k} from [0, 1] onto M. Moreover, there are compact subsets B_{k} of [\frac{2k−2}{2m−1}, \frac{2k−1}{2m−1}] as in Lemmas 2, 4 or 5 such that each B_{k} is of any pre-assigned Hausdorff dimension r, for 0 ≤ r ≤ 1, and g_{k}(B_{k}) fills up the whole of [0, 1]^{n}. In case 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)}, we can also assume that g_{k} is one-to-one over the image of a dense subset of B_{k} for each k = 1, 2, …, m.

We set

A_{1} =B_{1},
A_{2} =\overline{g_{2}^{−1}(h_{2}(U_{2}\U_{1})) ∩ B_{2}},
⋮
A_{n} =\overline{g_{n}^{−1}(h_{n}(U_{n}\(U_{1} ∪ ⋯ ∪ U_{n−1}))) ∩ B_{n}}.

Since h_{k} is a homeomorphism, we see that each C_{k} = g_{k}^{−1}(h_{k}(U_{k}\(U_{1} ∪ ⋯ ∪ U_{k−1}))) ∩ B_{k} is an open subset of B_{k} for k = 1, 2, …, n. Set A = ⋃_{k=1}^{n} A_{k} ⊆ [0, 1]. Then A is a compact set of Hausdorff dimension r such that f(A) = M. Moreover, the restriction of f to A is one-to-one over the image of a dense subset of A contained in ⋃_{k=1}^{∞} C_{k} provided 0 ≤ r ≤ \frac{log 2^{n}}{log(2^{n}+2)}.

3. Two examples

Example 7. A space-filling curve of the 3-dimensional cube [0, 1]^{3}.

A space-filling curve t ↦ (x(t), y(t), z(t)) of [0, 1]^{3} is given by writing

t = 0.t_{1}t_{2} ⋯ in base 10 expansion

and

\begin{matrix}x(t) = 0.x_{1}x_{2} ⋯ \\ y(t) = 0.y_{1}y_{2} ⋯ \\ z(t) = 0.z_{1}z_{2} ⋯ \end{matrix} \} in base 2 expansion

(in particular, t_{0} = x_{0} = y_{0} = 0) such that for k ≥ 1,

(x_{k}, y_{k}, z_{k}) = \{ \begin{matrix}(α, β, γ), & if 0 ≤ t_{k} − 1 = 4α + 2β + γ ≤ 7; \\ (x_{k−1}, y_{k−1}, z_{k−1}), & if t_{k} = 0 = t_{k−1} or t_{k} = 9 = t_{k−1}; \\ (1 − x_{k−1}, 1 − y_{k−1}, 1 − z_{k−1}), & if t_{k} = 0 ≠ t_{k−1} or t_{k} = 9 ≠ t_{k−1}. \end{matrix}

In general, the first k digits (in base 2) of x(t), y(t) and z(t) can be calculated in terms of the first k digits (in base 10) of t. The image of

A = \{∑_{k=1}^{∞} \frac{t_{k}}{10^{k}} : t_{k} = 1, 2, …, 8, k = 1, 2, … \}

fills up the entire cube [0, 1]^{3}. In this case, dim A = log 8/ log 10 and f is one-to-one over the image of a dense subset of the compact set A.

To have an idea how the Hilbert cube ℍ = ∏_{n=1}^{∞} [0, \frac{1}{2^{n−1}} ] is filled up, we re-scale our curve

to the one f(t) = (x(t), y(t)/2, z(t)/4). In Figure 1, we draw three polygons each of which approximates this space-filling curve within 1/2 (order 1), 1/4 (order 2), and 1/8 (order 3) uniformly in all x-, y- and z-directions, respectively. They are obtained by making linear interpolation for the sets of data consisting of first one, two, and three digits of t, x(t), y(t) and z(t), respectively, according to the methods described in [10] (in which we represent 1 = 0.99 ⋯ in base 10 for convenience).

Example 8. A space-filling curve of the ellipsoid E = \{(x, y, z) ∈ ℝ^{3} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1\} (a, b, c > 0).

We first construct a space-filling curve of the sphere S = {(x, y, z) ∈ ℝ^{3} : x^{2}+y^{2}+z^{2} = 1}. Let 0 < ϵ < 1 and

U_{1} = {(x, y, z) ∈ S^{2} : −1 ≤ z ≤ ϵ}
U_{2} = {(x, y, z) ∈ S^{2} : −ϵ ≤ z ≤ 1}.

Then {U_{1}, U_{2}} is a compact covering of S. We are going to define the homeomorphisms h_{i} from U_{i} onto [0, 1]^{2} for i = 1, 2.

Consider the stereographic projections P_{i} : U_{i} ⟶ D via the north pole (when i = 1) and the south pole (when i = 2), respectively, where

D = \{(a, b) ∈ ℝ^{2} : a^{2} + b^{2} ≤ \frac{1 + ϵ}{1 − ϵ}\}.

It is easy to see that

P_{1}(x, y, z) = (\frac{x}{1 − z} , \frac{y}{1 − z}) and P_{2}(x, y, z) = (\frac{x}{1 + z} , \frac{y}{1 + z}).

The next step is to consider the circle-to-square map

h′(a, b) = \begin{cases}\frac{∥(a,b)∥_{2}}{∥(a,b)∥_{∞}} (a, b) & if (a, b) ≠ (0, 0), \\ (0, 0) & if (a, b) = (0, 0), \end{cases}

where

∥(a, b)∥_{2} = \sqrt{a^{2} + b^{2}} and ∥(a, b)∥_{∞} = max\{|a|, |b|\}.

It is plain that the map

h(a, b) = \frac{1}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}}h′(a, b) + (\frac{1}{2}, \frac{1}{2})

is a homeomorphism from D onto [0, 1]^{2}. Consequently, h_{i} = h ∘ P_{i} is a homeomorphism from U_{i} onto [0, 1]^{2} for i = 1, 2.

Let g : [0, 1] ⟶ [0, 1]^{2} be a space-filling curve. For instance, we can take g to be the one given by Lemma 5 as in [5, Example 3]. More precisely, the space-filling curve g(t) = (x(t), y(t)) is given by writing

t = 0.t_{1}t_{2} ⋯ in base 6 expansion

and

\begin{matrix}x(t) = 0.x_{1}x_{2} ⋯ \\ y(t) = 0.y_{1}y_{2} ⋯ \end{matrix} \} in base 2 expansion

(in particular, t_{0} = x_{0} = y_{0} = 0) such that for k ≥ 1,

(x_{k}, y_{k}) = \{ \begin{matrix}(α, β), & if 0 ≤ t_{k} − 1 = 2α + β ≤ 3; \\ (x_{k−1}, y_{k−1}), & if t_{k} = 0 = t_{k−1} or t_{k} = 5 = t_{k−1}; \\ (1 − x_{k−1}, 1 − y_{k−1}), & if t_{k} = 0 ≠ t_{k−1} or t_{k} = 5 ≠ t_{k−1}. \end{matrix}

Then g(A) = [0, 1]^{2} for the compact set A = \{∑_{k=1}^{∞} \frac{t_{k}}{6^{k}} : t_{k} = 1, 2, 3, 4, k = 1, 2, … \} of Hausdorff dimension log 4/ log 6. Moreover, g is one-to-one over the image of a dense subset of A.

Let

f_{1} : [0, 1/3] → U_{1} and f_{2} : [2/3, 1] → U_{2}

be defined by

f_{1}(t) = h_{1}^{−1}(g(3t)) and f_{2}(t) = h_{2}^{−1}(g(3 − 3t)).

Following the proof of Theorem 6, we observe that

h_{1}f_{1}(1/3) = h_{2}f_{2}(2/3) = g(1) = (1, 1) ∈ [0, 1]^{2},
(\sqrt{1/2}, \sqrt{1/2}, 0) ∈ U_{1} ∩ U_{2},

and

h_{1}(\sqrt{1/2}, \sqrt{1/2}, 0) = h_{2}(\sqrt{1/2}, \sqrt{1/2}, 0) = (\frac{1}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}} + \frac{1}{2}, \frac{1}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}} + \frac{1}{2}) ∈ [0, 1]^{2}.

We can extend f_{1} from [0, 1/3] to [0, 1/2] and f_{2} from [2/3, 1] to [1/2, 1] by setting

f_{1}(\frac{1}{3} + λ\frac{1}{6}) = h_{1}^{−1}((1 − λ) · 1 + λ · (\frac{1}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}} + \frac{1}{2}), (1 − λ) · 1 + λ · (\frac{1}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}} + \frac{1}{2}))

= h_{1}^{−1}(1 − \frac{λ}{2} + \frac{λ}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}}, 1 − \frac{λ}{2} + \frac{λ}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}}),

and similarly,

f_{2}(\frac{2}{3} − λ\frac{1}{6}) = h_{2}^{−1}(1 − \frac{λ}{2} + \frac{λ}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}}, 1 − \frac{λ}{2} + \frac{λ}{2}\sqrt{\frac{1 − ϵ}{1 + ϵ}})

for 0 ≤ λ ≤ 1. In this way, f_{1}(1/2) = f_{2}(1/2) = (\sqrt{1/2}, \sqrt{1/2}, 0) and we have a continuous map f = f_{1} ∪ f_{2} from [0, 1] onto S. Suppose

f(t) = (x(t), y(t), z(t)) for t ∈ [0, 1].

Then, the map

g(t) = (ax(t), by(t), cz(t))

is a space-filling curve of the ellipsoid E. Moreover, g maps the \frac{log 4}{log 6}–dimensional compact set A onto E such that g is one-to-one over the image of a dense subset of A.

In Figure 2, we draw approximating polygons of g when a = 2b = 4c = 1 and ϵ = 0 for demonstration. To make the picture more easily to be visualized, only the lower hemiellipsoid is shown. Note that setting ϵ = 0 (for simplicity) in this case is still good enough for our task (either by direct observation or arguing by uniformity).

References

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[12] G. P. Scott and G. A. Swarup, Geometric finiteness of certain Kleinian groups, Proc. Amer. Math. Soc. 109, no. 3 (1990), 765–768.

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Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung, Taiwan, 804, Republic of China

E-mail address: linyf, wong@math.nsysu.edu.tw

Figure 1. Approximating polygons of order 1 (top), 2 (middle), and 3 (bottom) of a space-filling curve of [0, 1] × [0, \frac{1}{2}] × [0, \frac{1}{4}]. These figures are generated by Mathematica version 3.0 in SUN SPARC20-712.

Figure 2. Approximating polygons of order 2 (top), 3 (second), 4 (third), and 5 (bottom) of the lower half of a space-filling curve of the ellipsoid x^{2} + 4y^{2} + 16z^{2} = 1. These figures are generated by Mathematica version 3.0 in SUN SPARC20-712.