The problem of the existence of an equi-partition of a curve in R n has recently been raised in the context of computational geometry (see [2] and [3]). The problem is to show that for a (continuous) curve Γ : [0, 1] → R n and for any positive integer N, there exist points t 0 = 0 < t 1 < . . . < t N -1 < 1 = t N , such that d(Γ(t i-1 ), Γ(t i )) = d(Γ(t i ), Γ(t i+1 )) for all i = 1, . . . , N, where d is a metric or even a semi-metric (a weaker notion) on R n .
In fact, this problem, for R n replaced by any metric space (X, d) was given a positive solution by Urbanik [4] under the (necessary) constraint Γ(0) = Γ(1).
We show here that the existence of such points, in a broader context, is a consequence of Brower’s fixed point theorem (for reference see any intermediate level book on topology, e.g. [1]). Let ∆ n be the n-dimensional simplex and σ n be its boundary. Let τ be a permutation of {0, . . . , n}. We define a map ϕ τ : ∆ n → ∆ n in the following way: Every x ∈ ∆ n has a unique representation x = n i=0 α i (x)p i , where p i are the vertices of ∆ n and 0 ≤ α i (x) ≤ 1, n i=0 α i (x) = 1. We then define:
ϕ τ is, of course, continuous.
Lemma 1 Let τ be a cyclic permutation of {0, 1, . . . , n} and let Ψ : σ n → σ n be a map such that, for every proper face A of ∆ n , Ψ(A) ⊂ ϕ τ (A). Then Ψ has no fixed point.
Proof. Assume that Ψ(x) = x for some x ∈ σ n and let x = n i=0 α i (x)p i be the representation of x. Let A = face(x) be the minimal face of ∆ n containing x. Without loss of generality we may assume that A = conv{p 0 , . . . , p k } for some 0 ≤ k < n. That is, x = k i=0 α i (x)p i and α i (x) > 0 for i = 0, . . . , k. Now, by the assumption, Ψ(x
. By the uniqueness of the representation we have: {p τ (0) , . . . , p τ (k) } = {p 0 , . . . , p k } (and β i = α τ -1 (i) for all 0 ≤ i ≤ k). But, since τ is cyclic, no proper subset of {0, . . . , n} is mapped by τ on itself. This is a contradiction.
Let X be a non-empty set and Γ :
1] (this last property holds, for example, if X is a metric space, Γ is continuous and d is continuous on X × X). Associated with d as above, we denote, for
Theorem 2 which follows is motivated by the above context, but, as stated, is true for any continuous function D : [0, 1] ×[0, 1] → R + . In the particular case that D is constructed as above, with d a metric on X, Theorem 2 implies Urbanik’s result [4].
Then for every positive integer N there exist points
Moreover, if for a given N there are no points
Proof. If there are points
then the first claim is obvious. Thus let us assume that such a sequence does not exist and prove the “Moreover” assertion. Let S be the (N -1)-dimensional simplex
and let F : S → R N be defined by
Then, F is continuous and F (S) ⊂ R N + . Moreover, every proper face A of S is characterized as being the set of (N -1)-tuples (t 1 , . . . , t N -1 ) ∈ S such that, in the chain of inequalities 0
(here, and in the sequel, t 0 = 0 and t N = 1).
Let Σ be the (N -1)-dimensional simplex which is the convex hull in R N of the unit coordinate vectors e i , i = 1, . . . , N. Let G : R N + \ {0} → Σ be the radial projection with center 0 ∈ R N . The composed map GF maps S continuously into Σ (as we have excluded the possibility that 0 ∈ F (S)). Claim The relative interior of Σ is contained in GF (S). We remark that, by simple induction, it follows from the above Claim that, in fact, GF maps S onto Σ, but the Claim as is, is clearly sufficient to complete the proof of Theorem 2, since it implies that, for any positive α 1 , . . . , α N with N i=1 α i = 1, the ray L from the origin in R N with parametric equation L : x = (α 1 t, . . . , α N t), t > 0 , that intersects Σ at the point (α 1 , . . . , α N ), must, by the Claim, intersect also F (S). Proof of the Claim. Assume, for contradiction, that there exists a point P in the relative interior of Σ, such that P ∈ GF (S). Let H : Σ \ {P } → ∂Σ be the radial projection onto ∂Σ with center P (in the hyperplane containing Σ). H is continuous on Σ \ {P } and, since P ∈ GF (S), the map HGF : S → ∂Σ is continuous. H restricted to ∂Σ is the identity, thus, by the previous discussion, HGF (A) ⊂ Σ ∩ A for every proper face A of S. Let us identify now S with Σ by the natural affine homeomorphism that identifies every face A of S with Σ ∩ A. Via this identification, HGF induces a continuous map Φ : Σ → ∂Σ with the property that Φ(A) ⊂ A for every proper face A of Σ. Now let ϕ τ : Σ → Σ be the map associated with a cyclic permutation of the vertices of Σ, as in the context of Lemma 1, and let Ψ = ϕ τ Φ : Σ → ∂Σ .
Then Ψ restricted to ∂Σ satisfies the assumptions of Lemma 1, hence Ψ has no fixed point in ∂Σ. Considering Ψ as a map from Σ into Σ we get a continuous map from Σ into itself that has no fixed point. This contradicts Brower’s fixed point theorem.
Reminder: Brower’s fixed point theorem states that any continuous map from B K into B K has a fixed point (B K is the closed unit ball of R K but, of course, can be replaced by anything homeomorphic to it, like Σ here).
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