Problem Set Nine: Space Filling Curves
Definitions: Let S and T be metric spaces.
(1) A function h : S  T is a homeomorphism iff h is 1-1, onto, continuous and has a continuous inverse.
(2) S and T are homeomorphic (in symbols S~T) iff there is a homemorphism from one to the other.
(3) The middle third Cantor set will be denoted by C and { 0, 1 } N by  .
Theorem: Let S and T be metric spaces. If S is compact and h : S  T is 1-1, onto and continuous, then
the inverse function h  1 : T  S is continuous. In other words h is a homeomorphism.
Theorem (Problem 7.7): If h :   C is given by h(x)  2 
n
k 1
3
k
, then h is a homeomorphism.
x(k)
Definition. Let S and T be metric spaces and 0  p  1 . A function f : S  T is Lipschitz of order p iff
there is a constant c > 0 so that d  f(x), f(z)   c d(x, z) p for all x and z in S. These functions are uniformly
continuous and δ  (ε /c) 1/p in the epsilon-delta definition. The p-Lipschitz constant of f is
p
Lip (f)  inf c ,
c as in the defining inequality.
Theorem: For a and b integers with 0  b  a , define T :    by T(f)(k)  f( a k  b ) , i.e.,
T(f)   f ( a  b ), f ( 2 a  b ), f ( 3 a  b ), ... etc ...  .
Then for any f and g in  ,
d T(f), T(g)
  2 1  b/a d f, g  1/a .
Proof. Fix f and g in  and consider two cases.

 1/a
First, if 2  a  b  d f, g  then d T(f), T(g)   1  2 a  b d(f, g)
as required.
Second, if d f, g   2  a  b then there is a natural number n with 2  (n  1) a  b  d f, g   2  n a  b .
In this case f ( a k  b )  g ( a k  b ) for all k = 1, 2, …, n, for otherwise f ( a s  b )  g ( a s  b ) for some s =
1, 2, …, n would imply d(f, g)  f (a s  b)  g (a s  b) 2  (a s  b)  2  a s  b (1)  2  an  b .
Since f ( a k  b )  g ( a k  b ) for k = 1, 2, …, n,
d T(f), T(g)  

 k  n 1
2
n

 2
f(a k  b)  g(a k  b) 2
ab
d(f, g)

1/a
2
1  b/a
k

d(f, g)

 k  n 1 2
k
.
1/a
Theorem: For a natural number a let Δ a (respectively, C a ) be the product of a copies of Δ (resp., C).
Then let Δ ~ Δ a and C ~ C a . If fact there is a homoemorphism H : Δ  Δ a which is Lipschitz of
order 1/a and has a lipschitz inverse.
Proof: For each integer b with 0  b  a , define T b :    by T b (f)(k)  f( a k  b ) . Define
H :Δ
 Δ a by
H(f)  ( T 0 (f), T 1 (f), ... , T a  1 (f) ) . Clearly H is one-to-one and onto.
For f and g in Δ , d ( H(f), H(g) )  max { d( T b (f), T b (g) ) : 0  b  a }  2 d( f, g )1/a .
As to the inverse,
d (f, g) 
a 1

 b  0  k 1 2
a 1
b
a 1
b
 (a k  b)


 b  0 2  k 1 2

 b0 2
k
f(a k  b)  g(a k  b)
f(a k  b)  g(a k  b)
d(T b (f), T b (g))
a
 (2  1) max{ d(T b (f), T b (g)) : 0  b  a }
Theorem: For a natural number a let [0,1]
function ψ : [0,1]  [0,1]
a
a
be the product of a copies of [0,1]. There is continuous
which is onto. Such a function is called a space filling curve.
Proof: The main point is that there is a continuous function ψ 0 from the middle third Cantor set C onto
the unit square [0,1]
H :C  C
a
. For instance ψ 0 can be taken to be the composition ψ 0  [ f, f, ..., f ] o H with
a
any homeomorphism and f the Cantor function restricted to C. The composition is certainly
continuous and is onto since f maps C on [0,1] (Problem 8-4).
ψ0
can be extended to a continuous ψ : [0,1]  [0,1]
a
as follows. A point x not in C is in some
c
component of the complement of C, say x  (a, b)  C for points a and b in C. Write x  t a  (1  t)b
and define ψ(x)  t ψ 0 (a)  (1  t) ψ 0 (b) . This extension ψ is piecewise linear on the complement of C
and maps into [0,1]
a
since the latter is convex.
PROBLEMS
Problem 9-1. If 1  p and f : R  R satisfies f(x)  f(z)  c x  z
constant.
p
for all reals x and z, then f is
Problem 9-2. If f : [a, b]  R n is continuous on [a,b] and differentiable on (a,b), then f is Lipschitz of
order p  1 for the Euclidean metrics and Lip(f)  sup { f (x)
2
: a  x  b} .
Warning; the usual Mean
Value Theorem doesn’t quite hold.
Problem 9-3. Fix 0  p  1 and let f : (0,  )  (0,  ) be f(x)  x p . Show that f is Lipschitz of order p
with Lip p (f)  1 .
Problem 9-4. Let f : S  T be onto and Lipschitz of order p. Write m  Lip p (f). If CB( s k , ε ) ,
1 k  n
cover T.
, is a collection of closed balls covering S, then the closed balls CB( f(s k ), m ε p ) , 1  k  n ,
For the next two problems use 9-4 and any necessary/reasonable properties of length and area.
Problem 9-5. If  :[0,1]  [0,1]  [0,1] is onto and Lipschitz of order p for some 0  p  1 , then p  1/2 .
In particular  can’t be a smooth curve, i.e.,  can’t be differentiable on (0,1).
Problem 9-6. The Cantor function isn’t Lipschitz of order p for any p  ln(2)/ln(3 ) .
Problem 9-7. Write out a proof that [0,1] and [0,1]  [0,1] are not homeomorphic.
CONTINUITY OF THE EXTENSION. If you had trouble believing that the extension of ψ 0 from C to
[0,1] was continuous, here are the details. Read at your own risk.
Theorem. Let T be a convex subset of R n . If C is the middle third Cantor set and f : C  T is
~
continuous then there is a continuous f : [0,1]  T whose restriction to C is f.
~
~
Proof: Define f : [0,1]  T as follows. For x in C of course f (x)  f(x) . For x not in C let (a,b) be the
unique component of C c containing x. Then a and b in C, x  t a  (1  t) b for some 0  t  1 , and by
~
convexity f (x)  t f(a)  (1  t) f(b)  T.
~
c
c
f is linear on each component of C and thus is continuous on C . Let z  C and ε  0 . By the
continuity of f there is a δ  0 so that x  z  δ and x  C  f(x)  f(z)  ε /2 . We'll check the right
~
continuity of f at z. Consider two cases.
Case I. (z, z  δ)  C   . In this case (z, z  δ)  (z, b) for some component (z,b) of C c . Define
m  f(z)  f(b) / (b  z) . For x in (z,b) x  t z  (1  t) b for some 0  t  1 and
x  z  t z  (1  t) b  z  (1  t) (b  z) ,
~
~
f (x)  f (z)  t f(z)  (1  t) f(b)  f(z)  (1  t) f(z)  f(b)
~
~
f (x)  f (z)  m (x  z)
, and
.
~
If f(z) = f(b) then f is constant on (z,b). If f(z)  f(b) chose η  min { δ , ε /2 m } and note
z  x  z  η  f(z)  f(x)  ε
Case II.
If x  C
(z, z  δ)  C  
.
. Fix
w  (z, z  δ)  C
and z  x  z  η  z  δ then f(z)  f(x)
and let
 ε.
η  min { δ , w  z} .
On the other hand if x  C and z  x  z  η  w then x  (a, b) for some component of C c . Since
that component can't contain z or w, (a, b)  (z, w) . Write x  t a  (1  t) b . Using z  a, b  z  δ
~
~
f (x)  f (z)  t f(a)  (1  t) f(b)  f(z)  t [ f(a)  f(z) ]  (1  t) [ f(b)  f(z) ]
 t f(a)  f(z)  (1  t) f(b)  f(z)  t (ε /2)  (1  t) (ε /2)  ε