Re: Compact subsets of {0,1}^N

On Fri, 24 Jun 2005 23:23:53 -0700, William Elliot
<marsh@xxxxxxxxxxxxxxxxxx> wrote:

>Let K,L be two closed subsets of S = {0,1}^N without isolated points.
>
>How to show K and L are homeomorphic?

Probably you also want to assume that K and L are nonempty.

>{0,1} is understood to be a discrete subspace of R.
>S is zero dimensional compact Hausdorff space.
>K,L are uncountable zero dimensional compact subsets.
>
>Yes S is homeomorphic to the Cantor set, however without reference
>to the Cantor set and theorems about it, how can one show K and L
>are homeomorphic directly from the properties of S?

First note that we can assume that L = S (if we do this case
then in the general case it follows that K and L are both
homeomorphic to S, hence they are homeomorphic to each other.)

If a is a _finite_ sequence of 0's and 1's let S_a be the
set of all elements of S that begin with a. Note that
we're including the case where a is the empty sequence,
ie a sequence of length 0: if a is the empty sequence
then S_a = S.

Let B be the tree of all finite sequences of 0's and
1's, and let T be the tree of all finite sequences a
such that S_a intersect K is nonempty. Define a map
F from B onto T by induction:

If a is the empty sequence let F(a) = a. For each
a in B, of length n, let a_0 and a_1 be the two
extensions of a of length n + 1 (a_j consists of
a followed by a j.) Now if a is in B and F(a) has
already been defined, define F(a_0) and F(a_1)
as follows: Let b = F(a). If both b_0 and b_1
are elements of T then let F(a_0) = b_0 and
F(a_1) = b_1. On the other hand, if only one
of b_0, b_1 is in T, then let F(a_0) = F(a_1)
= either b_0 or b_1, whichever one is an element of T.

It follows that if a' is an extension of a then
F(a') is an extension of F(a), so F induces a
map f:S -> S in a natural way (given a in S,
f(a) is the unique element of S such that
f(every finite initial segment of a) is an
initial segment of f(a).

The fact that K is closed shows that f(S) is
a subset of K; the fact that F(B) = T shows
that in fact f(S) = K. It's easy to see that
f is continuous (given a and n there exists
m such that if a' agrees with a in the first
m places then f(a') agrees with f(a) in the
first n places). And the fact that K has no
isolated points shows that f is 1-1: If b is
in T then S_b intersect K must contain more
than one point, because if it contained only
one point that would be an isolated point
of K. So if a' <> a and b = f(a) then the
fact that b eventually splits in T (b has
some descendant in T which has two descendants
in T) shows that f(a') <> f(a). So f is a
homeomorphism of S onto K.

************************

David C. Ullrich
.

Relevant Pages

  • Re: Compact subsets of {0,1}^N
    ... >> where a is the empty sequence, ie a sequence of length 0: ... Since K has no isolated points and S_b is open, K intersect S_b ... such that b' is the only extension of b in T of length m ... >>So f is a homeomorphism of S onto K. ...
    (sci.math)
  • Re: Compact subsets of {0,1}^N
    ... > Probably you also want to assume that K and L are nonempty. ... > First note that we can assume that L = S (if we do this case ... > where a is the empty sequence, ie a sequence of length 0: ... So f is a homeomorphism of S onto K. ...
    (sci.math)
  • Re: Compact subsets of {0,1}^N
    ... > Probably you also want to assume that K and L are nonempty. ... > where a is the empty sequence, ie a sequence of length 0: ... > Let B be the tree of all finite sequences of 0's and 1's, ... > homeomorphism of S onto K. ...
    (sci.math)