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

On Sat, 25 Jun 2005, David C. Ullrich wrote:

>

Let K be a closed nonnul subset of {0,1}^N without isolated points.
How to show K homeomorphic to S = {0,1}^N ?

> 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).
>
I'm not understanding this correctly, for it seems to conclude
f(a) = a when a in K. Thus since f(a) in K, ff(a) = f(a) and were f
injection, which it's want to be, then f(a) = a for all a which is no no.

Would you clarify?

> 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.
>
.

Quantcast