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

A closed nonnul subset K without isolated points of S = {0,1}^N
is homeomorphic to S. I'm inclided to accept your recent
version, yet still have the details to rework and verify.

Theorem: A zero dimensional T0 space embeds in {0,1}^w(S)
where w(S) is the weight, ie smallest cardinality of a base.

A zero dimensional 2nd countable, compact T0 space without isolates
ie zero dimensional compact metric space without isolates
is homeomorphic to S because it embeds into S as closed subset.
It's also homeomorphic to the cantor set C, for it too is an infinite
zero dimensional compact metric space. Specifically, C homeomorphic S.

Whoops, except for the empty space. BTW, in a single
point space, is that point considered isolated? ;-)

Question: are two countable dense subsets of S homeomorphic?
Observation: A dense subset of S has no isolated points.

Well yes, so the real question is can this be shown within S
as was done with the first question?

Result: A countable, 2nd countable regular T0 space D without isolated
points, ie a countable metric space without isolated points, is
homeomorphic to Q. (Whoops, D not empty space)

Proof: D is zero dimensional T0; D embeds in S
cl D has no isolated points; cl D homeomorphic S; D dense subset S
Thus also Q dense subset D as Q is countable, 2nd countable etc.
Hence by question, D homeomorphic Q

Feeling up the question: For every finite binary sequence s,
there is a point p in D, for which s is the prefix.

pick any p and map it to 1/2
pick p0 with prefix 0 and map it to 1/4
pick p1 with prefix 1 and map it to 3/4

pick p00 with prefix 00 and map it to 1/8
pick p01 with prefix 01 and map it to 3/8

pick p10 with prefix 10 and map it to 5/8
pick p11 with prefix 11 and map it to 7/8

in an attempt to map D to the dyatic rationals Z[1/2] /\ (0,1)
which by theorem is order isomorphic, hence homeomorphic to Q.

Conclusion: ;-)
The dyatic cube is nicer and more fun than the Cantor set.

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