Re: Complement of zero dimensional space
- From: Denis Feldmann <denis.feldmann.sansspam@xxxxxxx>
- Date: Sun, 28 Jun 2009 06:12:06 +0200
G. A. Edgar a écrit :
In article <20090626221727.U8082@xxxxxxxxxxxxxxx>, William Elliot
<marsh@xxxxxxxxxxxxxxxx> wrote:
On Fri, 26 Jun 2009, G. A. Edgar wrote:<marsh@xxxxxxxxxxxxxxxx> wrote:A = [0,1] /\ Q and B = [1,2] /\ Q are two zero dimensional spaces.
Conjecture. If S is a zero dimensional subspace of R^2,A topological space has topological dimension n if and only if it can
then R^2 - S is a dense, path connected subspace.
be written as the union of n+1 sets of dimension zero.
Thus C = A \/ B = [1,2] /\ Q is one dimensional?
Hmmm... I was a bit off... How's this:
A topological space has topological dimension at most n if and only if
it can be written as the union of n+1 sets of dimension at most zero.
or
A topological space has topological dimension n if and only if it can
be written as the union of n+1 sets of dimension zero and cannot be
written as the union of n sets of dimension zero.
This is ridiculous. Which are the 3 sets (of dimension 0) whose union is R^2 ? Why dont you use *boundaries* ? (see http://en.wikipedia.org/wiki/Inductive_dimension )
.
R^2 has topological dimension 2.By what definition of dimension?
"topological dimension" ... In separable metric space, these all agree:
covering dimension, small inductive dimension, large inductive
dimension.
So: If S is a zero-dimensional subspace of R^2, then R^2 - S hasCorrection has been made as per your other post.
dimension at least 1. Furthermore, for any nonempty open set U in R^2,
also U - S has dimension at least 1. This proves R^2 - S is dense in R^2.
Let x,y be distinct points in R^2. Since R^2 has dimension 2, any setIf T is at lest two dimensional and connected and Z is
that separates x from y has dimension at least 1. So: R^2 - S is
connected.
This did not do path connected, though.
zero dimensional, is T - Z dense connected subset of T?
For the proof above we need more (every nonempty open set in T is at
least two dimensional) and less (no need to assume T is connected).
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