Re: Complement of zero dimensional space

In article <20090626221727.U8082@xxxxxxxxxxxxxxx>, William Elliot
<marsh@xxxxxxxxxxxxxxxx> wrote:

On Fri, 26 Jun 2009, G. A. Edgar wrote:
<marsh@xxxxxxxxxxxxxxxx> wrote:

Conjecture. If S is a zero dimensional subspace of R^2,
then R^2 - S is a dense, path connected subspace.

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.

A = [0,1] /\ Q and B = [1,2] /\ Q are two zero dimensional spaces.

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.


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 has
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.

Correction has been made as per your other post.

Let x,y be distinct points in R^2. Since R^2 has dimension 2, any set
that separates x from y has dimension at least 1. So: R^2 - S is
connected.

This did not do path connected, though.

If T is at lest two dimensional and connected and Z is
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).

--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
.

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