Re: Complement of zero dimensional space

On Sat, 27 Jun 2009, G. A. Edgar wrote:
<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 at most n if and only if
it can be written as the union of n+1 sets of dimension at most zero.

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.

Are the sets disjoint?

At first glance, the two seem equivalent.
Is this a theorem of dimension theory?

A space has dimension -1 iff it can be written
as the union of 0 sets of dimension zero.

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.

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 least 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).

Oh? Let T = [0,1]^2 \/ [2,3]^2.

If T is locally at least two dimensional and Z zero dimensional,
then T - Z is at least one dimensional.

Why not at least two dimensional?
R^2 = P^2 \/ Q^2 \/ (PxQ \/ QxP) and as
P^2 \/ Q^2 is path connected, it's one dimensional?
That's a weird one dimension. Is it's Hausdorff dimemsion a fraction?

If U is any open subset of T, then U - Z is at least one dimensional,
hence not empty. Thus T - Z is dense subset T.

If T - Z is not connected, then some open U,V
with empty (U /\ V)\Z, T - Z subset U \/ V.
U /\ V subset Z; T subset U \/ V \/ Z.

If U,V intersect, then some nonnul open, at least two dimensional W
with W subset U /\ V subset Z. Thus U,V disjoint.

Hence T /\ U subset U \/ Z/\U = U and T /\ V subset V
T /\ U\/V subset U\/V subset T and T = U \/ V is disconnected.

Does at least one dimensional imply locally path connected?
Then when T - Z is connected, it would be path connected.

If a space is at least n dimensional, isn't
it also locally at least n dimensional?

No. If a space is at least n dimensional at some point,
then isn't it at least n dimesnional?

A disk with a tail is 2 dimensional and locally one dimensional
along the tail. What's is it at the point of attachment to the
disk? Does it have a Hausdorff dimension there?
.

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