From: "Philip D. Loewen"
Newsgroups: sci.math.research
Subject: Re: Connectedness in T_2 Spaces
Date: Sun, 01 Nov 1998 19:45:30 -0800
Earlier this week, Julian Dontchev wrote:
> Let X be the Moore plane. The x-axis S is discrete and hence not connected
> (as a subspace). However, if you partition S into two sets A and B, then
> A and B cannot be contained in disjoint open sets in X. Clearly, X is
> Hausdorff.
The validity of this counterexample clearly rests on a precise
definition of "the Moore plane". If we take the definition given in
Stephen Willard's "General Topology" (Addison-Wesley, 1970), then it
doesn't withstand close scrutiny.
I have tried to "attach" a picture illustrating two disjoint open
subsets of the Moore plane whose union contains the entire horizontal
axis. Interested readers can scroll down for that, or review the theory
that follows.
[attached binary file deleted in interest of brevity -- djr]
According to Willard, the Moore plane consists of all points (x,y) in
the Cartesian plane with y>=0 , with a topology generated by the
"basic open sets" listed below:
(i) All open Euclidean balls in which every (x,y) has y>0;
(ii) Each set formed by taking an the union of an open Euclidean ball
tangent to the x-axis with its point of tangency.
There is a natural notion of "radius" for either type of basic open set.
Now for the advertised pair of disjoint open sets:
U = union over all x <= 0 of basic neighbourhoods with radius 1 for the
points (x,0). This looks like an infinite strip of height 2 running
horizontally to the left of the y-axis, with a semicircular end
protruding into the right half plane. The nonpositive x-axis is included
in U, but all other points on the semicircular arc and the top of the
strip are not.
V = union over all strictly positive x of suitable basic neighbourhoods
of the points (x,0). A basic neighbourhood of (x,0) is "suitable" if its
radius r satisfies r < minimum of (x^2/4) and 1. This is more
complicated to describe pictorially, but V is nonetheless a union of
basic open sets that are each disjoint from U, and this union contains
all points of the x-axis that aren't already in U.
Thanks to all who took an interest in my problem: my summary is forthcoming.
Philip D. Loewen
loew@math.ubc.ca
==============================================================================
From: "Philip D. Loewen"
Newsgroups: sci.math.research
Subject: Re: Connectedness in T_2 Spaces
Date: Sun, 01 Nov 1998 23:04:37 -0800
Thanks to these four colleagues who addressed my question on the fine
points of connectedness:
1. Julian Dontchev
2. Keith Ramsay
3. Paul Jan Szeptycki
4. Alain Verberkmoes
Ramsay [2] provided a Hausdorff topology on the upper half of the
complex plane in which the topology induced on the real axis is discrete
(about as disconnected as they come), but no pair of disjoint open sets
can be found that both meet the real axis and cover it with their
union. Almost simultaneously, Verberkmoes [4] provided a more general
construction that reduces to Ramsay's when one takes B as the real line
in its metric topology. Since [4] has been posted already, and
generalizes [2], I won't repeat the details here. (Both [2] and [4]
arrived in the same hour, well before my local news server had even put
up the question!)
Szeptycki [3] provided the following "do-it-yourself" solution:
If you can find a space X containing two disjoint connected closed
subsets A and B that can't be separated (in the sense that if U is open
and contains A and V open and contains B, then U and V are not disjoint)
then the subspace A u B would satisfy property [C] but not be
connected. To get such an X, consider any nonnormal T3 space and modify
so that the sets witnessing nonnormality are connected. E.g. take X to
be the deleted Tychonoff plank (example 87 in Steen and Seebach's
Counterexamples in Topology) and stick a unit interval between each
successive ordinal in omega_1 and omega.
I have replied to Dontchev's posting [1] separately.
Again, warm thanks to all who contributed.
Philip D. Loewen
loew@math.ubc.ca