Newsgroups: sci.math
From: chappell@symcom.math.uiuc.edu (Glenn Chappell)
Subject: A topology question
Date: Mon, 18 Mar 1991 04:58:30 GMT
Here's a scary little question I've run across lately, but have not
been able to figure out:
Does there exist a connected, T1 topological space (of more than one
point) such that no proper open set is connected?
Note: T1 = All points are closed.
GGC <><
==============================================================================
From: rld@function.mps.ohio-state.edu (Randall L Dougherty)
Newsgroups: sci.math
Subject: Re: A topology question
Date: 20 Mar 91 21:06:55 GMT
In article <1991Mar18.045830.3237@ux1.cso.uiuc.edu> chappell@symcom.math.uiuc.edu writes:
>Here's a scary little question I've run across lately, but have not
>been able to figure out:
>
>Does there exist a connected, T1 topological space (of more than one
>point) such that no proper open set is connected?
>
>Note: T1 = All points are closed.
It turns out that there is such a space, but it is not too easy to
construct (and quite hard to visualize). Here is a brief sketch.
Let Q be the space of rational numbers with the usual topology
(inherited from the reals). Let S be the set of finite sequences
of rational numbers. (Concatenation will be denoted by juxtaposition:
if s is a sequence and x is a number, then sx is s with x appended.)
Call a subset G of S open if it meets these three conditions:
(1) For any s in S, the set of x in Q such that sx is in G
is open in Q.
(2) For any s in G, sx is in G for all but finitely many x in Q.
(3) For any s in G and any x in Q,
sxy is in G for all sufficiently large y in Q.
(So we are using the ordering on Q as well as the topology.)
It is not hard to show that the collection of open subsets of S is
closed under finite intersections and arbitrary unions, so it is
a topology. Also, the complement of any singleton is open, so
the topology is T1.
To see that S is connected, suppose C is a nontrivial clopen subset
of S; we may assume that the null sequence is in C. Let t be a
sequence of minimal length which is not in C; then t = sx for some
s in C and x in Q. Let W be the set of y in Q such that sy is not in C;
then W is finite (by (2) for C), nonempty (since x
is in W), and open in Q (by (1) for the complement of C), which is
impossible. Therefore, C does not exist, and S is connected.
The next step is to show that, if s is in S, I is a nonempty clopen
_bounded_ subset of Q, and G(s,I) is the set of all t in S such that t
starts with sx for some x in I (this includes the sequences sx themselves),
then G(s,I) is open, and its closure is G(s,I)u{s} (u means union).
To see this, just verify that G(s,I) and S - G(s,I) - {s} satisfy
(1)-(3), but S - G(s,I) does not satisfy (2).
Now, suppose that A is a connected open subset of S; we will show
that, for any s in S and x in Q, sx is in A if and only if s is in A.
Applying this repeatedly shows that A must be either the whole space
(if the null sequence is in A) or the empty set (if the null sequence
is not in A), so we will be done.
Argue by contradiction. Suppose that there exist s in S and x in Q such
that sx is in A but s is not in A. By (1), there must be y in Q distinct
from X such that sy is in A. Let I be a clopen bounded subset of Q
which contains x but not y. Then the intersection of A with G(s,I) is
a nontrivial subset of A which is open in A (since G(s,I) is open) and
closed in A (since G(s,I)u{s} is closed and s is not in A), so
A is not connected, contradiction.
On the other hand, suppose there exist s in S and x in Q such that
sx is not in A but s is in A. Then, by (3), there is y in Q such that
sxy is in A, so we get a contradiction by applying the preceding case
to sx and y. This completes the proof.
[This construction doesn't have to be done with Q; I think it will work
using any zero-dimensional Hausdorff space which has no isolated points
and is not compact, but I haven't verified this.]
Randall Dougherty [Disclaimer out of order]
Internet: rld@function.mps.ohio-state.edu
UUCP: ...!{att,pyramid}!osu-cis!function.mps.ohio-state.edu!rld
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated." Poul Anderson, "Call Me Joe"
From linac!pacific.mps.ohio-state.edu!zaphod.mps.ohio-state.edu!function.mps.ohio-state.edu!rld Tue Mar 26 15:38:20 CST 1991
Article: 3645 of sci.math
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From: rld@function.mps.ohio-state.edu (Randall L Dougherty)
Newsgroups: sci.math
Subject: Re: A topology question
Message-ID: <1991Mar21.190720.7761@zaphod.mps.ohio-state.edu>
Date: 21 Mar 91 19:07:20 GMT
References: <1991Mar18.045830.3237@ux1.cso.uiuc.edu> <1991Mar20.210655.4257@zaphod.mps.ohio-state.edu> <1991Mar20.224105.17618@ms.uky.edu>
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In article <1991Mar18.045830.3237@ux1.cso.uiuc.edu> chappell@symcom.math.uiuc.edu (Glenn Chappell) writes:
>Does there exist a connected, T1 topological space (of more than one
>point) such that no proper open set is connected?
>
>Note: T1 = All points are closed.
In article <1991Mar20.210655.4257@zaphod.mps.ohio-state.edu> I gave
(with proof but without scaffolding) an example of such a space, which
is a bit long to include here (it's less than a day old as of this
writing, so it should still be available).
In article <1991Mar20.224105.17618@ms.uky.edu> ghot@ms.uky.edu (Allan Adler) writes:
>Can Randall Dougherty be induced to explain how he thought of this
>ingenious solution ?
OK, here goes. Of course, what is given below consecutively was actually
interrupted by fairly long periods of no progress, as well as time spent
on eating, sleeping, administering and grading final exams, and other such
nonproductive activities [1/2 :-)].
[I know that stream-of-consciousness writing is a Bad Thing, but when the
subject you are writing about is your stream of consciousness, what can
you do?]
When I saw Glenn Chappell's article, the problem looked interesting
(it is of the kind I like -- in terms of basic concepts, rather than
something where it requires months of study to understand the question).
The first step was to look for any sort of relevant example to see what
was going on. A particular case of the given requirements is that
removing any single point must disconnect the space; this reminded
me of the real line, which has this property. Of course, the
two pieces one gets by removing a point from the real line are
connected open sets, so this space does not solve the original problem.
In fact, it became clear that, in the desired space, deleting a point
could not break the set into only two pieces, because these would then
be connected open sets; instead, the two pieces would have to be
composed of smaller pieces, and so on, in the manner of the Cantor
set [the space of infinite 0-1 sequences, or the result of starting
with an interval within a line and iteratively deleting middle thirds].
So the next example to look at was a "Cantor fan": take a Cantor set
within a line and a point outside the line, to be called the base
point, and connect each point of the Cantor set to the base point
with a line segment. This set is clearly connected, but deleting
the base point causes it to fall apart in the manner the problem
seems to need.
But this only works for the base point; for the desired space, it has
to work for every point. In other words, every point should be the
base point of a fan. So let's start with one fan and, for each
non-base point, attach a new fan based there. Then the deletion of
any point in the original fan will cause this new space to fall apart.
But the newly added points will not have this property, so we will
have to attach fans to them, and iterate the process infinitely many
times, obtaining a "Cantor fan tree." This space does have the
property that deleting any point causes the space to fall apart
into infinitely (uncountably, actually) many pieces. (To reach any
point in this space from the original base point, one moves out
to somewhere in the original fan; then, maybe, one moves into the
fan based at this point; this can be repeated any finite number of
times. Hence, a point in the Cantor fan tree is specified by
a finite sequence of non-base points from the Cantor fan.)
OK, this looks promising; it's now time to check whether this space
is a solution to the original problem. Unfortunately, it isn't;
one can get a connected open subset by taking a small neighborhood
of the base point in the original fan and attaching all of the
subtrees based in this neighborhood. But this shouldn't be hard
to get around; instead of using the Cantor fan, use a pathological
space in which a base point is connected to a Cantor set without
needing intermediate points. (Oh, well, so much for the nice
path-connected metric space. But this didn't bother me much. After
all, Glenn Chappell must have had some reason for specifying T1
in particular; maybe he has a proof that there is no such T2 [Hausdorff]
space. [If not, this is probably the next question that should
be looked at.])
So, instead of the fan, let's use a space containing a base point
and a Cantor set (with its usual topology), so that any neighborhood
of the base point includes all points of the Cantor set. Oops, that's
not right; it won't be T1. OK, any neighborhood of the base point
includes a cofinite subset of the Cantor set. Now build up a tree
as before; this looks a little more like the usual discrete-math tree,
in that every point has a collection of children, forming a Cantor set.
Any neighborhood of a point must include cofinitely many of its
children, while any neighborhood of a child includes an open set of
its siblings; these statements became properties (2) and (1) in the
final writeup.
Does this space work? Again, no. The problem is that, if one deletes
a point (other than the root of the tree), then its descendants
fall apart nicely, but the rest of the space forms a connected open set.
Maybe we can get around this by changing the definition slightly --
instead of requiring a neighborhood of a point to include cofinitely
many of its children, require it to include cofinitely many of its
_descendants_. But this doesn't work either; now deleting a single
non-root point doesn't disconnect the space at all.
At this point I was wondering whether I was going in completely the
wrong direction; maybe no such space exists. To try to prove this,
I started examining how the components obtained by deleting one point
from such a space interact with the components obtained by deleting
a different point. (It turns out that they must all be disjoint
except for one component from each side, the component containing
the other chosen point.) This led to new possible example spaces to try.
For example, take two Cantor fans, and identify the base of each with
some end point of the other. Any nonempty connected open subset of this
space must include both base points. I tried more intricate
interconnections in the hope of forcing every point to be in the
connected open set, but it didn't work.
Going back to the first non-path-connected tree space, eventually I
decided that the problem was that a typical (non-root) point in the tree
has many neighbors (its children) in the form of a Cantor set, but it
also has one anomalous neighbor (its parent), which causes difficulties.
If I could somehow attach the parent of x to the children of x (the
grandchildren of the parent), then everything should work out. So
instead of making the children of x a Cantor set, let's make them a
Cantor set with one point missing, to be filled in by the parent of x.
A Cantor set with one point missing can be viewed as an infinite
sequence of Cantor sets converging to the missing point. At this
time, I decided to switch to the Baire space [the space of infinite
sequences of natural numbers], since it is already an infinite sequence
of copies of itself. A neighborhood of the parent of x should then
be forced to contain all but finitely many of the sequence of pieces
making up the children of x. It was easier to visualize this by
viewing the Baire space in its equivalent form as the space of
irrational real numbers; a neighborhood of the parent of x should
contain all sufficiently large irrationals among the children of x.
(The children of x thus form a broken line which somehow curves
around and aims for the parent of x.)
It was now time to look at the resulting space and see whether it
had the desired properties; everything seemed to work out this time.
It remained to clean things up, work out how to write down
the precise definition of the topology (I thought I would have to
give this in terms of a base or subbase, but it turned out to be
easier to just define "open" directly), and check again that
everything worked. [How does Santa get by with only checking his
list twice?] During this process, I realized that the only properties
of the irrationals I was using was that they are a dense codense
set of reals, so I decided to switch to the rationals instead (thus
making the space countable; why not?). This gave the final form
that I wrote up.
Anyway, that was (the relevant part of) my thought process. Maybe it
will make a good case study for something; certainly it gives many
pathological spaces which might be useful for other questions.
Randall Dougherty [Disclaimer out of order]
Internet: rld@function.mps.ohio-state.edu
UUCP: ...!{att,pyramid}!osu-cis!function.mps.ohio-state.edu!rld
"I have yet to see any problem, however complicated, that when looked at in the
right way didn't become still more complicated." Poul Anderson, "Call Me Joe"