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 Path: mp.cs.niu.edu!linac!pacific.mps.ohio-state.edu!zaphod.mps.ohio-state.edu!function.mps.ohio-state.edu!rld 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> Sender: usenet@zaphod.mps.ohio-state.edu Organization: Department of Mathematics, The Ohio State University Lines: 155 Nntp-Posting-Host: function.mps.ohio-state.edu 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"