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