Classic Textbook Topology Question

[tylerc1991]

Homework Statement

Prove that the topologist's comb is pathwise connected but not locally connected.

Homework Equations

For A = {1/n : n = 1,2,3...}, the topologist's comb (C) is defined as:

C = (I X {0}) U ({0} X I) U (A X I)

The Attempt at a Solution

Consider the point p = (1,0) in C and a neighborhood U of p. Since C is a subspace of R X R, the topology on C is such that any neighborhood around p contains points on the 'branch' ({0} X I) as well as points on the branch (I X {0}). (i.e. U intersect ({0} X I) intersect (I X {0}) =/= empty). => there DNE a connected neighborhood V of p such that V is a subset of U. (V is not connected because it can be expressed as the union of two separated sets) => C is not locally connected. I know that this is not as formal as one would like, but I believe the general idea is there and when I do write the proof formally it will be more technical.

Next consider two points p and p' that are on different 'branches' of C. The maximum distance that can separate p and p' is 3. (to see this consider p at the point (1,1) and p' at the point (0,1), then there is a path from (1,1) to (1,0), from (1,0) to (0,0), then finally a path from (0,0) to (0,1)). Then I could make a path product f_1 * f_2 = f_3 which goes from (1,1) to (0,0) and finally a path product f_3 * f_4 that goes from (1,1) to (0,1). Again, this is just a rough sketch of what I am trying to do and am looking for a little feedback. Thank you anyone for your help!

 

[Dick]

(I X {0}) U ({0} X I) U ({1} X I) is path connected AND locally connected. And that's the only subset of U you've mentioned so far. You aren't paying much attention to the (A X I) part. That's where the interesting action is. I think you could flesh out the second argument once you start paying attention to points besides (0,0), (1,0), (0,1) and (1,1). But I really don't get what you are up to with the first one.

 

[Deveno]

if you think of the comb as {{0}xI} being the "spine" and {{1/n}xI} being a "tooth", you can connect any two points on the comb by running down a tooth, across the spine, and back up another tooth.

note that ANY sufficiently small neighborhood of any point {0}x(0,1] always has a gazillion disconnected components, since the teeth become very dense near {0}x I (and if we're "up on the left-est tooth", we can't use the spine to connect all these in our neighborhood), so if the neighborhood U doesn't intersect the spine, there's just a bunch of "disconnnected teeth", we can't find a connected neighborhood V in the relative topology contained in U for such a point.

none of the other teeth have this problem, since we can always make a neighborhood small enough so it doesn't touch another tooth, and this neighborhood will be connected.