Graduate Is the Long Line Locally Euclidean, Hausdorff, and Path Connected?

[PsychonautQQ]

Let S be the minimal uncountable set. That is, for every m in S, there are countable many n s.t. n<m.

Let L = { S x [0,1) } \ $(a_0,0)$ where $a_0$ is the smallest element of S (S is well ordered so this element exists). Order L be dictionary order, and then give L the order topology.

Show that L is locally euclidean, Hausdorff and path connected.

_____________

I am first trying to understand why L is Hausdorff. If $a_0$ is the smallest element of S, then S/{a_0} will have smallest element a_1.

I am trying to find disjoint neighborhoods of points (a_1,0) and (a_2,0) which I believe to be the two smallest elements of L. Does each element of S have a discrete neighborhood in the order topology?

After I understand why it is Hausdorff then I will try to understand why it is locally euclidean and path connected.

Thanks PF!

 

[andrewkirk]

I am first trying to understand why L is Hausdorff. If a0a_0 is the smallest element of S, then S/{a_0} will have smallest element a_1.

L is Hausdorff if any two distinct points in L have disjoint neighbourhoods. Since we are using the order topology, we can assume WLOG for distinct x,y that x<y. We can find disjoint nbds of the two if we can find u such that x<u<y (what will the two disjoint nbds be?).

How can we find such a u? We may need to separately consider the cases where the x and y have the same first component, and where they don't.

Looking ahead: the locally Euclidean piece looks easy, but the path-connected piece looks hard.

By the way, terminology used in relation to the Long Line is non-standard and can be confusing. The set L above is sometimes called the Long Line but more often called the Long Ray. The Long Line - call it LL - is obtained by attaching a reversed copy of L to the unreversed L.

 

[PsychonautQQ]

L is Hausdorff if any two distinct points in L have disjoint neighbourhoods. Since we are using the order topology, we can assume WLOG for distinct x,y that x<y. We can find disjoint nbds of the two if we can find u such that x<u<y (what will the two disjoint nbds be?).

How can we find such a u? We may need to separately consider the cases where the x and y have the same first component, and where they don't.

Looking ahead: the locally Euclidean piece looks easy, but the path-connected piece looks hard.

By the way, terminology used in relation to the Long Line is non-standard and can be confusing. The set L above is sometimes called the Long Line but more often called the Long Ray. The Long Line - call it LL - is obtained by attaching a reversed copy of L to the unreversed L.

Hmm, okay, thanks for the heads up. So given an x and y we need to find a u between them and then we can find disjoint neighborhoods, that makes sense to me. The problem I was having was that if the set S has a smallest element, then it won't it have a second smallest element because it is well ordered? and a third smallest element then? So for (a_1,0) and (a_2,0) where a_1 and a_2 are the second and third smallest elements I don't see an element between them. My thinking must be flawed somewhere.

 

[andrewkirk]

@PsychonautQQ Sorry for some reason I didn't see your reply. I only just found it now.

What you say about a second-smallest element of S is correct. That's why we need to times S with [0,1) and apply the dictionary order. The resultant set, unlike S, is not well-ordered. To get an element between (a1,0) and (a2,0) use the second component of those pairs.