Topology, line with two origins

1. BrainHurts

98
1. The problem statement, all variables and given/known data

Let X be the set of all points (x,y)$\in$ℝ2 such that y=±1, and let M be the quotient of X by the equivalence relation generated by (x,-1)~(x,1) for all x≠0. Show that M is locally Euclidean and second-countable, but not Hausdorff.

2. Relevant equations

3. The attempt at a solution

So I haven't taken any topology and my understanding of it is to the basic. I've been reading a lot about quotient spaces and I'm really not understanding the phrase "let M be the quotient of X by the equivalence relation generated by (x,-1)~(x,1)"

Question: What is M? M={[x]$\in$X}={{v$\in$X:v~x}:x$\in$X}

Question: M is locally Euclidean because there is a one-to-one correspondence onto ℝ?

Question: M is second countable because the bases are open subsets of ℝ?

2. Dick

25,651
Even if you have studied topology this take some thought and possibly makes brain hurt. But the title of your thread says it all. It's a line with two distinct origins. Your original space was two distinct lines. All of the other points except for (0,1) and (0,-1) have been glued together. Away from those points it is just a single line. So it's obviously Euclidean except for those two points. Start with the Hausdorff question. You really have to think hard about the open sets are in the quotient space. Is there any open set containing (0,1) that is disjoint from (0,-1)? Then is there an open set containing (0,1) that does not contain (0,-1) that you might consider Euclidean?

Last edited: Feb 3, 2013
3. BrainHurts

98
I'm not too good with the formality here. So I see the gluing that you are talking about in my head. I think I'm having a hard time with the quotient space.

So M is the quotient space correct? I believe we are talking about the quotient map

∏: X→M

now I guess my question is this, elements in X are in ℝ2

M is the quotient of X by the equivalence relation (x,1)~(x,-1) for all x≠0

the problem is, when i'm thinking of this relation, I believe I see this and I don't know if this is correct,

let x be in X

the image of x under ∏, ∏(x) is a subset of ℝ, now there's a problem at the origin and this is why M is not Hausdorff because the inverse image of M will have problems. I know the last part is not very formal.

But from many cursory readings in topology, I've read that a line with two endpoints is homeomorphic to S1.

I'm thinking open sets of M, the quotient space, are the usual open sets in ℝ, and those are second countable.

I've been thinking about this problem all night and I'm going nowhere!

4. Dick

25,651
I think you should just start from the basics. ∏: X→M, X isn't all of R^2. It's just the union of the two lines y=1 and y=(-1). The points in the space M are going to be subsets of X, where the elements of the subsets are the points which are equivalent to each other under your identification. I'll give you some examples. ∏((1,1))={(1,1),(1,-1)}. ∏((1,-1))={(1,1),(1,-1)}. ∏((2,1))={(2,1),(2,-1)}, etc etc. This is because you are told to identify (x,1)~(x,-1) for all x≠0. Now, here's a trick question. What is ∏((0,1))?? If all of this seems too easily you are welcome to jump ahead and start telling me about the open sets in M. You'll be able to start answering your questions once you figure out what those are.

Last edited: Feb 3, 2013
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