# Topology, line with two origins

• BrainHurts
In summary, the conversation discusses a topology problem involving a quotient space M which is the result of identifying points in a set X by the equivalence relation (x,-1)~(x,1) for all x≠0. M is locally Euclidean and second-countable, but not Hausdorff due to the points (0,1) and (0,-1) being glued together. The conversation also delves into the basics of topology, discussing the quotient map and the elements of X being subsets of ℝ2. The conversation ends with a challenge to determine what ∏((0,1)) is and to discuss the open sets in M.
BrainHurts

## Homework Statement

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.

## 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?

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:
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.

BrainHurts said:
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 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:

## 1. What is topology?

Topology is a branch of mathematics that studies the properties of geometric objects that remain unchanged under continuous deformations, such as stretching, twisting, and bending.

## 2. What is a line with two origins?

A line with two origins is a topological space in which two distinct points are considered as origins or starting points, and all other points on the line are considered as belonging to either one of these origins.

## 3. Can a line with two origins be visualized in three-dimensional space?

Yes, a line with two origins can be visualized in three-dimensional space by considering it as a line that has been twisted or bent in a specific way to create two distinct starting points.

## 4. What are the key properties of a line with two origins?

The key properties of a line with two origins include the fact that it is a connected space, meaning that there is a continuous path between any two points on the line. It is also a Hausdorff space, which means that any two distinct points on the line have disjoint neighborhoods. Additionally, it is a non-Hausdorff space, as the two origins cannot be separated by disjoint neighborhoods.

## 5. What are some real-life applications of a line with two origins?

A line with two origins has various applications in physics, such as in the study of topological defects in crystals and the behavior of particles in two-dimensional materials. It can also be used in computer science, specifically in the development of data structures and algorithms for efficient storage and retrieval of data.

Replies
12
Views
2K
Replies
4
Views
296
Replies
10
Views
1K
Replies
1
Views
1K
Replies
8
Views
2K
Replies
4
Views
2K
Replies
3
Views
1K
Replies
2
Views
2K
Replies
3
Views
1K
Replies
1
Views
2K