A very simple question on a 1D manifold

[nrqed]

This is a very simple topology question. Consider two infinite lines crossing at one point. Now, I know that this is not a 1D manifold, and I know the usual argument (in the neighbourhood of the intersection, we don't have a a line, or that if we remove the intersection point, we end up with four disconnected segments, which shows that we don't have a 1D manifold).

But what I want is to see how to prove that it is not a 1D manifold using only the basic definition. By the way, for now I want to focus on topological manifolds, not differentiable manifolds.
If I understand correctly, one has a 1D manifold if around all points we can have open sets that can be mapped to the real line.
Now let me consider an open set going from -y to y along one of the lines (so it contains the intersection point, at x=0,y=0) and an open set going from -x to x.

Let me define phi_1 to map the open set (-x,x) to the real line, with the intersection point mapped to the origin.
Similarly let me define phi_2 to map the open set (-y,y) to the real line with the intersection point mapped to the origin.

Now let me define the map phi that maps the neighbourhood of the intersection point to the real line in the following way:

Phi is defined to be phi_1 for the open set (-x,x) and it is defined to be phi_2 for the open set (-y,y).

My question is: this seems to map the neighbourhood of the intersection point to R1.
Now I know that two lines intersection is not, topologically, a 1D manifold. But what is the flaw in the above ??

Thanks in advance.

 

[member 587159]

If I understand correctly, one has a 1D manifold if around all points we can have open sets that can be mapped to the real line.

Did not read your entire question, but more precisely we need that every point has a neighborhood that gets homeomorphically mapped onto an open subset of the real line. More concretely, the charts you defined must be homeomorphisms and the transition maps must be continuous.

You might want to check this.

 

[nrqed]

Did not read your entire question, but more precisely we need that every point has a neighborhood that gets homeomorphically mapped onto an open subset of the real line. More concretely, the charts you defined must be homeomorphisms and the transition maps must be continuous.

You might want to check this.

Thank you for your reply.

Yes, I was probably not very careful with my wording. But it seems to me that my map satisfies these requirements. But I am probably wrong and do not see the problem.

 

[PeroK]

Thank you for your reply.

Yes, I was probably not very careful with my wording. But it seems to me that my map satisfies these requirements. But I am probably wrong and do not see the problem.

A homeomorphism is 1-1, which your mapping clearly isn't.

 

[nrqed]

A homeomorphism is 1-1, which your mapping clearly isn't.

Ah! Good point, thanks!

 

[nrqed]

A homeomorphism is 1-1, which your mapping clearly isn't.

Me again. I still feel some confusion because it seems I could do the following:

If I am not wrong, a neighbourhood of a point is an open set that contains the point. Let's consider again the point which is at the intersection of the two lines. I don't want to use the topology induced by R2 here to define my open sets. So why can't I do the following:

I define two different neighbourhoods for the point of intersection:

For the first neighbourhood, I use for open set an open set along one of the line.
For the second neighbourhood, I use for open set an open set along the second line.

So here I have two neighbourhoods of the intersection point that can be mapped to an open set of R1.
In other words, for the neighbourhood of the intersection points I do not use the "cross".

So why can't I do this? I know there is a flaw in the reasoning, and maybe my understanding of what a neighbourhood is is actually incorrect. But I thought it was defined as simply an open set containing the point.

Thanks in advance!

 

[PeroK]

Me again. I still feel some confusion because it seems I could do the following:

If I am not wrong, a neighbourhood of a point is an open set that contains the point. Let's consider again the point which is at the intersection of the two lines. I don't want to use the topology induced by R2 here to define my open sets. So why can't I do the following:

I define two different neighbourhoods for the point of intersection:

For the first neighbourhood, I use for open set an open set along one of the line.
For the second neighbourhood, I use for open set an open set along the second line.

So here I have two neighbourhoods of the intersection point that can be mapped to an open set of R1.
In other words, for the neighbourhood of the intersection points I do not use the "cross".

So why can't I do this? I know there is a flaw in the reasoning, and maybe my understanding of what a neighbourhood is is actually incorrect. But I thought it was defined as simply an open set containing the point.

Thanks in advance!

If you don't use the usual topology on $\mathbb{R^1}$ and $\mathbb{R^2}$ then all bets are off. You're not free, however, to pick and choose which sets you want to be open. The family of open sets must obey the rules for a topology.

In this case if intervals along each line are considered open, then their intersection is open hence the point of intersection is an open set. That would mean that point was topologically disconnected from the four line segments.

Changing the topology also changes what functions are continuous, so in general problems in topology assume you are using a defined topology. Especially if you are trying to show two sets are homeomorphic. The question only makes sense if you are talking about a specific topology on each set.

The key point is that a topological space is the set along with its topology (defined family of open sets). Like a group is the set along with a defined binary operation.

 

[Infrared]

In fact, you can put a topology on your space for which those are open sets. Take two disjoint copies of $\mathbb{R}$ and identity $0$ from both copies. This space is locally Euclidean, and satisfies all manifold axioms except for being Hausdorff.

However, your exercise is presumably using the (subspace) topology inherited as a subset of $\mathbb{R}^2$. In this topology, every open set containing the intersection point will also contain points from both lines.

 

[nrqed]

Thank you Infrared and Perok for your replies.

I think I did not make my question clear.
First, I specifically do not want to use the topology inherited from R^2. Here is the topology I want to use: For the intersection point, I say that I can take open sets along one of the lines or open sets along the other line. These are just different open sets containing the intersection point.
This is Hausdorff, and I don't see any reason why this cannot be used as a topology (but I would be happy to be shown wrong!). Now, if this is allowed, then it seems that with this topology, this satisfies the definition of a 1D manifold.

I bet I am wrong, but I would like to know exactly where I failed the conditions for a 1D manifold.

Thanks!

 

[PeroK]

Thank you Infrared and Perok for your replies.

I think I did not make my question clear.
First, I specifically do not want to use the topology inherited from R^2. Here is the topology I want to use: For the intersection point, I say that I can take open sets along one of the lines or open sets along the other line. These are just different open sets containing the intersection point.
This is Hausdorff, and I don't see any reason why this cannot be used as a topology (but I would be happy to be shown wrong!). Now, if this is allowed, then it seems that with this topology, this satisfies the definition of a 1D manifold.

I bet I am wrong, but I would like to know exactly where I failed the conditions for a 1D manifold.

Thanks!

I think a good exercise for you would be to write out your proposition in mathematical language. E.g

Proposition: The set *define your set* endowed with the following topology *define your topology* is a 1D manifold.

And, then prove it if you can. I think we need a clear mathematical statement of your proposition.

 

[Infrared]

The intersection of two open sets should be open. In your topology, the intersection of two open intervals from different lines both containing your intersection point consists of only the intersection point. Are you sure you want this to be open? This implies that no neighborhood of the intersection point can be homeomorphic to an open subset of $\mathbb{R}^n$ because open subsets of $\mathbb{R}^n$ do not have singletons that are open.

Also, I got the 'line with a double point' backwards in my previous post. You should glue all corresponding points on the two lines together *except* 0.

 

[nrqed]

Thanks again.
Let me try to be more detailed.

The set I am using is defined by the two intersecting lines. Does that not define my set unambiguously?
The topology I am using is the union of metric topologies on both lines. And to answer a question, yes I consider the intersection point as an open set (is this not allowed?)

Unless I am wrong, a neighbourhood of a point is an open connected set containing the point.
If that is correct, then I can define a homeomorphism from any neighbourhood of a point set to R^1.
Therefore it seems to me that my set is a 1D manifold.

If this is wrong, what is the mistake (or mistakes)?

Thanks!

 

[Infrared]

You're allowed to make your intersection point $p$ an "open point" but it does stop your space from being a manifold. If $\varphi: U\to V$ is a homeomorphism from a neighborhood $U$ of $p$ to an open set $V\subset\mathbb{R}$, then $\varphi(p)$ has to be an open point in in $V$, which is impossible.

The map I think you're considering is not a homeomorphism because its inverse is not continuous. Consider $\psi=\varphi^{-1}:V\to U$. For $\psi$ to be continuous, $\psi^{-1}(\{p\})$ must be an open set in $V\subset\mathbb{R}$, but it is the singleton $\{\varphi(p)\}$ and so is not open (this is the same reasoning as in the previous paragraph).

 

[PeroK]

Thanks again.
Let me try to be more detailed.

The set I am using is defined by the two intersecting lines. Does that not define my set unambiguously?
The topology I am using is the union of metric topologies on both lines. And to answer a question, yes I consider the intersection point as an open set (is this not allowed?)

Unless I am wrong, a neighbourhood of a point is an open connected set containing the point.
If that is correct, then I can define a homeomorphism from any neighbourhood of a point set to R^1.
Therefore it seems to me that my set is a 1D manifold.

If this is wrong, what is the mistake (or mistakes)?

Thanks!

You can't just pick and choose which sets you want to be open. The intersection and union of open sets must also be open sets. See the definition of a topology. You cannot avoid the point of intersection being an open set. This is allowed but there are no single-point open sets in a 1D manifold.

You need to compare your posts (which are woolly hand-waving) with post 13 by @Infrared, which is pure mathematics.

 

[nrqed]

You're allowed to make your intersection point $p$ an "open point" but it does stop your space from being a manifold. If $\varphi: U\to V$ is a homeomorphism from a neighborhood $U$ of $p$ to an open set $V\subset\mathbb{R}$, then $\varphi(p)$ has to be an open point in in $V$, which is impossible.

Thank you, this is converging to what I am looking for, so let me focus on this part and ask a clarification. What is the precise definition of a manifold? I thought that it was that for every point, there must be at least one open neighbourhood that is homeomorphic to an open set of R^n. Is that the definition? Or is the definition that for ALL open neighbourhoods there must be a homeomorphism to an open set of R^n?

If the condition is only that there must be at least one neighbourhood, then in my topology if I consider the intersection point, I can take a neighbourhood that is not that single point.
But if we must consider all possible neighbourhoods, than this might be the key point I was missing.

Thank you for your patience!

 

[Infrared]

You have the right definition of a manifold (except that you should also require your space to be Hausdorff and second countable). There is no neighborhood of your intersection point that is homeomorphic to an open subset of $\mathbb{R}$, as I showed earlier.

 

[nrqed]

You can't just pick and choose which sets you want to be open. The intersection and union of open sets must also be open sets. See the definition of a topology. You cannot avoid the point of intersection being an open set. This is allowed but there are no single-point open sets in a 1D manifold.

You need to compare your posts (which are woolly hand-waving) with post 13 by @Infrared, which is pure mathematics.

I understand that I can't pick any open sets I want. I did define the open sets I was using and I did say that the point of intersection is an open set. I did not try to avoid that.

The definition I am using is that
"Each point p has an open neighbourhood U for which we can find a homeomorphism phi: U -> V to an open set V in R^n"

If this is not the accurate, let me know what is missing. If it is accurate, then I would like to understand how my suggestion fails without using extraneous information than this definition.
(for my intersection point, I can always use for open neighbourhood a segment along one of the lines. This is an open neighbourhood in my topology).

I am trying to be mathematically precise, which is why I want to apply the definition and see exactly where my case fails.

 

[Infrared]

for my intersection point, I can always use for open neighbourhood a segment along one of the lines. This is an open neighbourhood in my topology).

While this neighborhood "looks" like an interval in $\mathbb{R}$, it is not homeomorphic to one because of the weird topology you put on your space. See my post 11 or 13.

 

[PeroK]

I understand that I can't pick any open sets I want. I did define the open sets I was using and I did say that the point of intersection is an open set. I di not try to avoid that.

The definition I am using is that
"Each point p has an open neighbourhood U for which we can find a homeomorphism phi: U -> V to an open set V in R^n"

If this is not the accurate, let me know what is missing. If it is accurate, then I would like to understand how my suggestion fails without using extraneous information than this definition.
(for my intersection point, I can always use for open neighbourhood a segment along one of the lines. This is an open neighbourhood in my topology).

The problem, as described above, is finding the homeomorphism. A homeomorphism maps open sets to open sets. All "morphisms" are structure preserving. In this case, preserving the open set structure. If a point is an open set, then it's topologically disjoint from the rest of the set. So, your set (with your defined topology) can only be homeomorphic to another set if that set has a single disjoint point.

Effectively your set is four infinite intervals and a disjoint central point. By defining your topology you have disconnected the intersection point. But, when it suits your argument, you imagine that it is not disjoint and appeal to a physical argument that there is no gap around the intersection point.

The mistake is not to update the mental picture of your set on account of your new topology.

Anyway ... you can't find a homeomorphism from any neighbourhood of the intersection point because:

Any neighbourhood of the intersection point contains the point itself. So any homeomorphism from that neighbourhood must map the intersection point to an open set. But, there are no single-point open sets in $\mathbb{R}$.

 

[PeroK]

While this neighborhood "looks" like an interval in $\mathbb{R}$, it is not homeomorphic to one because of the weird topology you put on your space. See my post 11 or 13.

It's homeomorphic to, for example, the subset of $\mathbb{R}^2$ consisting of the x and y axes for $|x| >1$ and $|y| > 1$, plus the origin.

 

[Infrared]

The neighborhood in question lies on only one of the lines, so you should only have two intervals, but yes, so it's homeomorphic to something like $(-2,-1)\cup\{0\}\cup (1,2)$.

 

[nrqed]

You're allowed to make your intersection point $p$ an "open point" but it does stop your space from being a manifold. If $\varphi: U\to V$ is a homeomorphism from a neighborhood $U$ of $p$ to an open set $V\subset\mathbb{R}$, then $\varphi(p)$ has to be an open point in in $V$, which is impossible.

The map I think you're considering is not a homeomorphism because its inverse is not continuous. Consider $\psi=\varphi^{-1}:V\to U$. For $\psi$ to be continuous, $\psi^{-1}(\{p\})$ must be an open set in $V\subset\mathbb{R}$, but it is the singleton $\{\varphi(p)\}$ and so is not open (this is the same reasoning as in the previous paragraph).

AH! I see now, THANK YOU! I am sorry that I did not get it right away when I read your reply.
This indeed answered my question, thanks so much for your patience and for your help! And thank you to Perok too. This has been bugging me for so long and I never found an answer to my question (everything I was reading was using the induced topology from R^2.

Thank you!