Can a Topological Space be Both a 1-Manifold and an n-Manifold?

[PsychonautQQ]

This is not homework, self study, so I'm going to post here, where people know things :P.

a) Show that a topological space can't be both a 1-manifold and an n-manifold for any n>1.

If a topological space were both a 1-manifold and an n-manifold, then every point would have a neighborhood homeomorphic to a neighborhood of R and every point would also have a neighborhood homeomorphic to a 'hood of R^n. So this topological space would be locally euclidean of dimension 1 and n.

Really, I don't see why this couldn't happen; I mean, in a discrete topology, every point would have a hood homeomorphic to R^0 and perhaps could also have a neighborhood homeomorphic to R^n for some other n. (I'm sure there is something wrong with what I said here)

But yeah, I am confused why this is not possible.
b) Show that the union of the x-axis and y-axis in R^2 is not a manifold in the subspace topology

Well, this space is second countable and Hausdorff, so it must not be local euclidean for any n. It seems to me that it should be locally euclidean of dimension 1; every point other than the origin obviously has a 'hood homeomorphic to a hood in R^1.

So something must go wrong at the origin. However, I don't see why we can't have a neighborhood or the origin that's completely contained on the x or y-axis, and therefore would be homeomorphic to a hood in R^1. Anyone have insights on this?c) Show that any n-manifold is a disjoint union of countably many connected n-manifolds

So I'm imagining the set of disjoint countably many connected n-manifolds to be the components of some n-manifold. Am I on the right track here? If so, then I guess the trick would be showing that the number of components is countable; We know that each component has a countable basis, because manifolds are second countable.

Maybe the fact that the original n-manifold is second countable and therefore has a countable basis means that it has to have a countable number of components. Am I on the right track here?

d) define a topology on Z by declaring a set A to be open if and only if n in A implies -n is in A. Show that Z with this topology is second countable and limit point compact but not compact.

Well, Z is second countable and this topology is defined by an equivalence on Z. Every pair (a,-a) in this topology is a neighborhood and so taking one of these pairs for every integer and one for 0 would be a countable basis.

I'm trying to think about why this would be limit point compact but not compact, but seem to be going in circles in my brain without any real understanding. Can somebody give me some insight into this? PF IS THE BEST!

 

[andrewkirk]

For (a) would you be convinced if the manifold being both a 1- and a n-manifold implied that we could produce open sets $U\subseteq \mathbb R$ and $V\subseteq \mathbb R^n$ for $n>1$ such that $U,V$ are homeomorphic to one another?

I think it's reasonably straightforward to prove that that cannot happen. Think of a simple operation that disconnects an open set in $\mathbb R$ that would not disconnect an open set in $\mathbb R^n$.

If you can convince yourself of that, what remains to be shown is how to construct the subsets $U,V$ above, for a manifold that is both 1-D and n-D..

 

[andrewkirk]

(b) ... However, I don't see why we can't have a neighborhood or the origin that's completely contained on the x or y-axis, and therefore would be homeomorphic to a hood in R^1. Anyone have insights on this?

In the subspace topology, a neighbourhood of the origin must contain the intersection of $B_\epsilon(0)$, an open ball of radius $\epsilon$ centred at the origin, and the axis subspace $X$. What would $\epsilon$ have to be in order for that intersection to contain points from only one of the axes?

 

[PsychonautQQ]

In the subspace topology, a neighbourhood of the origin must contain the intersection of $B_\epsilon(0)$, an open ball of radius $\epsilon$ centred at the origin, and the axis subspace $X$. What would $\epsilon$ have to be in order for that intersection to contain points from only one of the axes?

I see what you are saying. My whole thing was that it could have a neighborhood U homeomorphic to a neighborhood in R^1 and neighborhood V homeomorphic to a neighborhood in R^n, where U and V wouldn't have to be homeomorphic to each other, as the defintion of locally euclidean says it has to have "a" neighborhood with the given property, so perhaps we could find two different neighborhoods each homemorphic to a different thing. I guess I'm still a little confused why that wouldn't work, although my intuition is saying it obviously wouldn't.

Hmm.. maybe the following arguement:
Let U be a 'hood homeomorphic to a hood of R^1 and V be a hood homemorphic to a hood of R^n.
Then U intersect V is a hood that is contained in U, so it will be homeomorphic to a hood in R^1, yet U and V is also the restriction of a map that is homeomorphic to a hood in R^n so it must be homeomorphic to a hood in R^n. This is a contradiction.
Does this work? Is there a simpler way t understand this?

 

[PsychonautQQ]

In the subspace topology, a neighbourhood of the origin must contain the intersection of $B_\epsilon(0)$, an open ball of radius $\epsilon$ centred at the origin, and the axis subspace $X$. What would $\epsilon$ have to be in order for that intersection to contain points from only one of the axes?

Oh duh, thanks, that helps a lot for (b) :D

 

[andrewkirk]

(c)... Maybe the fact that the original n-manifold is second countable and therefore has a countable basis means that it has to have a countable number of components. Am I on the right track here?

Yes. You'll need to use the Hausdorff property as well, to show that every base element is connected (why?).

I'm trying to think about why this would be limit point compact but not compact, but seem to be going in circles in my brain without any real understanding.

Of the two things to be proven for this part, I think proving the topology is not compact is easiest. Just find an infinite open cover of Z that doesn't have a finite subcover. The infinite open cover that seemed the most natural possibility to me, given the definition of the topology, turned out not to have a finite subcover.

To prove that the topology is limit point compact, it may help to note that the topology is not Hausdorff (why?).

 

[PsychonautQQ]

Yes. You'll need to use the Hausdorff property as well, to show that every base element is connected (why?).

Well, showing that every basis element is connected will show that every component is open, which would need to be true for the collection of disjoint components to be manifolds. Am on on the right track here?

 

[PsychonautQQ]

Cool, showing it is not compact sounds easy, thanks.
Noting the topology is not hausdorff.. I guess for starters this would mean that limit point compactness does not imply compactness perhaps? I'll have to review my text on the matter. Maybe If i go through the proof on why this implication fails in the case of a non-hausdorff space I can figure out why it is not limit point compact.

 

[PsychonautQQ]

To prove that the topology is limit point compact, it may help to note that the topology is not Hausdorff (why?).

Nevermind, what I said above is not correct. I will try to figure out how to use the non hausdorff-ness. It is non hausdorff because the elements n,-n in Z have no disjoint hood separating them.

edit: I think i know why it is limit point compact...

Given any infinite subset A of Z, every element n of A will be a limit point because every open neighborhood of n will also contain -n.

 

[andrewkirk]

Given any infinite subset A of Z, every element n of A will be a limit point because every open neighborhood of n will also contain -n.

Not quite. Consider the infinite subset that is all the positive integers. Does it contain any limit points? Does it have any limit points that are not in the set?

 

[PsychonautQQ]

every neighborhood of any negative number will intersect the set of all positive integers.

 

[andrewkirk]

Yes, so although the set of positive integers does not contain any limit points, it has an infinite number of limit points, being all the negative integers. This argument shows us that the proof strategy in post 9 needs to be generalised a bit, as sometimes some of the limit point(s) of a set are in the set and sometimes they are all outside it.

In fact we can prove a stronger result than limit point compactness: that every set with two or more elements has at least one limit point. I find that easier to prove than limit point compactness, even though limit point compactness follows from it.

 

[PsychonautQQ]

Sorry, i am being redundant here, but it's for my own clarity.

For a space to be limit point compact, every infinite set must contain a limit point.

I said: Let A be an infinite subset of Z. Then if n is going to be a limit point of A, every neighborhood of n must intersect A in some place other than itself. So what I said was not correct because maybe n is in A but not -n, we would have a neighborhood of n who's intersection with A is only n. However, if -n is not in A but n is in A then -n will be a limit point of A because every neighborhood of -n will have a nonempty intersection with A because n will be in it.

However, (and I'm sure you know this and just talked about the positive integers to build my understanding of the situation), EVERY infinite subset of Z must contain a limit point. So given an infinite subset of Z, call it A, then 3 cases are possible:
1) n is in A but -n isn't then -n is a limit point of A,
2) -n is in A but n isn't and so n is a limit point of A
3) both -n and n are in A and are both limit points of A

P.S. can you give me feedback as to post #7? You said I must use the Hausdorff property for every basis element in showing that every n-manifold is a product of a countable number of disjoint n-manifolds.

 

[andrewkirk]

EVERY infinite subset of Z must contain a limit point

Not contain, but have. The limit point need not be in the subset. From wiki, the definition of limit point compact is:

a topological space X is said to be limit point compact[1] or weakly countably compact if every infinite subset [A] of X has a limit point in X.

The limit point is in X, but need not be in the subset A. Note that in your case (1) the limit point -n is not in A, but limit point compactness is still satisfied because there is still a limit point.

can you give me feedback as to post #7? You said I must use the Hausdorff property for every basis element in showing that every n-manifold is a product of a countable number of disjoint n-manifolds.

I'll revert later on post 7. Gotta dash right now.

 

[andrewkirk]

Re post 7, for (c), we need to prove that the n-manifold, call it M, is a disjoint union of countably many connected n-manifolds.

To do the first part, we consider a separation of M into a collection C of connected components. If E is a countable topological base for M, can you construct a surjection from E into C, using the fact that connected components are by definition open, and any open set can be expressed as a union of base elements? From that, it is a few short steps to proving that C is countable.

Then we need to prove that an arbitrary component U of C is a n-manifold, ie Hausdorff, second-countable and locally Euclidean. The first two are easy. the second one will involve considering homeomorphisms of an open nbd of $x\in U$ point to an open set in $\mathbb R^n$.

On reflection, I think my comment in post 6 about needing to use the Hausdorff property to show that every base element is connected is wrong, and also unnecessary. It is not necessary for all base elements to be connected. We deal with connectedness simply by starting with a partition of M into connected components. I think we only need the Hausdorff property to prove that each connected component is Hausdorff.

 

[WWGD]

You can also use the fact that a diffeomeomorphism $f$ between manifolds M,N ( Here $\mathbb R^1) , \mathbb R^n$ gives rise to an isomorphism between respective tangent spaces at $p, f(p)$ ; this last in the " category" of vector spaces. So the diffeomorphism would require to have a 1-dimensional vector space to be ( vector-space) isomorphic to a 9d vector space, which cannot happen for many reasons. Not clean/deep, but does the job.

 

[PsychonautQQ]

You can also use the fact that a diffeomeomorphism $f$ between manifolds M,N ( Here $\mathbb R^1) , \mathbb R^n$ gives rise to an isomorphism between respective tangent spaces at $p, f(p)$ ; this last in the " category" of vector spaces. So the diffeomorphism would require to have a 1-dimensional vector space to be ( vector-space) isomorphic to a 9d vector space, which cannot happen for many reasons. Not clean/deep, but does the job.

hahaaaa yeah okay bro i'll get back to you on that :P

 

[WWGD]

hahaaaa yeah okay bro i'll get back to you on that :P

Sorry if I over did it. I don't know if a simpler answer, i.e., one that does not use heavy machinery like invariance of domain.