- #1
PsychonautQQ
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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!
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!