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A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.What's the definition of boundary ?

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- #26

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A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.What's the definition of boundary ?

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fresh_42

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The crucial question is what are in- and outside? Considered as a topological manifold in itself, you will have to use charts to define that. Considered as a subset of an Euclidean space things are different.A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.

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Are there ever any points outside a manifold? Unless it's explicitly a sub-manifold.A boundary point P is a point such that any neighberhood centered around P contains a point inside the set and a point outside the set.

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I don't think so. Are there any points outside the 2-D Euclidean plane (the coordinate system)?Are there ever any points outside a manifold? Unless it's explicitly a sub-manifold.

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By your definition, therefore, no manifold has a boundary.I don't think so. Are there any points outside the 2-D Euclidean plane (the coordinate system)?

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As far as I know topological spaces are preserved via homeomorphism. I doubt that a metric is required for a manifold.By your definition, therefore, no manifold has a boundary.

- #32

fresh_42

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The termAs far as I know topological spaces are preserved via homeomorphism. I doubt that a metric is required for a manifold.

You gave the definition of the latter whereas I assume the former is meant.

- #33

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The topological definition of the boundary of a subset is its closure minus its interior. The interior of a subset is the largest open set contained within it. An entire topological space is open by definition and is also closed by definition so its closure minus its interior is the empty set. So one might say that the topological boundary of a topological space in itself is the empty set which is not nothing so to be super precise an entire topological space (rather than a subset)always has empty boundary.

In particular, the topological boundary of a manifold is empty.

Manifolds are defined by the condition that around each point there is an open subset that is homeomorphic to an open subset of Euclidean space of a fixed dimension. A manifold with boundary is defined to be locally homeomorphic to an open subset of the closed upper half space of Euclidean space. Its boundary is those points that are mapped to the boundary of the upper half space by some local homeomorphism.

If a local homeomorphism maps a point to the interior of this upper half space(interior when it is considered as a subset of Euclidean space) then no other can map it to the boundary of the half space. Conversely if a local homeomorphism maps a point to the boundary of the half space, then no other local homeomorphism can map it to the interior. This is a theorem that requires proof and it guarantees that the boundary of a manifold is well-defined. This is a theorem worth understanding. As an exercise try the proof for a smooth manifold.

If one considers the open subset of a manifold with boundary of those points that are mapped to the interior of the upper half space then the boundary of this subset is the boundary of the manifold. Perhaps this is what can cause some confusion.

In particular, the topological boundary of a manifold is empty.

Manifolds are defined by the condition that around each point there is an open subset that is homeomorphic to an open subset of Euclidean space of a fixed dimension. A manifold with boundary is defined to be locally homeomorphic to an open subset of the closed upper half space of Euclidean space. Its boundary is those points that are mapped to the boundary of the upper half space by some local homeomorphism.

If a local homeomorphism maps a point to the interior of this upper half space(interior when it is considered as a subset of Euclidean space) then no other can map it to the boundary of the half space. Conversely if a local homeomorphism maps a point to the boundary of the half space, then no other local homeomorphism can map it to the interior. This is a theorem that requires proof and it guarantees that the boundary of a manifold is well-defined. This is a theorem worth understanding. As an exercise try the proof for a smooth manifold.

If one considers the open subset of a manifold with boundary of those points that are mapped to the interior of the upper half space then the boundary of this subset is the boundary of the manifold. Perhaps this is what can cause some confusion.

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- #34

mathwonk

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" if a local homeomorphism maps a point to the boundary of the half space, then no other local homeomorphism can map it to the interior. This is a theorem that requires proof"

a point of the boundary has a neighborhood basis of contractible punctured sets, but no point of the interior does. I think that proves it. ( I may have said this before, since I thought of it so quickly.) does this scan?

added later: oh yes, one has to prove that a punctured ball is not contractible, and my proof of that uses some tools, like homology or homotopy, i.e. degree theory, e.g. via simplicial triangulation.

a point of the boundary has a neighborhood basis of contractible punctured sets, but no point of the interior does. I think that proves it. ( I may have said this before, since I thought of it so quickly.) does this scan?

added later: oh yes, one has to prove that a punctured ball is not contractible, and my proof of that uses some tools, like homology or homotopy, i.e. degree theory, e.g. via simplicial triangulation.

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- #35

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@mathwonk

Some questions about your proof since you gave a sketch.

The punctured open ball has the homology of a sphere while a half ball with a boundary point removed is contractible. So an open ball and a half open ball are not homeomorphic. OK.

The argument about basis I think then wants to use this to say that if point ##p## is mapped to the boundary of the upper half space by homeomorphism ##φ:U→R^{n}_{+}## of an open set ##U## then ##U## can not contain any open subset that both contains ##p## and is homeomorphic to an open subset of ##R^{n}##. Such a subset could not be punctured and still be contractible. Correct?

But what about the other way around? Suppose ##p## is mapped to the interior of the upper half space by a local homeomorphism of an open set ##U##. How does your proof show that ##U## can not contain an open subset containing ##p## that can be mapped homeomorpically to an open subset of the boundary of ##R^{n}_{+}## and take ##p## to the boundary?

Some questions about your proof since you gave a sketch.

The punctured open ball has the homology of a sphere while a half ball with a boundary point removed is contractible. So an open ball and a half open ball are not homeomorphic. OK.

The argument about basis I think then wants to use this to say that if point ##p## is mapped to the boundary of the upper half space by homeomorphism ##φ:U→R^{n}_{+}## of an open set ##U## then ##U## can not contain any open subset that both contains ##p## and is homeomorphic to an open subset of ##R^{n}##. Such a subset could not be punctured and still be contractible. Correct?

But what about the other way around? Suppose ##p## is mapped to the interior of the upper half space by a local homeomorphism of an open set ##U##. How does your proof show that ##U## can not contain an open subset containing ##p## that can be mapped homeomorpically to an open subset of the boundary of ##R^{n}_{+}## and take ##p## to the boundary?

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- #36

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https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took ##S=[0,1]^2\cap\mathbb{Q}^2##

Is the following correct :

##\partial S=[0,1]^2##

##\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])##

##\partial\partial\partial S=\emptyset## ?

- #37

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How do you justify each these three steps?

https://en.m.wikipedia.org/wiki/Boundary_(topology)

I took ##S=[0,1]^2\cap\mathbb{Q}^2##

Is the following correct :

##\partial S=[0,1]^2##

##\partial\partial S=([0,1]\times\{0,1\})\cup (\{0,1\}\times[0,1])##

##\partial\partial\partial S=\emptyset## ?

- #38

mathwonk

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"But what about the other way around? Suppose p is mapped to the interior of the upper half space by a local homeomorphism of an open set U. How does your proof show that U can not contain an open subset containing p that can be mapped homeomorpically to an open subset of the boundary of R+n and take p to the boundary?"

By shrinking around the boundary point, and composing, it seems this would give a homeomorphism from a half disc H around the boundary point y, to a bounded connected open set W in the interior, taking y to some point x in W. Then the complement of y in the half disc, i.e. the punctured half disc H - {y}, would be homeomorphic to the punctured set W-{x}. But one of them is contractible and the other, namely W - {x}, is not. I.e. a small sphere centered at x wraps once around x, and hence cannot be deformed off it, as a contraction would do.

I.e. as stated originally, y has arbitrarily small contractible punctured neighborhoods in H, while x has no contractible punctured nbhds in W. So H cannot be homeomorphic to W, with y going to x.

Perhaps I should say (H,y) and (W,x) have different local homology groups, i.e. the local homology of H at y is zero, unlike that of W at x.

believe that? sorry, I am a bit rusty on topology. But I am using excision here to compute the local homology of W at x, by replacing W by a ball centered at x. So even though originally I am trying to compare a punctured half ball to a punctured open set in R^n, excision lets me just compare a punctured half ball to a punctured ball.

By shrinking around the boundary point, and composing, it seems this would give a homeomorphism from a half disc H around the boundary point y, to a bounded connected open set W in the interior, taking y to some point x in W. Then the complement of y in the half disc, i.e. the punctured half disc H - {y}, would be homeomorphic to the punctured set W-{x}. But one of them is contractible and the other, namely W - {x}, is not. I.e. a small sphere centered at x wraps once around x, and hence cannot be deformed off it, as a contraction would do.

I.e. as stated originally, y has arbitrarily small contractible punctured neighborhoods in H, while x has no contractible punctured nbhds in W. So H cannot be homeomorphic to W, with y going to x.

Perhaps I should say (H,y) and (W,x) have different local homology groups, i.e. the local homology of H at y is zero, unlike that of W at x.

believe that? sorry, I am a bit rusty on topology. But I am using excision here to compute the local homology of W at x, by replacing W by a ball centered at x. So even though originally I am trying to compare a punctured half ball to a punctured open set in R^n, excision lets me just compare a punctured half ball to a punctured ball.

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