# What is the boundary of a surface?

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What is the boundary of a surface???

A surface is a two dimensional manifold. I would like to know what constitute the boundary of a surface. (wiki is nor clear enough for me)

HallsofIvy
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?? The boundary of a surface is exactly what you might think it is from everyday experience. In terms of manifolds, you can only have a "boundary" if the set is embedded in a larger set. A point is on the boundary if and only if any neighborhood of the point contains some points in the set and some not in the set.

Of course, a surface may not have a boundary- or, more correctly, its boundary is the empty set. An entire plane does not have a boundary and a sphere (the surface of a ball) does not have a boundary. Both of those, by the way, are examples of a basic theorem in topology- the boundary of a boundary is always the empty set. (But the converse is not neccesarily true- thinking more deeply on that leads to algebraic topology.)

mathwonk
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look at a piece of typing paper. the boundary is the edges, the sharp parts where you cut yourself.

those are points which have a nbhd looking like the upper half of the unit disc plus the diameter as a nbhd of the origin, as opposed to looking like the whole open unit disc as a nbhd of the origin.

i.e. boundary points look like (0,0) looks in relation to the set of those (x,y) where y is non negative. i.e. it is on the edge of that set of points.

and i believe halls is confusing topological boundary with manifold boundary. i am defining manifold boundary.

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I know from regular analysis that for a set in R^n, its boundary is any point such that there exists an open ball centered on that point, part of which is not in the set. (That is probably the definition of topological boundary)

But by this definition, every point of, say, a sphere is part of the boundary. So this is not the definition of boundary that we're talking about here.

monk, could you write the definition of manifold boundary in mathematical language please?

It cannot simply be "is a boundary point of a manifold a point such that the image of that point by homeomorphism of a neighborhood is a boundary point (in the topological sense) of the image of the whole neighborhood." beause if the neighborhood chosen is open, then the image by homeomorphism is open, which is boundaryless.

HallsofIvy
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I know from regular analysis that for a set in R^n, its boundary is any point such that there exists an open ball centered on that point, part of which is not in the set. (That is probably the definition of topological boundary)

But by this definition, every point of, say, a sphere is part of the boundary. So this is not the definition of boundary that we're talking about here.
Yes, it is. Because you are talking about a sphere (rather than a ball), your neighborhoods are themselves are in the sphere. Since no neighborhood of any point contains any points that are not in sphere, it has no boundary.

monk, could you write the definition of manifold boundary in mathematical language please?

It cannot simply be "is a boundary point of a manifold a point such that the image of that point by homeomorphism of a neighborhood is a boundary point (in the topological sense) of the image of the whole neighborhood." beause if the neighborhood chosen is open, then the image by homeomorphism is open, which is boundaryless.

First you will have to define "manifold with boundary" which is different from just "manifold". An n-manifold with boundary is a set, M, with a Hausdorf topology, having a collection of pairs (U, fU) such that the collection of all U's covers M and the corresponding fU is a homeomorphism from U to the closed half space Rn+: the subset of Rn with first coordinate non-negative. Then p is a boundary point of M if and only if, for all U containing p, fU(U) contains some points of Rn with first coordinate 0.

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Hey Halls, thanks for the explanation. I read it numerous times, but again, there must be something I'm missing because

1) Every manifold can be made into a manifold with boundary by changing the atlas by defining new homeomorphisms by composing them with translations in R^n such that all the $f_U(U)$ are now in $\mathbb{R}_+^n$.

2) In the same way, every manifold with boundary can be made boundaryless by a translation of the $f_U(U)$ such that no point in them has first coordinate equal to 0.

While this certainly does not pose any real problem, as by changing the atlas I change the manifold itself, it certainly seems weird to me that the property of possessing a boundary or not is not independant of the choice of atlas...

Yes, it is. Because you are talking about a sphere (rather than a ball), your neighborhoods are themselves are in the sphere. Since no neighborhood of any point contains any points that are not in sphere, it has no boundary.

I simply don't get this part. Consider any point on the sphere and any open ball of R^n centered on that point. Then there is only a small fraction of the ball that intersects the sphere.

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mathwonk
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a topological space X is an n manifold with boundary if every point p of X has a nbhd homeomorphic to an open set in the half space { x in R^n, and x(n) non negative}.

the boundary points are those points of X that do not have an open nbhd homeomorphic to an open set in the open half space where x(n)>0.

i.e. a boundary point of X is a point p having an open nbhd which is homeomorphic to a nbhd of the origin in R^n, which is open as a subset of the closed half space where x(n) is non negative.

i.e. if C is the closed half space in R^n, p is a boundary point of X if some nbhd of p in X is homeomorphic to a nbhd of 0 in C.

p is a non boundary point if p has a nbhd in X which is homeo to a nbhd of 0 which is open in R^n itself.

this is the same definition i gave above, where "looks like" meant "homeomorphic to".

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Hey Halls, thanks for the explanation. I read it numerous times, but again, there must be something I'm missing because

1) Every manifold can be made into a manifold with boundary by changing the atlas by defining new homeomorphisms by composing them with translations in R^n such that all the $f_U(U)$ are now in $\mathbb{R}_+^n$.
First, note that you are looking at all points in R^n s.t. their last coordinate is GREATER THAN OR EQUAL TO 0. If you have an open set in R^n and translate it so that its in the upper half space then it will still be open, it will not have any point with last coordinate = 0. Hence the boundary of the manifold will be empty.

2) In the same way, every manifold with boundary can be made boundaryless by a translation of the $f_U(U)$ such that no point in them has first coordinate equal to 0.
(Let's work on R^2) for a while) If you take something that has a neighborhood that is an upper half disc (say centered at 0) and includes the diameter, (This is basically what mathwonk said before) this is an open neighborhood w.r.t to the topology of the upper half plane. If you translate this, then it will no longer be an open neighborhood.

While this certainly does not pose any real problem, as by changing the atlas I change the manifold itself, it certainly seems weird to me that the property of possessing a boundary or not is not independent of the choice of atlas...
It is in dependent of choice of atlas.

I simply don't get this part. Consider any point on the sphere and any open ball of R^n centered on that point. Then there is only a small fraction of the ball that intersects the sphere.
Lets consider the 2-sphere sitting in R^3. Now suppose you want to find an open neighborhood of the north pole. You do not do this by taking a 3-dimensional ball. A neighborhood of the point is supposed to be two dimensional and on the sphere. Basically what you are after are images of 2-dimensional (open) balls on the sphere.

If instead you consider the upper hemisphere of the 2 dimensional sphere including the equator, then we get a manifold with boundary.

Also, even if you just consider the sphere as a topological space, then you see that the open sets are not what you are thinking of. The way you get open sets in a subspace of a particular space is to look at the open sets in the bigger space and intersect with the subspace you are considering. So you need to be thinking of balls centered at a point on the sphere intersected with the sphere. (Again this is if you are just looking at topology, for the manifold concept you need to really consider what mathwonk has been saying)

arildno
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HallsofIvy said:
Both of those, by the way, are examples of a basic theorem in topology- the boundary of a boundary is always the empty set. (But the converse is not neccesarily true- thinking more deeply on that leads to algebraic topology.)

Hmm-- are you sure of this?
Shouldn't it be of measure zero, rather?

Let S be the set of irrationals in [0,1], and R/S the complement to S.

then, the boundary B of S should be the intersection of the closures cl(S) and cl(R/S).

From what I can see, B=[0,1].
Thus, the boundary of B should be the intersection of B and R/B which should be the point set $\{0,1\}$

A single-point set should be its own boundary as well, I think.

HallsofIvy
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Sorry- I should have said "manifolds" explicitely rather than "topology".

arildno
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Sorry- I should have said "manifolds" explicitely rather than "topology".

Ok, then! Everybody stop all these topological manifold nonesense and simplify your life. An oriented surface has two sides, inside and outside, if you can only go from one side to the other by piercing through the surface, then the surface has no boundary (sphere, closed cube, torus, cylinder closed at both ends, closed cone, etc). If the surface has a boundary, then you can only go from inside to outside of the surface through the boundary, in other words if the surface is closed, then its boundary is empty.

sylas

Everybody stop all these topological manifold nonesense and simplify your life. An oriented surface has two sides, inside and outside, if you can only go from one side to the other by piercing through the surface, then the surface has no boundary (sphere, closed cube, torus, cylinder closed at both ends, closed cone, etc). If the surface has a boundary, then you can only go from inside to outside of the surface through the boundary, in other words if the surface is closed, then its boundary is empty.

The Klein bottle is a counter example to this notion.

The Klein bottle is a counter example to this notion.
Not orientable.

sylas

Not orientable.

Quite so. Good point; I withdraw the example. Thank you.

Even so... this continues to illustrate why we do have the formal mathematical definitions, IMO.

His intuitive explanation is actually a theorem, if you precisely define your assumptions (i.e. a compact connected oriented surface S without boundary in R^3 separates R^3\S into two components), whereas a surface with boundary embedded in R^3 can't possibly have an interior, since the boundary of a boundary is empty.

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