# Talk:Boundary (topology)

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Field: Topology

## Boundary of a boundary

Note sure about this section. Consider a closed disk {(x,y) : x^2+y^2<=1}. Here the complement is {(x,y) : x^2+y^2>1} and the closure of the complement is {(x,y):x^2+y^2>=1}. The intersection of the two is {(x,y):x^2+y^2=1} the circle.

I think the fact that the boundary of a boundary of manifolds is the empty set has more to do with soothness and differential constraints rather than the fact that its not a topological space. --Salix alba (talk) 11:23, 28 January 2006 (UTC)

Ah I see what your getting at. Your considering the disk as the lying in R^3, in which case topoligical boundary will be the whole disk. However we should be careful with the notion of embedding. Manifold do not have to be embedded in a particular euclidean space, a manifold != the embedding of a manifold. Consider projective plane often though of as a set in R^4, but works fine as a topological 2-manifold defined by its charts but not necessarily embedded in R^4.

As manifold is now a non technical article maybe we need better definition of manifold with bounday here. --Salix alba (talk) 11:42, 28 January 2006 (UTC)

Heres a simple non rigrious proof of why the boundary of a boundary of a smooth manifold is the empty set. Note this does require the intrinsic view of a manifold, rather than an embedded manifold.

Let M be a smooth n-manifold with boundary and let N be its boundary. Consider a point p on the boundary of M, there is an open set $A\subset M$ containing p and an open set $B \subset [0,\infty) \times R^{n-1}$ containing the origin and a chart f:A->B. Now f maps $N \cap A$ onto $C = B \cap [0] \times R^{n-1}$ which is isomorphic to an open set in $R^{n-1}$. The boundary of C is empty. QED. I don't get why the boundary of C is empty. Like, suppose B is the filled upper hemisphere of a sphere. Then C is a disc, which has a nonempty boundary. You only discover the empty boundary when you patch together all the neighborhoods, you cannot discover it locally. The easiest way to prove that the boundary of the boundary of a manifold is empty is to realize that the boundary of the boundary is given in the subspace topology. Every set has empty boundary in the subspace topology. -lethe talk 20:54, 28 January 2006 (UTC)

Filled upper hemisphere of sphere is not a smooth (i.e. differential) manifold with boundary the charts around the equator are not diffeomorphic to $[0,\infty) \times R^{n-1}$ instead they are diffeomorphic to $[0,\infty) \times [0,\infty) \times R^{n-2}$. --Salix alba (talk) 21:41, 28 January 2006 (UTC)
You're mistaken about that. The filled hemisphere is of course a smooth manifold with boundary. But more importantly to my argument, it is a neighborhood in [0,∞)×R2. -lethe talk 22:53, 28 January 2006 (UTC)

Looking back at the revison history there was a much better treatment of manifold with boundary a while back, rather than the slighlty confused and possible incorect version we have now. --Salix alba (talk) 18:49, 28 January 2006 (UTC)

You're probably talking about this version. We decided that it rambled on too long, I have a bad habit of doing that sometimes. But why do you think the version we have now is incorrect? -lethe talk 20:56, 28 January 2006 (UTC)
I think tere is some stuff which could go back, at very least we need a link to [1] which is probably the best treatment. The link to boundary of a manifold just redirects back to manifold. Currently its saying they are different but we are not going to tell you how. Add the appropriate restrictions (differential manifold?) and topological def of boundary gives the same as boundary of manifold and also implies that the boundary of a manifold with boundary is empty. (sorry fuzzy on my terms, will think later) --Salix alba (talk) 21:41, 28 January 2006 (UTC)

The top half of a filled sphere is an interesting example, highlighting the difference between a topological and differentiable manifold. Viewed as a topological manifold it is a topological ball so all points on the boundary (including those on the equator) are homeomorphic to $[0,\infty) \times R^{n-1}$.

Now if we view it as a differentiable manifold we require that the charts need to be differentiable. Around the equator with its sharp edge differentiability fails for a chart onto open sets in $[0,\infty) \times R^{n-1}$. Hence we need we need charts which map onto open sets in $[0,\infty) \times [0,\infty) \times R^{n-2}$. --Salix alba (talk) 21:50, 29 January 2006 (UTC)

### Clearly Wrong

This section is clearly wrong. Indeed, one can characterize the nowhere dense sets as those sets whose closures are contained in a boundary set. Recall that a set is nowhere dense if the interior of its closure is empty. Since boundary sets are closed, their closure is trivially contained in a boundary set. So every boundary set is nowhere dense. This implies that the interior of a boundary set is empty, again because boundary sets are closed. So the topological boundary operator is in fact idempotent. —The preceding unsigned comment was added by 12.154.214.29 (talkcontribs) 21:49, 17 August 2006 (UTC)

I'm not quite sure what you're trying to say here: specifically, I'm not sure what you mean by "sets whose closures are contained in a boundary set". In any case, the boundary of a set can certainly be dense: for example, in the usual topology of the real line, the boundary of the set of rational numbers is the entire real line, which is obviously everywhere dense. —Ilmari Karonen (talk) 23:26, 18 August 2006 (UTC)
I think I became confused by two distinct but related notions of "boundary sets". The notion I had in mind is that a set $A$ is a boundary set if the closure of its complement cl$(A^c)$ is dense in the ambient space. It is used primarily in Descriptive Set Theory and other related branches of topology and analysis. My confusion explains my difficulties proving a theorem in the field, and is very irksome. —The preceding unsigned comment was added by 12.154.214.29 (talkcontribs) 23:56, 25 August 2006 (UTC2)
Since "boundary of a set" and "boundary set" are two entirely different things, maybe there should be a separate Wikipedia page for the latter. In his influential 1922 paper "Sur l'Operation Ā de l'Analysis Situs" (Fundamenta Mathematicae), Kuratowski defined a boundary set as any subset with empty interior, or equivalently, any subset A such that A equals the intersection of itself with the closure of its complement. I have seen this definition applied elsewhere as well. The preceding commenter clearly meant to say that $A$ is a boundary set if just its complement (not the closure of its complement) is dense in the ambient space. This definition is equivalent to Kuratowski's. The first commenter meant to say that nowhere dense sets are those whose closures are boundary sets (not just contained in them). This created a little confusion. — Preceding unsigned comment added by Mathematrucker (talkcontribs) 22:44, 28 June 2012 (UTC)

## Venn Diagram

I'm wondering about the diagram illustrating the relationship between points in S, limit points (A) and boundary points (B), along with isolated and interior points. Is this diagram only about subsets of $R^n$ (the bullet just above the diagram references $R^n$, although it isn't clear that the diagram only pertains to this point) with usual ball topology ? Otherwise there are topological spaces where there exist isolated points which aren't boundary points, no ? Take X with a single point "x" along with its trivial topology and consider the subset S=X itself: "x" is an isolated point, but not a boundary point; the boundary of X is empty.

Similarly, an interior point is not necessarily a limit point, for example if X has the discrete topology, since singletons are open sets then for every point "s" of a subset S there exists neighborhoods completely contained in S, hence all points of S are interior. Those same points are not limit points but isolated points (since these singletons sets are also neighborhoods of points "s" containing only "s".) 70.81.15.136 (talk) 02:37, 27 February 2008 (UTC)

I have added that S must be a subset of $R^n$, now the text should be correct. But I don't find this diagram very interesting nor particularly clear, in my opinion it could be removed. Daniele 84.220.23.101 (talk) 19:49, 1 March 2008 (UTC)

The Venn Diagram is incorrect, period. Every isolated point of a set is in its interior and not its boundary. Proof: For x in S, if {x} is open in X (x is isolated), Then x is in the union of all open sets in X which are in S, the definition of the interior. Also, {x} cannot intersect both S and S^c, so an isolated point (meaning {x} is a neighborhood of x) is never in the boundary of S.

## The boundary

    Two media may be located so close to one another that there is no third medium between the two. Nevertheless there is a boundary between them except that the boundary is Nothingness which is 'something' but not space or time. An observer wanting to cross from one medium to the other has to cross the boundary. Change of one medium for the other creates the duality of exitence non-existence because in the boundary one medium ends existence and the other starts. If the boundary is perfect the chasnge is quantittive. If it is imperfect, and it has magnitude either spatial or temporsal, there is transformation through the boundary. There is no observation in the boundary if it cannot accommodate observer's 'now' containing consciousness. Two units 'now' are required to observe a difference. It is the boundary and not 'now' that contains change from existence to non-existence and the other way round. Sleep is the boundary in whcih we are not conscious of existence of anything.
There is only one medium common for all the space times. It is the observer who creates from this medium different space times and boundaries between them. The medium in which creation takes place is the Nothingness which is neither space nor time. Without space time only the Nothingness is left as the boundary between the existence and non-existence of the observer's 'self'. A very important boundary is the one between the material and the immaterial space times. Human observer, located in the material world uses 'now' of constant magnitude. The result is that all forms of change in his world are measured with that static unit 'now', including changes in that unit 'now' itself. Each different space time uses a different unit 'now'. To change his location from the material world to the immaterial, the observer can use the continuous method when he changes his unit 'now' until it is small enough to cross the boundary. The other way is to change the unit 'now' quantitatively. In either case the observer has to change from existence to non-existence in the material world before he can exist in the immaterial space time. KK (78.146.162.196 (talk) 11:49, 3 April 2010 (UTC))


This is irrelevant —Preceding unsigned comment added by 24.59.177.33 (talk) 16:10, 4 September 2010 (UTC)

## Dubious property

Let $n = 1$, so consider the real line. Let $A = [-1,1]$ be the unit closed circle. A is closed. Suppose it is the boundary of some set of the reals, say $B$. Then $B - A$ is open. Furthermore, it is the boundary of $X = \mathbb R - B$, which is open too. Hence $X$ consist of two open components, say $O1, O2$. But as $Y = {x real, x > 1}$ is open, both $Y \cap O1$, $Y \cap O2$ are open. Furthermore, as $X = O1 \cup O2$, the union of these two open intersections is the whole of Y. But Y is connected. Contradiction.

Hence the last stated property, that each closed set of $R^n$ is the boundary of some subset of $R^n$ is clearly wrong. 131.220.99.58 (talk) 14:07, 9 July 2010 (UTC)

I didn't write the mentioned property, but it is obviously true for the example you provided. Let B= (Q intersect [-1,1]) I dont see why you said R-B is open. 216.80.140.25 (talk) 22:33, 17 August 2010 (UTC)

The property is correct an cited at Willard problem 3B page 29 (at least for n=2 but then it clearly works for any n). I can try and solve the exercise with an obvious generalization of the above solution for the case A=[-1,1]. if A is a closed set we can take B to be the union of the complement of A with the points in A that have rational coordinates. I think that A is the boundary of B. —Preceding unsigned comment added by 79.183.41.125 (talk) 09:59, 21 August 2010 (UTC)

The property is indeed correct. The above 'proof' is flawed because the set $B - A$ may be empty (as in the noted counterexample), in which case the argument is invalid, because connectedness needs components to be non-empty. —Preceding unsigned comment added by 212.163.33.165 (talk) 15:52, 22 September 2010 (UTC)

I am removing the dubious tag, since indeed, if C is a closed subset of R^n, then C is the boundary of (C \ Q^n) union (boundary C). --kundor (talk) 00:15, 4 November 2010 (UTC)