Multivariable Advanced Calculus

294 INTEGRATION OF DIFFERENTIAL FORMSTheorem 11.1.4 Let ∂Ω and int (Ω) be as defined above. Then int (Ω) is openin Ω and ∂Ω is closed in Ω. Furthermore, ∂Ω ∩ int (Ω) = ∅, Ω = ∂Ω ∪ int (Ω), and forn ≥ 1, ∂Ω is an n − 1 dimensional manifold for which ∂ (∂Ω) = ∅. The property ofbeing in int (Ω) or ∂Ω does not depend on the choice of atlas.Proof: It is clear that Ω = ∂Ω∪int (Ω). First consider the claim that ∂Ω∩int (Ω) =∅. Suppose this does not happen. Then there would exist x ∈ ∂Ω ∩ int (Ω). Therefore,there would exist two mappings R i and R j such that R j x ∈ R n 0 and R i x ∈ R n < withx ∈ U i ∩ U j . Now consider the map, R j ◦ R −1i , a continuous one to one map from R n ≤to R n ≤ having a continuous inverse. By continuity, there exists r > 0 small enough that,R −1i B (R i x,r) ⊆ U i ∩ U j .Therefore, R j ◦ R −1i (B (R i x,r)) ⊆ R n ≤ and contains a point on Rn 0 , R j x. However,this cannot occur because it contradicts the theorem on invariance of domain, Theorem10.4.3, which requires that R j ◦ R −1i (B (R i x,r)) must be an open subset of R n andthis one isn’t because of the point on R n 0 . Therefore, ∂Ω ∩ int (Ω) = ∅ as claimed. Thissame argument shows that the property of being in int (Ω) or ∂Ω does not depend onthe choice of the atlas.To verify that ∂ (∂Ω) = ∅, let S i be the restriction of R i to ∂Ω ∩ U i . ThusS i (x) = (0, (R i x) 2, · · · , (R i x) n)and the collection of such points for x ∈ ∂Ω ∩ U i is an open bounded subset of{u ∈ R n : u 1 = 0} ,identified with R n−1 . S i (∂Ω ∩ U i ) is bounded because S i is the restriction of a continuousfunction defined on R m and ∂Ω ∩ U i ≡ V i is contained in the compact set Ω. Thusif S i is modified slightly, to be of the formS ′ i (x) = ((R i x) 2− k i , · · · , (R i x) n)where k i is chosen sufficiently large that (R i (V i )) 2− k i < 0, it follows that {(V i , S ′ i )}is an atlas for ∂Ω as an n − 1 dimensional manifold such that every point of ∂Ω is sentto to R n−1< and none gets sent to R n−10 . It follows ∂Ω is an n − 1 dimensional manifoldwith empty boundary. In case n = 1, the result follows by definition of the boundaryof a 0 dimensional manifold.Next consider the claim that int (Ω) is open in Ω. If x ∈ int (Ω) , are all points ofΩ which are sufficiently close to x also in int (Ω)? If this were not true, there wouldexist {x n } such that x n ∈ ∂Ω and x n → x. Since there are only finitely many chartsof interest, this would imply the existence of a subsequence, still denoted by x n and asingle map, R i such that R i (x n ) ∈ R n 0 . But then R i (x n ) → R i (x) and so R i (x) ∈ R n 0showing x ∈ ∂Ω, a contradiction to int (Ω) ∩ ∂Ω = ∅. Now it follows that ∂Ω is closedin Ω because ∂Ω = Ω \ int (Ω). This proves the theorem. Definition 11.1.5 An n dimensional manifold with boundary, Ω is a C k manifoldwith boundary for some k ≥ 0 ifR j ◦ R −1i ∈ C k ( R i (U i ∩ U j ); R n)and R −1i ∈ C ( k R i U i ; R m) . It is called a continuous manifold with boundary if themappings, R j ◦ R −1i , R −1i , R i are continuous. In the case where Ω is a C k , k ≥ 1