A point p in Rn is said to be a boundary point of A ⊂ Rn if for any > 0, (1) B(p, ) ∩ A 6= ∅, and
(2) B(p, ) ∩ Ac6= ∅.
The set of all boundary points of A is denoted by ∂A. It follows from the definition that
∂A = A ∩ Ac.
Theorem 0.1. Let A be a subset of Rn. Then A is closed if and only if ∂A ⊂ A.
Proof. Suppose that A is closed. Then Rn\ A is open. For each p ∈ Rn\ A, p is an interior point of Rn\ A. There exists > 0 such that B(p, ) ⊂ Rn\ A. Hence B(p, ) ∩ A = ∅. We see that p is not a boundary point of A. Therefore p ∈ Rn\ ∂A. We prove that Rn\ A is a subset of Rn\ ∂A. This implies that ∂A is a subset of A.
Suppose ∂A ⊂ A. To show A is closed, we need to show that Rn\ A is open. Let p ∈ Rn\ A, then p is not a boundary point of A. Thus there exists > 0 such that B(p, ) ∩ A = ∅. Hence B(p, ) ⊂ Rn\ A. This shows that p is an interior point of Rn\ A. Since p is an arbitrarily chosen point of Rn\ A, we find every point of Rn\ A is an interior point of A. Thus Rn\ A is open. We prove that A is closed.
We have two method to show that D = {(x, y) ∈ R2: x2+ y2≤ 1} is closed. At first, we observe that Dc= {(x, y) ∈ R2: x2+y2> 1}. Since Dcis open, D is closed. Secondly, since the boundary of D is ∂D = {(x, y) ∈ R2: x2+ y2= 1} and D contains ∂D, D is closed. Notice that the set of all exterior points of D is ext(D) = Dc and the set of all interior points of D is B = {(x, y) ∈ R2: x2+ y2< 1}.
Then R2has a decomposition into a disjoint union of sets:
R2= Ba
∂Ba ext(D).
Here disjoint union means: the union of B and ∂B and ext(D) is R2and {B, ∂B, ext(D)} are pairwise disjoint.
In general, for any subset A of Rn, we have the decomposition of disjoint union of sets:
Rn = int(A)a
∂Aa ext(A).
(1) int(A) ∩ ∂A = ∅, and (2) ∂A ∩ ext(A) = ∅ and (3) int(A) ∩ ext(A) = ∅.
The proof of the above equations are left to the readers.
This allows us to obtain that
A = int(A)a
∂A = A ∪ ∂A.
Proof. The inclusion ∂A ∪ A ⊂ A is obvious. For p ∈ A, p is never an exterior point of A. Hence
p ∈ int(A) ∪ ∂A ⊂ A ∪ ∂A. We find A ⊂ A ∪ ∂A.
Theorem 0.2. Let A be a subset of Rn. Then int(A) is the largest open set contained in A, i.e.
(1) int(A) is open, and
(2) if V is an open set, V ⊂ A, then V ⊂ int(A).
Proof. Let U = int(A). To show that U is open, we show that int(U ) = U. Notice that int(U ) ⊂ U follows from the definition. Let p ∈ U = int(A). We can choose > 0 so that B(p, ) ⊂ A. For each q ∈ B(p, ), we can choose r > 0 so that B(q, r) ⊂ B(p, ). Hence B(q, r) ⊂ A. This shows that q is an interior point of A. Hence B(p, ) ⊂ int(A) = U. This shows that p ∈ int(U ). We conclude that U ⊂ int(U ). Thus U = int(U ).
Let V be an open set with V ⊂ A. For each p ∈ V, we can choose > 0 so that B(p, ) ⊂ V. Since
V ⊂ A, B(p, ) ⊂ A. We see that q ∈ int(A). Thus V ⊂ int(A).
Lemma 0.1. Let A and B be subsets of Rn with A ⊂ B. Then A ⊂ B.
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Proof. Let p ∈ A. Then B(p, ) ∩ A 6= ∅ for any > 0. For each > 0, we choose q ∈ B(p, ) ∩ A.
Then q∈ B(p, ) and q∈ A. Since A ⊂ B, q ∈ B. We find that q∈ B(p, ) ∩ B. This shows that B(p, ) ∩ B is nonempty for any > 0. Thus p ∈ B.
Theorem 0.3. Let A ⊂ Rn. Then A is the smallest closed subset of A containing A, i.e.
(1) A is closed and
(2) if B is a closed set B ⊃ A, then B ⊃ A.
Proof. Since Rn\ A = ext(A) = int(Ac) is open, A is closed.
If B is closed, B ⊃ A, then
B = B ⊃ A.
We find B ⊃ A.
Example 0.1. The set S = {(x, y) : 0 ≤ x ≤ 1 or x = 2} is closed.
Proof. We find
Sc= {(x, y) : x < 0, 1 < x < 2, x > 2} = {(x, y) : x < 0}∪({(x, y) : x > 1}∩{(x, y) : x < 2})∪{(x, y) : x > 2}
is open. Hence S is closed.
We can show that ∂S = {(x, y) : x = 0} ∪ {(x, y) : x = 1} ∪ {(x, y) : x = 2}. Since S contains ∂S, S is closed.
