Let A be a subset of Rn

1. Supplement Note

Let A be a subset of Rⁿ. The set of all adherent points of A is denoted by A called the closure of A. We say that A is closed if A = A.

Example 1.1. The empty set ∅ and Rⁿ are both closed sets.

Theorem 1.1. Let A be a subset of Rⁿ. Then A is closed if and only if Rⁿ\ A is open.

Proof. When A is empty, the statement is obvious. We assume that A is nonempty.

Suppose A is a closed subset of Rⁿ. Then A = A. Let us prove that Rⁿ\ A is open. Let p be a point of Rⁿ\ A. Since A contains all of its adherent points, p is never an adherent point of A. We can find  > 0 so that B(p, ) ∩ A = ∅. For q ∈ B(p, ), q 6∈ A. Thus q ∈ Rⁿ\ A. This proves that B(p, ) is contained in Rⁿ\ A. (It is because every element of B(p, ) is an element of Rⁿ\ A.) Hence p is an interior point of Rⁿ\ A. We find that Rⁿ\ A is open. (Since p is an arbitrary choice of point of Rⁿ\ A and p is an interior point of Rⁿ\ A.)

Let us assume that Rⁿ\ A is open. To show that A = A, we only need to show that A ⊂ A. This is because A already contains A. To show A is a subset of A, it is equivalent to show that Rⁿ\ A is a subset of Rⁿ\ A. Let p be a point of Rⁿ\ A. Since Rⁿ\ A is open, we can find  > 0 so that B(p, ) is contained in Rⁿ\ A. In this case, B(p, ) ∩ A = ∅. Therefore p is not an adherent point of

A and hence p 6∈ A, i.e. p ∈ Rⁿ\ A. 

Lemma 1.1. (de Morgan’s law) Let A be any set and {Ai: i ∈ I} be a family of subsets of A. Then (1) A \S

i∈IAi=T

i∈I(A \ Ai).

(2) A \T

i∈IAi=S

i∈I(A \ Ai).

Proof. The proof is left to the readers. The readers should work out this lemma by themselves.

 Corollary 1.1. In Rⁿ,

(1) ∅, Rⁿ are both closed.

(2) Any intersection of closed sets is closed.

(3) Any finite union of closed sets is closed.

Proof. See class notes or Bartles’ book.

 Example 1.2. The set D = {(x, y) ∈ R²: x²+ y²≤ 1} is closed.

The complement of D is U = {x²+ y²> 1}. We leave it to the reader to verify that U is open.

Example 1.3. Is S = {(x, y) : 0 < x ≤ 1, 0 ≤ y ≤ 2} closed?

The set S is not closed. We prove that S is not closed by showing that S 6= S. Let p = (0, 1/2).

Then p 6∈ S. If we can show that p ∈ S, then p ∈ S \ S.

For 0 <  < 1, we choose q= (/2, 1/2). Then q∈ S and d(p, q) = 

2 < .

Hence q∈ B(p, ) ∩ S for 0 <  < 1. We find that B(p, ) ∩ S is nonempty for any 0 <  < 1. For

 > 1, we choose q⁰_= (1, 1/2). Then q_⁰ ∈ S and

d(p, q⁰_) = 1 < .

Hence q⁰_∈ B(p, )∩S. We find that B(p, )∩S is nonempty for any  > 1. We conclude that B(p, )∩S is nonempty for any  > 0. We see that p is an adherent point of S i.e. p ∈ S.

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