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 q0 ∈ 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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