## Thursday, September 14, 2006

### Open Sets

Today, I talk about open and closed sets. These definitions are needed as part of the foundation of calculus. For example, I will use these properties in my future blog on compactness.

Definition 1: Open Set

A set S that is a subset of the real numbers R is open if and only if for all x ∈ S, there exists ε greater than 0 such that (x + ε, x - ε) is a subset of S.

Lemma 1: ∅ is open

Proof:

This is true since there are no elements of . In other words, for all x ∈ S (none), there exists ε greater than 0 such that (x + ε, x - ε) is a subset of S.

QED

Lemma 2: The set of real numbers R is open

Proof:

(1) Let x be any real number.

(2) Let ε be any real number greater than 0.

(3) (x + ε, x - ε) is a subset of S since for all y ≤ ε, x + y and x - y ∈ R. [By the property of closure, see here for properties of real numbers]

QED

Lemma 3: The union of open sets is open

Proof:

(1) Let { Ai } be a set of open sets.

(2) Let A = ∪ { Ai }

(3) Let x be an element of A.

(4) There must exist an i such that x ∈ Ai

(5) By definition of open sets, there exists ε such that (x + ε, x - ε) is a subset of Ai

(6) But then (x + ε, x - ε) is a subset of A since Ai is a subset of A.

QED

Lemma 4: The intersection of a finite number of open sets is open.

Proof:

(1) Let { Ai } be a finite set of open sets.

(2) Let A ∩ { Ai }

(3) If A = ∅, then A is open, [See Lemma 1 above] so we can assume that A is not .

(4) So there exists x such that x ∈ A

(5) for all i, x ∈ Ai.

(6) Since all Ai are open, for each Ai, there must exist an εi greater than 0, such that (x + εi, x - εi) is a subset of Ai

(7) Let ε = min({εi})

(8) Since (x + ε, x - ε) ⊂ (x + εi, x - εi) for all i, it follows that (x + ε, x - ε) ∈ A = ∩ {Ai}

QED

Lemma 5: (a,b) = { x : a is less than x is less than b } is open

Proof:

(1) Let x be any real number greater than a and less than b.

(2) Let ε = min(abs(a - x),abs(b-x))

(3) So we can see that x + ε ≤ b and a ≤ x - ε.

(4) We can further see that (x + ε/2, x - ε/2) is a subset of (a,b)

QED

Lemma 6: (a, ∞) = { x : a is less than x } is open

Proof:

(1) Let x be any real number greater than a.

(2) Let ε = abs(x - a)/2

(3) It is clear that (x - ε, x + ε) is a subset of (a, ∞)

QED

Definition 2: Complement A'

A' is the complement of A if and only if A' includes all the points that are not in A.

For example if A = negative integers, then A' = nonnegative integers.

Definition 3: Closed Set

A subset of the real numbers R is closed if its complement is open.

Lemma 7: [a,b]= { x : a ≤ x ≤ b } is closed

Proof:

(1) To prove this, we need to show that the complement of [a,b] is open.

(2) So that we have complement[a,b] = { x: x is less than a or x is greater than b }

(3) if x is less than a, then let ε = (a - x)/2. Clearly, { x - ε, x + ε } is a subset of { x : x is less than a or x is greater than b }

(4) if x is greater than b, then let ε = (x - b)/2. Clearly, { x - ε, x + ε } is a subset of { x : x is less than a or x is greater than b }

QED

Lemma 8: [a, ∞ ) = { x : x ≥ a } is closed

Proof:

(1) The complement of this is (-∞, a) = { x : x is less than a }

(2) Since x is less than a, let ε = (a - x)/2.

(3) Clearly, { x - ε, x + ε } is a subset of { x : x is less than a }

QED

References