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
- Charles R. MacCluer, Honors Calculus, Princeton University Press, 2006
- Bob Gardner, Introductory Level Analysis: Synthesizing R, Rn, Metric Spaces and Topological Spaces.
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