## Friday, September 15, 2006

### Compactness

The following content is taken directly from Bert Mendelson's Introduction to Topology. The definitions are used in the proof for the existence of the definite integral (to be added later)

Definition 1: Covering

Let K be a subset of the real numbers R. Let C be a collection of subsets of K. C is a covering of K if and only if the union of all subsets of K in C = K.

An open covering is a covering that is also an open set. [See Definition 1, here for definition of an open set]

Definition 2: Subcovering

S is a subcovering of a set X if S is a subset of C which is a covering of X and S itself is a covering of X.

Definition 3: Finite Subcovering

A subcovering S is a finite subcovering of X if it consists of a finite collection of subsets of X.

Definition 4: Compact

A subset K of the real numbers R is said to be compact if every open covering of K has a finite subcovering.

Theorem: [0,1] is compact

Proof:

(1) Let {Oα}α∈I be a covering of [0,1] by open sets. [See Definition 1 above for definition of covering, see Definition 1, here for definition of open sets]

(2) Assume that there is no finite subcovering of {Oα}α∈I [See Definition 3 above for definition of finite subcovering]

(3) This means that at least one of the two closed intervals [0, 1/2] or [1/2, 1] cannot be covered by a finite subcollection of {Oα}α∈I.

(4) Let [a1, b1] denote one of these two intervals of length 1/2 such that it cannot be covered by a finite subcovering.

(5) We may now divide [a1, b1 ] into two subintervals of length 1/4:

[ a1, (a1 + b1)/2 ]

and

[(a1 + b1)/2, b1 ]

(6) We know that at least one of these subintervals cannot be covered by a finite subcovering.

(7) Let [ a2, b2 ] denote the one of the two subintervals in step #5 that cannot be covered by a finite subcovering.

(8) We can in this way define a sequence of intervals such as:

[a0, b0], [a1, b1], [a2, b2 ], ... , [ an, bn ]

(9) Assume that for i = 0, 1, 2, ..., n we have defined intervals [ai, bi] such that:

(a) [a0, b0 ] = [0,1]

(b) bi - ai = 1/(2i) for i = 0, 1, ..., n

(c) [ai+1, bi+1] = [ai, (ai+bi)/2] or [(ai+bi)/2,bi]

(d) for each i = 0, 1, ..., n, no finite subcovering of {Oα}α∈I covers [ai, bi]

(10) We know that bi ≥ ai since bi - ai = 1/(2i) for i = 0, 1, ..., n

(11) We further know that:

ai ≤ ai+1 ≤ bi+1 ≤ bi

since each time, we the average of (ai + bi)/2 ≥ ai.

(12) From step #11, it follows that for each pair of positive integers m and n, am ≤ bn.

(13) Thus, each bn is an upper bound of the set { a0, a1, a2, ... }

(14) Let a be the least upper bound of the set { a0, a1, a2, ... }

(15) Then a ≤ bn for each n and a is the lower bound of the set {b0, b1, b2, ... }

(16) Let b be the greatest lower bound of the set {b0, b1, b2, ... }

(17) We therefore have a ≤ b

(18) By the definition of a and b, we have:

an ≤ a ≤ b ≤ bn for all n.

(19) But this means that b - a ≤ (1/2n) for all n.

(20) Since n can be any size, (1/2n) can get arbitrarily small, so that b -a = 0 so that b=a.

(21) a = b ∈ [a,b] [this follows from how we defined a,b]

(22) Since {Oα}α∈I covers [0,1], it follows that there exists β ∈ I such that a ∈ Oβ.

(23) By definition, Oβ is an open set, so there exists a positive real number ε such that (a - ε, a + ε) is a subset of Oβ [See Definition 1, here for definition of Open Set]

(24) Let us choose a positive integer N that is large enough so that:

1/(2N) is less than ε

(25) Using step #9, we know then that there exists aN, bN such that:

bN - aN = 1/(2N) which is less than ε

(26) Now, a,b ∈ [aN, bN] by step #18 so that we have:

a - aN is less than bN - aN = 1/(2N) which is less than ε

a - bN = b - bN is less than bN - aN = 1/(2N) which is less than ε

(27) Consequently, [aN, bN] is a subset of (a - ε, a + ε ) which is a subset of Oβ

(28) Thus [aN, bN] may be covered by a finite subcollection (namely, one!) of {Oα}α∈I.

(29) This contradicts with step #9d so assumption that no finite subcollection of {Oα}α∈I can cover the interval [an,bn] so we have a contradiction that emerged from our assumption in step #2.

(30) We therefore reject our assumption in step #2 and conclude that the closed interval [0,1] is compact.

QED

References