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
- Bert Mendelson, Introduction to Topology, Dover Edition, 1990
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