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 {O

_{α}}

_{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 {O

_{α}}

_{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 {O

_{α}}

_{α∈I}.

(4) Let [a

_{1}, b

_{1}] 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 [a

_{1}, b

_{1}] into two subintervals of length 1/4:

[ a

_{1}, (a

_{1}+ b

_{1})/2 ]

and

[(a

_{1}+ b

_{1})/2, b

_{1}]

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

(7) Let [ a

_{2}, b

_{2}] 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:

[a

_{0}, b

_{0}], [a

_{1}, b

_{1}], [a

_{2}, b

_{2}], ... , [ a

_{n}, b

_{n}]

(9) Assume that for i = 0, 1, 2, ..., n we have defined intervals [a

_{i}, b

_{i}] such that:

(a) [a

_{0}, b

_{0}] = [0,1]

(b) b

_{i}- a

_{i}= 1/(2

^{i}) for i = 0, 1, ..., n

(c) [a

_{i+1}, b

_{i+1}] = [a

_{i}, (a

_{i}+b

_{i})/2] or [(a

_{i}+b

_{i})/2,b

_{i}]

(d) for each i = 0, 1, ..., n, no finite subcovering of {O

_{α}}

_{α∈I }covers [a

_{i}, b

_{i}]

(10) We know that b

_{i}≥ a

_{i}since b

_{i}- a

_{i}= 1/(2

^{i}) for i = 0, 1, ..., n

(11) We further know that:

a

_{i}≤ a

_{i+1}≤ b

_{i+1}≤ b

_{i}

since each time, we the average of (a

_{i}+ b

_{i})/2 ≥ a

_{i}.

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

_{m}≤ b

_{n}.

(13) Thus, each b

_{n}is an upper bound of the set { a

_{0}, a

_{1}, a

_{2}, ... }

(14) Let a be the least upper bound of the set { a

_{0}, a

_{1}, a

_{2}, ... }

(15) Then a ≤ b

_{n}for each n and a is the lower bound of the set {b

_{0}, b

_{1}, b

_{2}, ... }

(16) Let b be the greatest lower bound of the set {b

_{0}, b

_{1}, b

_{2}, ... }

(17) We therefore have a ≤ b

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

a

_{n}≤ a ≤ b ≤ b

_{n}for all n.

(19) But this means that b - a ≤ (1/2

^{n}) for all n.

(20) Since n can be any size, (1/2

^{n}) 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 {O

_{α}}

_{α∈I }covers [0,1], it follows that there exists β ∈ I such that a ∈ O

_{β}.

(23) By definition, O

_{β}is an open set, so there exists a positive real number ε such that (a - ε, a + ε) is a subset of O

_{β}[See Definition 1, here for definition of Open Set]

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

1/(2

^{N}) is less than ε

(25) Using step #9, we know then that there exists a

_{N}, b

_{N}such that:

b

_{N}- a

_{N}= 1/(2

^{N}) which is less than ε

(26) Now, a,b ∈ [a

_{N}, b

_{N}] by step #18 so that we have:

a - a

_{N}is less than b

_{N}- a

_{N}= 1/(2

^{N}) which is less than ε

a - b

_{N}= b - b

_{N}is less than b

_{N}- a

_{N}= 1/(2

^{N}) which is less than ε

(27) Consequently, [a

_{N}, b

_{N}] is a subset of (a - ε, a + ε ) which is a subset of O

_{β}

(28) Thus [a

_{N}, b

_{N}] may be covered by a finite subcollection (namely, one!) of {O

_{α}}

_{α∈I}.

(29) This contradicts with step #9d so assumption that no finite subcollection of {O

_{α}}

_{α∈I}can cover the interval [a

_{n},b

_{n}] 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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