General Topology — Part 5 (Compact Set)

8 min readOct 29, 2023

Extreme Value Theorem

When you studied calculus in high school, you may have learnt the Extreme Value Theorem. This theorem states that if a real-valued function f is continuous on the closed interval [a,b], then f must have at least one maximum and one minimum. That is, there exist numbers c and d in [a,b] such that:

f(c)≥f(x)≥f(d)

This theorem is intuitive for you to understand when you see the above graph. But if you want to prove it rigorously, you will need topological concepts. We are now going to study those concepts that can help us to prove this theorem.

Cover, Compact Space and Compact Set

Let (X, τ) be the topological space for set X.

The Cover C of X is a collection of subsets of X whose union is all of X.
C = {Uα| α ∈ A, Uα ⊂ X} where subset Uα is indexed by set A.
∪Uα = X

The collection C will be called Open Cover if each of its member is an Open Set of (X, τ).

If Y is a subset of X, the Open Cover of the subset Y or the induced Subspace Topology (Y,τY), is a collection of subsets of X whose union contains Y:
∪Uα ⊇ Y

The sub-collection F of C will be called Subcover if the union of its members still covers X.
F = {U[i]| i ∈ N, U[i] ⊂ X} where subset U[i] is indexed by natural number N.
∪U[i] = X

A topological space (X, τ) is called the Compact Space if every Open Cover of X has a finite Subcover:
∪Uα = X ⇒ ∪U[i] = X

The subset Y is called a Compact Set if every Open Cover of Y has a finite Subcover:
∪Uα ⊇ Y ⇒ ∪U[i] ⊇ Y

For a singleton set {a}, it is compact since every open cover of a singleton set has at least one open set that contains element a:
{a} ⊆ ∪Uα ⇒ ∃U ∈ C s.t. a ∈ U ⇒ {a} ⊆ U

Properties of Compact Set

CP1. A closed subset of a compact set is compact
Proof-
Let S be the closed subset of the compact set Y where
S ⊆ Y ⊂ X

For any open cover {Uα| α ∈ A} of S,
∪Uα ⊇ S
⇒ ∪Uα ∪ (X\S) ⊇ S ∪ (X\S) = X ⊃ Y
⇒ {Uα| α ∈ A} ∪ {X\S} is a open cover of Y [ since X\S is open ]
⇒ {U[i]| i ∈ N} ∪ {X\S} is a finite subcover of Y [ since Y is compact ]
⇒ ∪U[i] ∪ (X\S) ⊃ Y ⊇ S
⇒ ∪U[i] ⊃ S
⇒ {U[i]| i ∈ N} is a finite subcover of S ⇒ S is compact

CP2. Finite union of compact set is compact
Proof-
Let {S[i]| i ∈ N} be the collection of the compact sets
For any open cover {Uα| α ∈ A} of ∪S[i],
∪Uα ⊇ ∪S[i]
⇒ ∪Uα ⊇ S[i] ∀i
⇒ {Uα| α ∈ A} is a open cover of S[i] ∀i
⇒ {U[j]| j ∈ N} is a finite subcover of S[i] ∀i [ since Si is compact ]
⇒ {U[j]| j ∈ N} is a finite subcover of ∪S[i] ⇒ ∪S[i] is compact

CP3. The continuous mapping on the compact set is compact
Proof-
Let f:X->Y be the continuous mapping
Let A be the compact set in X and B be image of f on A
f(A) = B
For any open cover {Uα} of B
B ⊆ ∪Uα
⇒ A ⊆ f-1(B) ⊆ f-1(∪Uα) = ∪f-1(Uα)
⇒ {f-1(Uα)} is open cover of A
⇒ {f-1(U[i]) | i ∈ N} is a finite subcover of A [ since A is compact ]
⇒ A ⊆ ∪f-1(U[i])
⇒ B = f(A) ⊆ f(∪f-1(U[i])) = ∪f(f-1(U[i])) ⊆ ∪U[i]
⇒ {U[i]} is a finite subcover of B ⇒ B is compact

Let Y be a subspace of X and for subset A ⊆ Y ⊆ X:

CP4. A is compact in X iff A is compact in Y
Proof-
(only if)
For any open cover {oα| α ∈ A, oα is open set in Y} of A,
⇒A ⊆ ∪oα = ∪{Y∩wα | wα is open set in X and oα=Y∩wα} = Y ∩ ∪wα
⇒A ⊆ Y ∩ ∪w[i] [ A is compact in X and ∪wα is cover of A ]
⇒A ⊆ ∪(Y∩w[i]) = ∪{o[i] | o[i] is open set in Y }
⇒{o[i]| i ∈ N, o[i] is open set in Y} is a finite subcover of A
⇒A is compact in Y

(if)
For any open cover {oα| α ∈ A, oα is a open set in X} of A,
⇒A ⊆ ∪oα
⇒A∩Y ⊆ ∪oα ∩ Y = ∪(oα ∩ Y) = ∪{wα | wα is open set in Y and wα=oα ∩ Y}
⇒A ⊆ ∪wa [ Since A ⊆ Y ]
⇒A ⊆ ∪w[i] ⊆ ∪{o[i] | o[i] is open set in X and w[i]=o[i]∩Y}
[ A is compact in Y and ∪wα is cover of A ]
⇒{o[i]| i ∈ N, o[i] is open set in X} is a finite subcover of A
⇒A is compact in X

Hausdorff Space and Compact Set

Another special type of topological space is Hausdorff Space. A topological space (X, τ) is called the Hausdorff Space if for every p, q ∈ X, p ≠ q, there exist open set U and V such that p ∈ U, q ∈ V and U ∩ V = ∅.

HD1. The compact set in the Hausdorff Space is closed
Proof-
Let Y be the compact set in the Hausdorff Space, Y ⊂ X,
∀ p ∈ X\Y
⇒ ∀ q ∈ Y, ∃ Uq and Vq s.t. p ∈ Uq and q ∈ Vq and Uq∩Vq= ∅ [ By definition of Hausdorff Space and p ≠ q ]
⇒ ∪Vq ⊇ Y
⇒ {Vq} is an open cover of Y and {Uq} is a collection of neighbourhood of p
⇒ {V[i]} is a finite subcover of Y and {U[i]} is a collection of neighbourhood of p [Since Y is compact ]
⇒ ∪V[i] ⊇ Y and p ∈ ∩U[i] is open [ By topology axiom T3, see Part 1 ]
⇒ ∃U=∩U[i] s.t. p ∈ U and U ∩ Y ⊆ U ∩ ∪V[i] = ∪((∩U[j]) ∩ V[i]) = ∅ [ U[i] ∩ V[i] = ∅]
⇒ ∃U s.t. p ∈ U ⊆ X\Y [ since U ∩ Y = ∅ ]
⇒ X\Y is a neighbourhood of p
⇒ X\Y is a neighbourhood of all its element ⇒ X\Y is open [ By NB1, see Part 3 ] ⇒ Y is closed

HD2. Rn is a Hausdorff Space
Proof-
Rn is an Euclidean topology.
∀ p, q ∈ Rn, p ≠ q,
let r = 0.5 * |p-q|, consider open ball B(p, r) and B(q, r):
p ∈ B(p, r), q ∈ B(q, r) [ B(p, r)={x∈Rn| |x-p| < r} which is an open set ]
Suppose B(p, r)∩B(q, r) ≠ ∅
⇒ ∃x ∈ Rn s.t. x ∈ B(p, r)∩B(q, r)
⇒ ∥p-x∥ < r and ∥q-x∥ < r
⇒ ∥p-q∥ < ∥p-x∥+∥x-q∥ < 2r = ∥p-q∥ ⇒ contradiction
⇒ B(p, r)∩B(q, r) = ∅
⇒ Rn is Hausdorff Space

Heine–Borel theorem

This theorem, named after Eduard Heine and Émile Borel, states that any closed and bounded set in Rn can be approximated by a finite number of open sets, but we will use this theorem in another way to help us to prove Extreme Value Theorem. The Heine-Borel theorem states the following feature of compact set:

HB. A is a compact subset in Rn iff A is closed and bounded

Proof-
(only if)
A is compact and A ⊂ Rn
⇒ A is a compact set in the Hausdorff Space [ By HD2, Rn is Hausdorff ]
⇒ A is closed [ By HD1 ]

Let { B(x, 1) | x ∈ A} be the open cover of A
⇒ C={ B(x[i], 1) | x[i] ∈ A, 0≤i≤L } be the finite subcover of A. [ Since A is compact, finite subcover exists ]
∀p, q ∈ A, p ≠ q
⇒ ∃B(x[m], 1), B(x[n], 1) ∈ C s.t. p ∈ B(x[m], 1) and q ∈ B(x[n], 1)
⇒ ∥p-q∥ ≤ ∥p-x[m]∥+∥x[m]-x[n]∥+∥x[n]-q∥
⇒ ∥p-q∥ ≤ 1 + ∥x[m]-x[n]∥ + 1 [ since p ∈ B(x[m], 1) and q ∈ B(x[n], 1) ]
⇒ ∥p-q∥ ≤ M + 2 [ since {∥x[m]-x[n]∥ | 0≤m,n≤L} is finite and thus is bounded above by a constant M ]
⇒ A is bounded

(if)
A is bounded
⇒ A ⊆ [-M/2, M/2]^n where M is a constant

Let T0 = [-M/2, M/2]^n
Let C be an open cover of T0.

Assume on the contrary, C does not have finite subcover.

T0 can be bisected into 2^n smaller cube with size M/2 and at least one of them must have no finite subcover. Otherwise, T0 can be covered by finite subcover. Let this box be T1.

Using bisection method, the cube can be divided infintely since no cube with finite subcover exists. Thus, the following nested interval sequence {Ti} exists:

T0 ⊇ T1 ⊇ T2 …

Now, the sequence (Ti) has these properties:
1. Ti is closed
2. ∀i∈N, Ti ⊇ Ti+1
3. ∀i∈N, |Ti| = M/(2^i)
⇒ (Ti) converges to some real number x ∈ A [ By Nested Interval Theorem ]
⇒ x ∈ U where U ∈ C [ x ∈ A and A is covered by C ]
⇒ x ∈ B(x, r) ⊆ U for some r > 0 [ By Ba1 where B(x, r) is a base for Rn ]
⇒ ∃ k∈N s.t. |Tk| = M/2^k < 2r
⇒ Tk can be covered by U
⇒ T0 can be covered by finite cover
⇒ C has finite subcover ⇒ contradiction

T0 is compact
⇒ A is compact [ By CP1 where A is a closed subset of T0 ]

Finally, we can prove the Extreme Value Theorem
Let f: [a,b] → R where f is continuous on [a,b]

[a,b] is closed and bounded
⇒ [a,b] is compact [ By Heine–Borel theorem ]
⇒ f([a,b]) is compact set on R [ By CP3 ]
⇒ f([a,b]) is bounded set on R [ By Heine–Borel theorem ]
⇒ f([a,b]) has at least one maximum and one minimum [ By Greatest Lower Bound Axiom ]

Conclusion

In this post, we have studied the definition of compact set and its properties.

In the coming next article, I will show you more properties of compact set under Hausdorff Space, Compact Hausdorff Space and Locally Compact Hausdorff Space.