Compactness

Compactness is a property in metric spaces. Before discussing the compactness of metric spaces, we must know what a cover, subcover, and finite is. The definition of compactness is based on these concepts.

Cover of a metric space (X, d) means, for collection C = {G_𝛼 | 𝛼 ∈ I; I is an index set} of subsets of X such that ⋃_{𝛼 ∈ I} G_𝛼 = X, the C is called the cover of X.

If C’ is a sub-collection of C, such that C’ itself covers X then C’ is said to be a subcover of C. If C’ is a collection of finite elements then it is called finite subcover.

The collection C = {G_𝛼 | 𝛼 ∈ I; I is an index set} is such that every G_𝛼 is an open set, then C is called open cover of X.

Definition of Compactness

The compactness of a metric space is defined as, let (X, d) be a metric space such that every open cover of X has a finite subcover.

A non-empty set Y of X is said to be compact if it is compact as a metric space.

For example, a finite set in any metric space (X, d) is compact. In particular, a finite subset of a discrete metric (X,d) is compact.

Sequentially Compact: A metric space (X, d) is said to be sequentially compact if every sequence in X has a subsequence that converges in X.

With the concept of compactness and sequential compactness, there is a significant result.

Properties of Compactness

Let (X, d) be a metric space with metric d such that (X, d) is a compact metric space, then

• If Y is a closed subset of X, Y is also compact.
• The union of two compact subsets of a metric space is compact.
• Every compact metric space has the Bolzano-Weierstrass Property (BWP). A metric space is said to have Bolzano-Weierstrass Property if every infinite subset of X has a limit point. That is, every sequence within that infinite subsets converges to a point in it.
• A compact subset of a metric space is closed and bounded.

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Solved Examples on Compactness

Example 1:

Prove that the usual metric space (R, d) is not compact.

Solution:

We have to prove that the usual metric space (R, d) is not compact, where R is the set of real numbers.

Consider C = {(-n, n) | n ∈ N}. Then C is an open cover of R as

(i) each element in C is an open set.

(ii) if x ∈ R, then there exist n such that x ∈ (-n, n), so

{x} ⊆ (-n, n), n ∈ N

⇒ ∪ {x} ⊆ ∪ (-n, n)

⇒ R ⊆ ∪ (-n, n)

Again, (-n, n) ⊆ R, ∀ n ∈ N so ∪ (-n, n) ⊆ R

Hence R = ∪ (-n, n)

Now to prove (R, d) is not we shall show that there does not exist any finite subcover of R.

Let {(-n_i, n_i) | 1 ≤ i ≤ p} be a finite sub-collection of C and let n’ = max {n₁, n₂, …, n_p}

Then n’ ∈ N ⊆ R be such that

n’ ∉ ⋃ (-n_i, n_i)

Thus, there is no finite subcollection of C which covers R.

Hence, R is not compact.

Example 2:

Check whether the usual metric space (X, d) where X = {1, 2, 3, …, N} is compact.

Solution:

Given metric space (X, d) where X = {1, 2, 3, …, N} then X is finite.

A finite set is compact.

Thus, (X, d) is a compact space.