Completeness is a property of metric spaces. Before discussing the completeness of a metric space, it is essential to know about the Cauchy sequence within a metric space.

Let (X, d) be a metric space with d as a metric, if {x_n} be a sequence in X. Then {x_n} is a Cauchy sequence if and only if for every ε > 0, there exist a positive integer n_o, where the value of n_o depends on ε, that is, n_o = n_o(ε) such that;

∀ m, n ≥ n_o ⇒ d(x_m, x_n) ≤ ε

A significant result is associated with the Cauchy sequences: “every convergent sequence in a metric space is a Cauchy sequence”. However, this result does not ensure the convergence of a Cauchy sequence. Later on, we see that the completeness of a metric space proves this thing.

With this definition of the Cauchy sequence, we shall define the completeness of a metric space.

Definition of Completeness

A metric space (X, d) is a complete metric space if and only if every Cauchy sequence in X converges to a point in X.

We can note the following points:

• A complete metric space suggests that if d(x_m, x_n) → 0 as m, n → ∞, then there exist x_o in X such that d(x_n, x_o) → 0 as n → ∞, that is, the sequence {x_n} converges to x_o in X.
• Every Cauchy sequence in a complete metric space has a convergent subsequence. Since a sequence is convergent if it has a convergent subsequence.

Completeness of Real Metric Space R

To prove this, we must show that every Cauchy sequence in R converges to a point in R. Let us consider {x_n} be a Cauchy sequence in R, we can choose the ε > 0 as ε = 1/2^q, for this ∃ n_q ∈ N such that

m, n ≥ n_q ⇒ d(x_m, x_n) < 1/(2^{q + 1}) ⇒ |x_m – x_n| < 1/(2^{q + 1}) ∀ n, m ≥ n_o and x_m, x_n ∈ X

Then we can have, |x_n – x_nq| < 1/(2^{q + 1})

If I_q is the closed interval such that n_q ∈ I_q

where I_q = [x_nq – 2^–q, x_nq + 2^–q] and since |x_nq – x_{nq + 1}| < 1/(2^{q + 1})

we get, I_{q + 1} ⊂ I_q

The length of I_q = 2/2^q = 1/2^{q – 1} → 0 if q → ∞, that means, |I_q| → 0 as k → ∞.

Hence by the Nested Interval theorem,

“I₁, I₂, I₃, …, I_n, … be a collection of nested closed intervals such that I₁ ⊇ I₂ ⊇ I₃ ⊇ …. ⊇ I_n ⊇ …, then the intersection of this collection is non-empty and equal to a singleton set.”

Thus, the intersection of all I_q consists of exactly one point of R, say s.

So s ∈ I_q ∀ q ∈ N such that

|s – x_nq| < 1/(2^{q + 1})

Hence for all n ≥ n_q, we have

|x_n – s| = |x_n – x_nq – s + x_nq| ≤ |x_n – x_nq | + |s – x_nq| < 1/(2^{q + 1}) + 1/(2^{q + 1}) = 1/2^q

⇒ |x_n – s| < 1/2^q

From the above, it follows that the Cauchy sequence {x_n} converges to the point s in R.

Thus, every Cauchy sequence in R converges to a point in R.

Therefore, R is complete.

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Solved Examples on Completeness

Example 1:

Let X = C[0, 1] be the space of all continuous functions defined on the closed interval [0, 1]. Show that the metric space (X, d), where d(f, g) = Max _{0 ≤ x ≤ 1} |f(x) – g(x)| ∀ f, g ∈ X is complete.

Solution:

Let {f_n(x)} be a Cauchy Sequence in the metric space (X, d). Then for a given ε > 0, there exist n_o ∈ N depending on ε such that

m, n ≥ n_o ⇒ d(f_m, f_n) < ε ∀ m, n ≥ n_o

⇒ Max _{0 ≤ x ≤ 1} |f_m(x) – f_n(x)| < ε ; ∀ m, n ≥ n_o and ∀ f, g ∈ X

⇒ |f_m(x) – f_n(x)| < ε ; ∀ m, n ≥ n_o and ∀ x ∈ [0, 1]

Hence, by Cauchy’s Principle of uniform convergence, “Let {f_n} be sequence of real-valued function over a set E then it will be uniformly convergent on E if and only if for a given ε > 0, ∈ n_o ∈ N such that for n, m ≥ n_o ⇒ |f_m(x) – f_n(x)| < ε; ∀ m, n ≥ n_o and ∀ x ∈ E.”

Therefore, the sequence {f_n} converges to f in X as n → ∞.

Now, fix n and let m → ∞, we have

|f_n(x) – f(x)| < ε ; ∀ n ≥ n_o and ∀ x ∈ [0, 1]

⇒ Max _{0 ≤ x ≤ 1} |f_n(x) – f(x)| < ε ; ∀ n ≥ n_o and ∀ x ∈ [0, 1]

⇒ d(f_n, f) < ε ; ∀ n ≥ n_o

Thus, the Cauchy sequence {f_n} converges to f in metric space (X, d). (X, d) is complete.

Example 2:

A metric space (X, d) with d as a discrete metric is complete.

Solution:

Let {x_n} be a Cauchy sequence in metric space (X, d). Then we can choose ε = 1/2 > 0 accordingly there exist n_o depending on ε such that

For n, m ≥ n_o ⇒ d(x_m, x_n) < ½

⇒ d(x_m, x_n) = 0 ∀ n, m ≥ n_o (since d is a discrete metric)

⇒ x_n = x_{n + 1} = x_{n + 2} = … = x (say)

Therefore, d(x_n, x) = 0 < 1/2 ∀ n ≥ n_o

That means the Cauchy sequence {x_n} converges to a point x in X.

Thus, every Cauchy sequence in (X, d) converges which makes (X, d) a complete metric space.