# 3.4. Sequences#

Let $$(X, d)$$ be a metric space.

## 3.4.1. Sequences#

Definition 3.32

A sequence in a metric space $$(X,d)$$ is a function $$x : \Nat \to X$$. It maps every natural number to an element in the set $$X$$.

If $$A$$ is a subset of $$X$$, then a sequence of $$A$$ means a function $$x: \Nat \to A$$.

A sequence can be thought of as an ordered (countable) list of points in $$X$$. We often write a sequence $$x : \Nat \to X$$ as $$\{ x_n \}$$ which means the list $$\{x_1, x_2, x_3, \dots \}$$ where $$x_n = x(n)$$. In this sense, $$x_n$$ denotes the $$n$$-th entry in the sequence.

1. A sequence need not enumerate all elements of $$X$$.

2. An element may be repeated multiple times in a sequence.

## 3.4.2. Convergence#

Definition 3.33 (Convergence)

A sequence $$\{ x_n \}$$ in a metric space $$(X,d)$$ is said to converge to $$x \in X$$ if for every $$\epsilon > 0$$, there exists a natural number $$n_0$$ (depending upon $$\epsilon$$) such that

$d (x_n,x) < \epsilon \Forall n > n_0.$

The point $$x$$ is called the limit of the sequence $$\{ x_n \}$$, and we write $$x_n \to x$$ or $$x = \lim x_n$$.

In other words, $$x_n \in B(x, \epsilon)$$ for all $$n > n_0$$.

Remark 3.5

The sequence $$\{x_n\}$$ gives a sequence of real numbers $$\{ y_n \}$$ where $$y_n = d(x_n, x)$$.

Thus, $$\{ x_n \}$$ converges if $$\lim d(x_n, x) = 0$$.

### 3.4.2.1. Limit Uniqueness#

Theorem 3.30 (Sequence Limit Uniqueness)

A sequence of points can have utmost one limit.

Proof. If a sequence doesn’t converge, then there is nothing to prove. Otherwise, suppose a sequence $$\{ x_n \}$$ converges to two limits $$x$$ and $$y$$. Thus, for every $$\epsilon > 0$$, there exist $$n_1, n_2 \in \Nat$$ such that $$d(x_n,x) < \epsilon \Forall n > n_1$$ and $$d(x_n, y) < \epsilon \Forall n > n_2$$. Now, choose $$n_0 = \max (n_1, n_2)$$. Then, by triangle inequality, for every $$n > n_0$$

$0 \leq d(x, y) \leq d(x, x_n) + d(x_n, y) < \epsilon + \epsilon = 2\epsilon.$

Since this is true for all $$\epsilon > 0$$, hence $$d(x,y)=0$$. This means that $$x = y$$ (Identity of indiscernibles).

This result and proof is adapted from Theorem 2.4.

### 3.4.2.2. Choice of Metric#

Let $$X$$ be a set and $$d_1$$ and $$d_2$$ be two different metrics defined on $$X$$. A sequence which converges in the metric space $$(X, d_1)$$ may not converge in the metric space $$(X, d_2)$$ unless the metrics are equivalent.

Example 3.11 (Sequence convergence on different metrics)

Consider the space of all continuous real valued functions defined on the interval $$[0,1]$$ denoted by $$C[0,1]$$.

We introduce two different metrics on $$C[0,1]$$.

1. The metric $$d_1$$ is defined as

$d_1(f, g) = \int_0^1 |f(x) - g(x)| d x.$
2. The metric $$d_{\infty}$$ is defined as

$d_{\infty}(f, g) = \sup_{x \in [0,1]} \{|f(x) - g(x)| \}.$

We now introduce a sequence which converges in $$(C[0,1], d_1)$$ but doesn’t in $$(C[0,1], d_{\infty})$$.

1. Consider the sequence of functions $$\{ f_n \}$$ where $$f_n(x) = e^{-n x}$$.

2. It is clear that $$f_n \in C[0,1]$$.

3. The sequence of functions converges point-wise to a function $$f_p$$ given by

$\begin{split} f_p(x) = \begin{cases} 0 & x \neq 0 ; \\ 1 & x = 0. \end{cases} \end{split}$
4. As we can see that $$f_p \notin C[0,1]$$ as the function is not continuous (from the right) at $$x=0$$.

5. We introduce the zero function $$\bzero : [0,1] \to \RR$$ defined as

$\bzero(x) = 0 \quad \Forall x \in [0,1].$
6. Clearly $$\bzero \in C[0,1]$$.

7. Note that

$d_1(f_n, \bzero) = \int_0^1 |f_n(x) - 0| d x = \int_0^1 e^{-n x} d x = \frac{1 - e^{-n}}{n}.$
8. Clearly $$d_1(f_n, \bzero) \to 0$$ as $$n \to \infty$$.

9. Hence $$\{ f_n \}$$ converges to $$\bzero$$ in $$(X, d_1)$$.

We now show that $$\{ f_n \}$$ doesn’t converge to any function in $$(C[0,1], d_{\infty})$$.

1. We first note that

$d_{\infty} (f_n, \bzero) = \sup_{x \in [0, 1]}\{ |f_n(x) - 0 | \} = \sup_{x \in [0, 1]}\{ e^{-n x} \} = 1.$
2. Hence $$\{ f_n \}$$ doesn’t converge to $$\bzero$$.

3. For contradiction, let $$f$$ be the limit of $$\{ f_n \}$$.

4. Since $$f$$ is not identically zero and is continuous, hence it must be non-zero throughout an open interval $$(c,d)$$.

5. Thus $$|f(x)| \geq M \Forall x \in (c,d)$$ for some $$M > 0$$.

6. Since for any fixed $$x \in (c,d)$$, $$\lim e^{-n x} = 0$$, hence there exists $$n_1 \in \Nat$$ such that $$|e^{-n x}| < M$$ for every $$x \in (c, d)$$.

7. Now,

$|f(x) - e^{-n x}| \geq | |f(x)| - |e^{-nx}| |.$
8. For $$n > n_1$$, we get

$|f(x) - e^{-n x}| \geq |f(x)| - e^{-nx}.$
9. Therefore

$\sup_{x \in (c,d)} \{ |f(x) - e^{-n x}| \} \geq M - e^{- n c} \Forall n > n_1.$
10. Taking the limit

$\lim_{n \to \infty}\sup_{x \in (c,d)} \{ |f(x) - e^{-n x}| \} \geq M.$
11. Therefore

$\lim_{n \to \infty} d_{\infty} (f_n , f) = \lim_{n \to \infty}\sup_{x \in (c,d)} \{ |f(x) - f_n(x) | \} \geq M.$
12. We arrive at a contradiction as the distance doesn’t approach the limit to 0.

13. Thus the sequence $$\{ f_n \}$$ doesn’t converge to any function $$f \in C[0, 1]$$.

### 3.4.2.3. Closure Points#

Theorem 3.31 (Characterization of closure points as limits)

A point $$x \in X$$ is a closure point of $$A \subseteq X$$ if and only if there is a sequence $$\{ x_n \}$$ of $$A$$ such that $$\lim x_n = x$$.

Proof. Assume $$x$$ is a closure point of $$A$$. We construct a sequence which converges to $$x$$.

1. For each $$n \in \Nat$$, we can pick a point $$x_n \in A$$ such that $$d(x, x_n) < \frac{1}{n}$$. This is possible since $$B(x, \frac{1}{n}) \cap A \neq \EmptySet$$ for every $$n \in \Nat$$.

2. Form the sequence $$\{ x_n \}$$.

3. Since $$\lim d(x, x_n) = 0$$, hence $$\{x_n \}$$ converges to $$x$$.

Assume a sequence $$\{x_n\}$$ of $$A$$ converges to $$x$$.

1. For each $$r > 0$$, there exists some $$k$$ such that $$d(x, x_n) < r$$ for all $$n > k$$.

2. Thus, $$B(x, r) \cap A \neq \EmptySet$$ for every $$r > 0$$.

3. Thus, $$x$$ is a closure point of $$A$$.

If $$x \in A$$, we can simply pick the constant sequence $$\{ x_n = x \}$$. It’s more challenging only when $$x \in \closure A \setminus A$$.

### 3.4.2.4. Accumulation Points#

Theorem 3.32 (Accumulation point and distinct sequences)

Let $$x$$ be an accumulation point of $$A$$. Then, there exists a sequence $$\{x_n \}$$ of $$A$$ with distinct terms, that converges to $$x$$.

Proof. We assume that $$x$$ is an accumulation point of $$A$$.

1. For every $$r> 0$$, $$B(x,r) \cap (A \setminus \{ x \}) \neq \EmptySet$$.

2. Let $$r=1$$. We can pick $$x_1 \in A$$ distinct from $$x$$ from the set $$B(x,1) \cap (A \setminus \{ x \})$$ which is not empty.

3. Assume inductively that distinct $$x_1, x_2, \dots, x_n$$ have been chosen from $$A$$ (all different from $$x$$).

4. Let $$r = \min \{\frac{1}{n+1}, d(x, x_n) \}$$.

5. Pick $$x_{n+1}$$ from the set $$B(x,r) \cap (A \setminus \{ x \})$$.

6. By construction, $$x_{n+1}$$ is distinct from previously chosen points.

7. By induction, we can construct a sequence $$\{ x_n \}$$ such that each term in the sequence is distinct and $$\lim d(x, x_n)=0$$.

8. Thus, the sequence converges to $$x$$.

### 3.4.2.5. Closedness#

Theorem 3.33 (Closedness = Convergence of sequences)

Let $$A$$ be a subset of $$X$$. $$A$$ is closed if and only if every convergent sequence of $$A$$ converges in $$A$$.

Proof. Assume $$A$$ to be closed.

1. Let $$\{x_n\}$$ be a convergent sequence of $$A$$.

2. By Theorem 3.31 $$x = \lim x_n$$ is a closure point of $$A$$.

3. Since $$A$$ is closed, hence it contains all its closure points.

4. Thus, $$\{x_n\}$$ converges in $$A$$.

Assume that every convergent sequence of $$A$$ converges in $$A$$.

1. Let $$x$$ be a closure point of $$A$$.

2. By Theorem 3.31, there exists a convergent sequence $$\{ x_n \}$$ of $$A$$ that converges to $$x$$.

3. But since, $$\{ x_n \}$$ converges in $$A$$, hence $$x \in A$$.

4. Thus, $$A$$ contains all its closure points.

5. Hence, $$A$$ is closed.

### 3.4.2.6. Distance between Sequences#

Theorem 3.34 (Sequence distance in the limit)

If $$\lim x_n = x$$ and $$\lim y_n = y$$, then

$\lim_{n \to \infty} d(x_n, y_n) = d(x, y).$

Proof. Recall from the triangle inequality:

$| d(x,z) - d(z, y) | \leq d(x,y).$

Now

\begin{split} \begin{aligned} | d(x_n, y_n) - d(x, y) | &\leq | d(x_n, y_n) - d(x, y_n) | + | d(x, y_n) - d (x, y) |\\ &\leq d (x_n, x) + d(y_n, y). \end{aligned} \end{split}

Choose $$n_0$$ such that for all $$n > n_0$$,

$d(x_n, x) < \frac{\epsilon}{2} \text { and } d(y_n, y) < \frac{\epsilon}{2}.$

Then $$| d(x_n, y_n) - d(x, y) | < \epsilon$$.

Thus, for every $$\epsilon > 0$$, there exists $$n_0$$ such that for all $$n > n_0$$, $$| d(x_n, y_n) - d(x, y) | < \epsilon$$ holds. Thus,

$\lim_{n \to \infty} d(x_n, y_n) = d(x, y).$

## 3.4.3. Cauchy Sequences#

One issue of working with convergence of a sequence is that one has to know the limit of convergence to show that the sequence indeed converges to the limit. This may become problematic. We should have some way to qualify sequences in which the points are coming closer to a limit as the sequence progresses without knowing the limit. One way is to check if all the points of the sequence beyond a certain point are close enough to each other. Sequences in which the points come closer to each other at higher indices are known as Cauchy sequences. However it is not necessary that every Cauchy sequence is convergent. Yet such sequences are of great relevance.

Definition 3.34 (Cauchy sequence)

A sequence $$\{x_n\}$$ of $$X$$ is called a Cauchy sequence if for every $$\epsilon > 0$$, there exists $$n_0$$ (depending on $$\epsilon$$) such that

$d(x_n, x_m) < \epsilon \text{ for every } n, m > n_0.$

Proposition 3.19

Every convergent sequence is a Cauchy sequence.

Proof. Let $$\{x_n\}$$ be a convergent with the limit $$\lim x_n = x$$. Thus, for every $$\epsilon > 0$$, there exists $$n_0$$ such that $$d(x_n, x) < \epsilon / 2$$ for all $$n > n_0$$.

Then, for all $$m, n > n_0$$

$d(x_n, x_m) \leq d(x_n, x) + d(x, x_m) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

Thus, $$\{ x_n\}$$ is a Cauchy sequence.

Example 3.12 (Not every Cauchy sequence is convergent)

Consider $$X = \RR_{++}$$ with metric $$d(x,y) = |x-y|$$. Consider the sequence $$x_n = \frac{1}{n}$$. The sequence doesn’t converge in $$X$$ (Its limit 0 doesn’t belong to $$X$$). At the same time, $$\{ x_n \}$$ is Cauchy.

1. Let $$\epsilon > 0$$.

2. Let $$n_0$$ be any natural number larger than $$\frac{2}{\epsilon}$$.

3. Note that $$n_0 > \frac{2}{\epsilon} \iff \epsilon > \frac{2}{n_0}$$.

4. Then, for $$m, n > n_0$$:

$d(x_n, x_m) = |x_n - x_m | \leq x_n + x_m = \frac{1}{n} + \frac{1}{m} < \frac{1}{n_0} + \frac{1}{n_0} = \frac{2}{n_0} < \epsilon.$
5. Thus, for every $$\epsilon > 0$$, there exists $$n_0$$ such that for all $$m, n > n_0$$, $$d(x_n, x_m) < \epsilon$$.

6. Hence $$\{ x_n\}$$ is Cauchy.

## 3.4.4. Subsequences#

Recall from Definition 1.95 that a subsequence of a sequence $$\{ x_n \}$$ is a sequence $$\{ y_n \}$$ for which there exists a strictly increasing sequence $$\{ k_n \}$$ of natural numbers (i.e. $$1 \leq k_1 < k_2 < k_3 < \ldots)$$ such that $$y_n = x_{k_n}$$ holds for each $$n$$.

A natural question that arises is that if a subsequence converges then does the sequence also converge. Alternatively, if a sequence converges then do all of its subsequences converge?

It turns out that if a subsequence converges then it is not necessary that the sequence itself will converge. However, if a sequence converges, then all its subsequences converge to the same limit.

Theorem 3.35 (Subsequence convergence)

Subsequences of a convergent sequence converge to the same limit as the original sequence. If $$\lim_{n \to \infty} x_n = x$$, then $$\lim_{n \to \infty} y_n = x$$ for every subsequence $$\{ y_n \}$$ of $$\{ x_n \}$$.

Conversely, if two different subsequences of $$\{ x_n \}$$ converge to different limits, then the sequence $$\{ x_n \}$$ does not converge.

This result is a generalization of Theorem 2.16 for metric spaces.

Proof. Let $$\{x_n\}$$ be a convergent sequence of $$X$$ and Let $$\{ y_n\}$$ be a subsequence of $$\{ x_n\}$$.

1. Since $$\lim_{n \to \infty} x_n = x$$, for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that $$d (x, x_n) < \epsilon \Forall n > n_0$$.

2. Since $$\{ y_n \}$$ is a subsequence, there exists a strictly increasing sequence $$\{ k_n \}$$ of natural numbers (i.e. $$1 \leq k_1 < k_2 < k_3 < \ldots)$$ such that $$y_n = x_{k_n}$$ holds for each $$n$$.

3. Thus, there exists a $$k_0 > 0$$ such that $$k_n \geq n_0 \Forall n > k_0$$. Then,

4. $$d(x,y_n) < \epsilon \Forall n > k_0$$.

5. Thus, $$\{ y_n \}$$ converges to $$x$$ too.

## 3.4.5. Dense Sets#

Theorem 3.36

A subset $$A$$ is dense in $$X$$ if and only if for every $$x \in X$$, there exists a sequence $$\{ x_n \}$$ of $$A$$ such that $$\lim x_n = x$$.

This is a direct application of Theorem 3.31.

## 3.4.6. Equivalent Metrics#

Theorem 3.37 (Metric equivalence and convergent sequences)

Let $$d_a$$ and $$d_b$$ be two different metrics on the same set $$X$$. Then, $$d_a$$ and $$d_b$$ are equivalent if and only if they lead to identical set of convergent sequences; i.e., a sequence is convergent in $$(X, d_a)$$ if and only if it is also convergent in $$(X, d_b)$$ and it has same limit in both metric spaces.

In other words, a sequence $$\{ x_n \}$$ of $$X$$ satisfies $$\lim d_a(x_n, x) = 0$$ if and only if $$\lim d_b(x_n, x) = 0$$.

Proof. Assume that the two metric spaces $$(X, d_a)$$ and $$(X, d_b)$$ have same topology. Thus, they have same open sets.

1. Let $$\{x_n \}$$ be a convergent sequence of $$(X, d_a)$$ converging to $$x$$.

2. Now, let $$\epsilon > 0$$ be arbitrary and consider the open ball $$B_b(x, \epsilon)$$.

3. By Theorem 3.26, there exists an $$r > 0$$ such that $$B_a(x, r) \subseteq B_b(x, \epsilon)$$.

4. Since $$\{ x_n \}$$ is convergent in $$(X, d_a)$$, hence there exists $$n_0 \in \Nat$$ such that $$x_n \in B_a(x, r) \subseteq B_b(x, \epsilon)$$ for all $$n > n_0$$.

5. Thus, for every $$\epsilon > 0$$, there exists $$n_0 \in \Nat$$ such that $$x_n \in B_b(x, \epsilon)$$ for all $$n > n_0$$.

6. Thus, $$\{x_n \}$$ is convergent in $$(X, d_b)$$ with limit $$x$$.

7. Similar reasoning shows that if a sequence is convergent in $$(X, d_b)$$ then it is convergent in $$(X, d_a)$$ too.

8. Thus, the convergent sequences in both metric spaces are identical and have same limits.

Now, assume that the convergent sequences in both metric spaces $$(X, d_a)$$ and $$(X, d_b)$$ are identical and have same limits.

1. Let $$C$$ be a closed set of $$(X, d_a)$$.

2. Let $$x \in C$$. Then, $$x$$ is a closure point of $$C$$ in $$(X, d_a)$$.

3. Then, there exists a sequence $$\{ x_n \}$$ of $$C$$ such that $$\lim x_n = x$$ in $$(X, d_a)$$.

4. But by our hypothesis, convergent sequences are identical in both metric spaces.

5. Hence $$\lim x_n = x$$ in $$(X, d_b)$$ also.

6. Hence, $$x$$ is a closure point of $$C$$ in $$(X, d_b)$$ too.

7. Thus, every element of $$C$$ is a closure point of $$C$$ in $$(X, d_b)$$.

8. Thus, $$C$$ is closed in $$(X, d_b)$$.

9. A similar argument shows that if $$C$$ is closed in $$(X, d_b)$$ then it is closed in $$(X, d_a)$$ too.

10. Thus, both metrics determine the same set of closed sets on $$X$$.

11. Thus, both metrics determine the same set of open sets on $$X$$.

12. Thus, they determine the same topology.

13. Thus, the two metrics are equivalent.

Procedure to show that two metrics are equivalent.

• Choose an arbitrary sequence $$\{x_n\}$$ which converges in $$(X, d_a)$$ to a limit (say $$x$$).

• Show that $$\lim d_b(x_n, x) = 0$$.

• Now, choose an arbitrary sequence $$\{x_n\}$$ which converges in $$(X, d_b)$$ to a limit (say $$x$$).

• Show that $$\lim d_a(x_n, x) = 0$$.