## Section7.3Sequences and convergence

Note: 1 lecture

### Subsection7.3.1Sequences

The notion of a sequence in a metric space is very similar to a sequence of real numbers. The related definitions are essentially the same as those for real numbers in the sense of Chapter 2, where $$\R$$ with the standard metric $$d(x,y)=\abs{x-y}$$ is replaced by an arbitrary metric space $$(X,d)\text{.}$$

#### Definition7.3.1.

A sequence in a metric space $$(X,d)$$ is a function $$x \colon \N \to X\text{.}$$ As before we write $$x_n$$ for the $$n$$th element in the sequence and use the notation $$\{ x_n \}\text{,}$$ or more precisely

\begin{equation*} \{ x_n \}_{n=1}^\infty . \end{equation*}

A sequence $$\{ x_n \}$$ is bounded if there exists a point $$p \in X$$ and $$B \in \R$$ such that

\begin{equation*} d(p,x_n) \leq B \qquad \text{for all } n \in \N. \end{equation*}

In other words, the sequence $$\{x_n\}$$ is bounded whenever the set $$\{ x_n : n \in \N \}$$ is bounded.

If $$\{ n_j \}_{j=1}^\infty$$ is a sequence of natural numbers such that $$n_{j+1} > n_j$$ for all $$j\text{,}$$ then the sequence $$\{ x_{n_j} \}_{j=1}^\infty$$ is said to be a subsequence of $$\{x_n \}\text{.}$$

Similarly we define convergence. Again, we cheat a little and use the definite article in front of the word limit before we prove that the limit is unique. See Figure 7.9, for an idea of the definition.

#### Definition7.3.2.

A sequence $$\{ x_n \}$$ in a metric space $$(X,d)$$ is said to converge to a point $$p \in X$$ if for every $$\epsilon > 0\text{,}$$ there exists an $$M \in \N$$ such that $$d(x_n,p) < \epsilon$$ for all $$n \geq M\text{.}$$ The point $$p$$ is said to be the limit of $$\{ x_n \}\text{.}$$ We write

\begin{equation*} \lim_{n\to \infty} x_n := p . \end{equation*}

A sequence that converges is convergent. Otherwise, the sequence is divergent.

Let us prove that the limit is unique. The proof is almost identical (word for word) to the proof of the same fact for sequences of real numbers, Proposition 2.1.6. Proofs of many results we know for sequences of real numbers can be adapted to the more general settings of metric spaces. We must replace $$\abs{x-y}$$ with $$d(x,y)$$ in the proofs and apply the triangle inequality correctly.

#### Proof.

Suppose the sequence $$\{ x_n \}$$ has limits $$x$$ and $$y\text{.}$$ Take an arbitrary $$\epsilon > 0\text{.}$$ From the definition find an $$M_1$$ such that for all $$n \geq M_1\text{,}$$ $$d(x_n,x) < \nicefrac{\epsilon}{2}\text{.}$$ Similarly find an $$M_2$$ such that for all $$n \geq M_2\text{,}$$ we have $$d(x_n,y) < \nicefrac{\epsilon}{2}\text{.}$$ Now take an $$n$$ such that $$n \geq M_1$$ and also $$n \geq M_2\text{,}$$ and estimate

\begin{equation*} \begin{split} d(y,x) & \leq d(y,x_n) + d(x_n,x) \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{split} \end{equation*}

As $$d(y,x) < \epsilon$$ for all $$\epsilon > 0\text{,}$$ then $$d(x,y) = 0$$ and $$y=x\text{.}$$ Hence the limit (if it exists) is unique.

The proofs of the following propositions are left as exercises.

#### Example7.3.7.

Take $$C([0,1],\R)$$ be the set of continuous functions with the metric being the uniform metric. We saw that we obtain a metric space. If we look at the definition of convergence, we notice that it is identical to uniform convergence. That is, $$\{ f_n \}$$ converges uniformly if and only if it converges in the metric space sense.

#### Remark7.3.8.

It is perhaps surprising that on the set of functions $$f \colon [a,b] \to \R$$ (continuous or not) there is no metric that gives pointwise convergence. Although the proof of this fact is beyond the scope of this book.

### Subsection7.3.2Convergence in euclidean space

In the euclidean space $$\R^n\text{,}$$ a sequence converges if and only if every component converges:

#### Proof.

Suppose the sequence $$\{ x_j \}_{j=1}^\infty$$ converges to $$y = (y_1,y_2,\ldots,y_n) \in \R^n\text{.}$$ Given $$\epsilon > 0\text{,}$$ there exists an $$M\text{,}$$ such that for all $$j \geq M\text{,}$$ we have

\begin{equation*} d(y,x_j) < \epsilon. \end{equation*}

Fix some $$k=1,2,\ldots,n\text{.}$$ For all $$j \geq M\text{,}$$

\begin{equation*} \bigl\lvert y_k - x_{j,k} \bigr\rvert = \sqrt{{\bigl(y_k - x_{j,k} \bigr)}^2} \leq \sqrt{\sum_{\ell=1}^n {\bigl(y_\ell-x_{j,\ell}\bigr)}^2} = d(y,x_j) < \epsilon . \end{equation*}

Hence the sequence $$\{ x_{j,k} \}_{j=1}^\infty$$ converges to $$y_k\text{.}$$

For the other direction, suppose $$\{ x_{j,k} \}_{j=1}^\infty$$ converges to $$y_k$$ for every $$k=1,2,\ldots,n\text{.}$$ Given $$\epsilon > 0\text{,}$$ pick an $$M\text{,}$$ such that if $$j \geq M\text{,}$$ then $$\bigl\lvert y_k-x_{j,k} \bigr\rvert < \nicefrac{\epsilon}{\sqrt{n}}$$ for all $$k=1,2,\ldots,n\text{.}$$ Then

\begin{equation*} d(y,x_j) = \sqrt{\sum_{k=1}^n {\bigl(y_k-x_{j,k}\bigr)}^2} < \sqrt{\sum_{k=1}^n {\left(\frac{\epsilon}{\sqrt{n}}\right)}^2} = \sqrt{\sum_{k=1}^n \frac{{\epsilon^2}}{n}} = \epsilon . \end{equation*}

That is, the sequence $$\{ x_j \}$$ converges to $$y = (y_1,y_2,\ldots,y_n) \in \R^n\text{.}$$

#### Example7.3.10.

As we said, the set $$\C$$ of complex numbers $$z = x+iy$$ is considered as the metric space $$\R^2\text{.}$$ The proposition says that the sequence $$\{ z_j \}_{j=1}^\infty = \{ x_j + iy_j \}_{j=1}^\infty$$ converges to $$z = x+iy$$ if and only if $$\{ x_j \}$$ converges to $$x$$ and $$\{ y_j \}$$ converges to $$y\text{.}$$

### Subsection7.3.3Convergence and topology

The topology—the set of open sets of a space—encodes which sequences converge.

#### Proof.

Suppose $$\{ x_n \}$$ converges to $$x\text{.}$$ Let $$U$$ be an open neighborhood of $$x\text{,}$$ then there exists an $$\epsilon > 0$$ such that $$B(x,\epsilon) \subset U\text{.}$$ As the sequence converges, find an $$M \in \N$$ such that for all $$n \geq M\text{,}$$ we have $$d(x,x_n) < \epsilon\text{,}$$ or in other words $$x_n \in B(x,\epsilon) \subset U\text{.}$$

Let us prove the other direction. Given $$\epsilon > 0\text{,}$$ let $$U := B(x,\epsilon)$$ be the neighborhood of $$x\text{.}$$ Then there is an $$M \in \N$$ such that for $$n \geq M\text{,}$$ we have $$x_n \in U = B(x,\epsilon)\text{,}$$ or in other words, $$d(x,x_n) < \epsilon\text{.}$$

A closed set contains the limits of its convergent sequences.

#### Proof.

Let us prove the contrapositive. Suppose $$\{ x_n \}$$ is a sequence in $$X$$ that converges to $$x \in E^c\text{.}$$ As $$E^c$$ is open, Proposition 7.3.11 says that there is an $$M$$ such that for all $$n \geq M\text{,}$$ $$x_n \in E^c\text{.}$$ So $$\{ x_n \}$$ is not a sequence in $$E\text{.}$$

To take a closure of a set $$A\text{,}$$ we take $$A\text{,}$$ and we throw in points that are limits of sequences in $$A\text{.}$$

#### Proof.

Let $$x \in \widebar{A}\text{.}$$ For every $$n \in \N\text{,}$$ Proposition 7.2.22 guarantees a point $$x_n \in B(x,\nicefrac{1}{n}) \cap A\text{.}$$ As $$d(x,x_n) < \nicefrac{1}{n}\text{,}$$ we have $$\lim\, x_n = x\text{.}$$

For the other direction, see Exercise 7.3.1.

### Subsection7.3.4Exercises

#### Exercise7.3.1.

Finish the proof of Proposition 7.3.13: Let $$(X,d)$$ be a metric space and $$A \subset X\text{.}$$ Let $$x \in X$$ be such that there exists a sequence $$\{ x_n \}$$ in $$A$$ that converges to $$x\text{.}$$ Prove that $$x \in \widebar{A}\text{.}$$

#### Exercise7.3.2.

1. Show that $$d(x,y) := \min \{ 1, \abs{x-y} \}$$ defines a metric on $$\R\text{.}$$

2. Show that a sequence converges in $$(\R,d)$$ if and only if it converges in the standard metric.

3. Find a bounded sequence in $$(\R,d)$$ that contains no convergent subsequence.

#### Exercise7.3.5.

Suppose $$\{x_n\}_{n=1}^\infty$$ converges to $$x\text{.}$$ Suppose $$f \colon \N \to \N$$ is a one-to-one function. Show that $$\{ x_{f(n)} \}_{n=1}^\infty$$ converges to $$x\text{.}$$

#### Exercise7.3.6.

Let $$(X,d)$$ be a metric space where $$d$$ is the discrete metric. Suppose $$\{ x_n \}$$ is a convergent sequence in $$X\text{.}$$ Show that there exists a $$K \in \N$$ such that for all $$n \geq K\text{,}$$ we have $$x_n = x_K\text{.}$$

#### Exercise7.3.7.

A set $$S \subset X$$ is said to be dense in $$X$$ if $$X \subset \widebar{S}$$ or in other words if for every $$x \in X\text{,}$$ there exists a sequence $$\{ x_n \}$$ in $$S$$ that converges to $$x\text{.}$$ Prove that $$\R^n$$ contains a countable dense subset.

#### Exercise7.3.8.

(Tricky)   Suppose $$\{ U_n \}_{n=1}^\infty$$ is a decreasing ($$U_{n+1} \subset U_n$$ for all $$n$$) sequence of open sets in a metric space $$(X,d)$$ such that $$\bigcap_{n=1}^\infty U_n = \{ p \}$$ for some $$p \in X\text{.}$$ Suppose $$\{ x_n \}$$ is a sequence of points in $$X$$ such that $$x_n \in U_n\text{.}$$ Does $$\{ x_n \}$$ necessarily converge to $$p\text{?}$$ Prove or construct a counterexample.

#### Exercise7.3.9.

Let $$E \subset X$$ be closed and let $$\{ x_n \}$$ be a sequence in $$X$$ converging to $$p \in X\text{.}$$ Suppose $$x_n \in E$$ for infinitely many $$n \in \N\text{.}$$ Show $$p \in E\text{.}$$

#### Exercise7.3.10.

Take $$\R^* = \{ -\infty \} \cup \R \cup \{ \infty \}$$ be the extended reals. Define $$d(x,y) := \bigl\lvert \frac{x}{1+\abs{x}} - \frac{y}{1+\abs{y}} \bigr\rvert$$ if $$x, y \in \R\text{,}$$ define $$d(\infty,x) := \bigl\lvert 1 - \frac{x}{1+\abs{x}} \bigr\rvert\text{,}$$ $$d(-\infty,x) := \bigl\lvert 1 + \frac{x}{1+\abs{x}} \bigr\rvert$$ for all $$x \in \R\text{,}$$ and let $$d(\infty,-\infty) := 2\text{.}$$

1. Show that $$(\R^*,d)$$ is a metric space.

2. Suppose $$\{ x_n \}$$ is a sequence of real numbers such that for every $$M \in \R\text{,}$$ there exists an $$N$$ such that $$x_n \geq M$$ for all $$n \geq N\text{.}$$ Show that $$\lim\, x_n = \infty$$ in $$(\R^*,d)\text{.}$$

3. Show that a sequence of real numbers converges to a real number in $$(\R^*,d)$$ if and only if it converges in $$\R$$ with the standard metric.

#### Exercise7.3.11.

Suppose $$\{ V_n \}_{n=1}^\infty$$ is a sequence of open sets in $$(X,d)$$ such that $$V_{n+1} \supset V_n$$ for all $$n\text{.}$$ Let $$\{ x_n \}$$ be a sequence such that $$x_n \in V_{n+1} \setminus V_n$$ and suppose $$\{ x_n \}$$ converges to $$p \in X\text{.}$$ Show that $$p \in \partial V$$ where $$V = \bigcup_{n=1}^\infty V_n\text{.}$$

#### Exercise7.3.13.

Let $$(X,d)$$ be a metric space and $$\{ x_n \}$$ a sequence in $$X\text{.}$$ Prove that $$\{ x_n \}$$ converges to $$p \in X$$ if and only if every subsequence of $$\{ x_n \}$$ has a subsequence that converges to $$p\text{.}$$

#### Exercise7.3.14.

Consider $$\R^n\text{,}$$ and let $$d$$ be the standard euclidean metric. Let $$d'(x,y) := \sum_{\ell=1}^n \abs{x_\ell-y_\ell}$$ and $$d''(x,y) := \max \{ \abs{x_1-y_1},\abs{x_2-y_2},\cdots,\abs{x_n-y_n} \}\text{.}$$

1. Use Exercise 7.1.6, to show that $$(\R^n,d')$$ and $$(\R^n,d'')$$ are metric spaces.

2. Let $$\{ x_j \}_{j=1}^\infty$$ be a sequence in $$\R^n$$ and $$p \in \R^n\text{.}$$ Prove that the following statements are equivalent:

1. $$\{ x_j \}$$ converges to $$p$$ in $$(\R^n,d)\text{.}$$

2. $$\{ x_j \}$$ converges to $$p$$ in $$(\R^n,d')\text{.}$$

3. $$\{ x_j \}$$ converges to $$p$$ in $$(\R^n,d'')\text{.}$$

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