8.3: Sequences and Convergence

Sequences

The notion of a sequence in a metric space is very similar to a sequence of real numbers.

A sequence in a metric space $(X,d)$ is a function $x \colon {\mathbb{N}}\to X$. As before we write $x_n$ for the $n$th element in the sequence and use the notation $\{ x_n \}$, or more precisely $$\{ x_n \}_{n=1}^\infty .$$

A sequence $\{ x_n \}$ is bounded if there exists a point $p \in X$ and $B \in {\mathbb{R}}$ such that $$d(p,x_n) \leq B \qquad \text{for all $n \in {\mathbb{N}}$.}$$ In other words, the sequence $\{x_n\}$ is bounded whenever the set $\{ x_n : n \in {\mathbb{N}}\}$ is bounded.

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

Similarly we also define convergence. Again, we will be cheating a little bit and we will use the definite article in front of the word limit before we prove that the limit is unique.

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

A sequence that converges is said to be convergent. Otherwise, the sequence is said to be divergent.

Let us prove that the limit is unique. Note that the proof is almost identical to the proof of the same fact for sequences of real numbers. In fact many results we know for sequences of real numbers can be proved in the more general settings of metric spaces. We must replace $\left\lvert {x-y} \right\rvert$ with $d(x,y)$ in the proofs and apply the triangle inequality correctly.

[prop:mslimisunique] A convergent sequence in a metric space has a unique limit.

Suppose that the sequence $\{ x_n \}$ has the limit $x$ and the limit $y$. Take an arbitrary $\epsilon > 0$. From the definition we find an $M_1$ such that for all $n \geq M_1$, $d(x_n,x) < \nicefrac{\epsilon}{2}$. Similarly we find an $M_2$ such that for all $n \geq M_2$ we have $d(x_n,y) < \nicefrac{\epsilon}{2}$. Now take an $n$ such that $n \geq M_1$ and also $n \geq M_2$ $$\begin{split} d(y,x) & \leq d(y,x_n) + d(x_n,x) \\ & < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . \end{split}$$ As $d(y,x) < \epsilon$ for all $\epsilon > 0$, then $d(x,y) = 0$ and $y=x$. Hence the limit (if it exists) is unique.

The proofs of the following propositions are left as exercises.

[prop:msconvbound] A convergent sequence in a metric space is bounded.

[prop:msconvifa] A sequence $\{ x_n \}$ in a metric space $(X,d)$ converges to $p \in X$ if and only if there exists a sequence $\{ a_n \}$ of real numbers such that $$d(x_n,p) \leq a_n \quad \text{for all $n \in {\mathbb{N}}$},$$ and $$\lim_{n\to\infty} a_n = 0.$$

Convergence in euclidean space

It is useful to note what convergence means in the euclidean space ${\mathbb{R}}^n$.

[prop:msconveuc] Let $\{ x^j \}_{j=1}^\infty$ be a sequence in ${\mathbb{R}}^n$, where we write $x^j = \bigl(x_1^j,x_2^j,\ldots,x_n^j\bigr) \in {\mathbb{R}}^n$. Then $\{ x^j \}_{j=1}^\infty$ converges if and only if $\{ x_k^j \}_{j=1}^\infty$ converges for every $k$, in which case $$\lim_{j\to\infty} x^j = \Bigl( \lim_{j\to\infty} x_1^j, \lim_{j\to\infty} x_2^j, \ldots, \lim_{j\to\infty} x_n^j \Bigr) .$$

For ${\mathbb{R}}= {\mathbb{R}}^1$ the result is immediate. So let $n > 1$.

Let $\{ x^j \}_{j=1}^\infty$ be a convergent sequence in ${\mathbb{R}}^n$, where we write $x^j = \bigl(x_1^j,x_2^j,\ldots,x_n^j\bigr) \in {\mathbb{R}}^n$. Let $x = (x_1,x_2,\ldots,x_n) \in {\mathbb{R}}^n$ be the limit. Given $\epsilon > 0$, there exists an $M$ such that for all $j \geq M$ we have $$d(x,x^j) < \epsilon.$$ Fix some $k=1,2,\ldots,n$. For $j \geq M$ we have $$\bigl\lvert x_k - x_k^j \bigr\rvert = \sqrt{{\bigl(x_k - x_k^j\bigr)}^2} \leq \sqrt{\sum_{\ell=1}^n {\bigl(x_\ell-x_\ell^j\bigr)}^2} = d(x,x^j) < \epsilon .$$ Hence the sequence $\{ x_k^j \}_{j=1}^\infty$ converges to $x_k$.

For the other direction suppose that $\{ x_k^j \}_{j=1}^\infty$ converges to $x_k$ for every $k=1,2,\ldots,n$. Hence, given $\epsilon > 0$, pick an $M$, such that if $j \geq M$ then $\bigl\lvert x_k-x_k^j \bigr\rvert < \nicefrac{\epsilon}{\sqrt{n}}$ for all $k=1,2,\ldots,n$. Then \[d(x,x^j) = \sqrt{\sum_{k=1}^n {\bigl(x_k-x_k^j\bigr)}^2} < \sqrt{\sum_{k=1}^n {\left(\frac{\epsilon}{\sqrt{n}}\right)}^2} = \sqrt{\sum_{k=1}^n \frac

ParseError: invalid DekiScript (click for details)

Callstack:
    at (Bookshelves/Analysis/Introduction_to_Real_Analysis_(Lebl)/08:_Metric_Spaces/8.03:_Sequences_and_Convergence), /content/body/div[2]/p[5]/span, line 1, column 1

{n}} = \epsilon .\] The sequence $\{ x^j \}$ converges to $x \in {\mathbb{R}}^n$ and we are done.

Convergence and topology

The topology, that is, the set of open sets of a space encodes which sequences converge.

[prop:msconvtopo] Let $(X,d)$ be a metric space and $\{x_n\}$ a sequence in $X$. Then $\{ x_n \}$ converges to $x \in X$ if and only if for every open neighborhood $U$ of $x$, there exists an $M \in {\mathbb{N}}$ such that for all $n \geq M$ we have $x_n \in U$.

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

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

A set is closed when it contains the limits of its convergent sequences.

[prop:msclosedlim] Let $(X,d)$ be a metric space, $E \subset X$ a closed set and $\{ x_n \}$ a sequence in $E$ that converges to some $x \in X$. Then $x \in E$.

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

When we take a closure of a set $A$, we really throw in precisely those points that are limits of sequences in $A$.

[prop:msclosureapprseq] Let $(X,d)$ be a metric space and $A \subset X$. If $x \in \overline{A}$, then there exists a sequence $\{ x_n \}$ of elements in $A$ such that $\lim\, x_n = x$.

Let $x \in \overline{A}$. We know by that given $\nicefrac{1}{n}$, there exists a point $x_n \in B(x,\nicefrac{1}{n}) \cap A$. As $d(x,x_n) < \nicefrac{1}{n}$, we have that $\lim\, x_n = x$.

Exercises

Let $(X,d)$ be a metric space and let $A \subset X$. Let $E$ be the set of all $x \in X$ such that there exists a sequence $\{ x_n \}$ in $A$ that converges to $x$. Show that $E = \overline{A}$.

a) Show that $d(x,y) := \min \{ 1, \left\lvert {x-y} \right\rvert \}$ defines a metric on ${\mathbb{R}}$. b) Show that a sequence converges in $({\mathbb{R}},d)$ if and only if it converges in the standard metric. c) Find a bounded sequence in $({\mathbb{R}},d)$ that contains no convergent subsequence.

Prove

Prove

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

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

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

Suppose $\{ U_n \}_{n=1}^\infty$ be 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 $p \in X$. Suppose that $\{ x_n \}$ is a sequence of points in $X$ such that $x_n \in U_n$. Does $\{ x_n \}$ necessarily converge to $p$? Prove or construct a counterexample.

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

Take ${\mathbb{R}}^* = \{ -\infty \} \cup {\mathbb{R}}\cup \{ \infty \}$ be the extended reals. Define $d(x,y) := \frac{\left\lvert {x-y} \right\rvert}{1+\left\lvert {x-y} \right\rvert}$ if $x, y \in {\mathbb{R}}$, define $d(\infty,x) := d(-\infty,x) = 1$ for all $x \in {\mathbb{R}}$, and let $d(\infty,-\infty) := 2$. a) Show that $({\mathbb{R}}^*,d)$ is a metric space. b) Suppose that $\{ x_n \}$ is a sequence of real numbers such that for $x_n \geq n$ for all $n$. Show that $\lim x_n = \infty$ in $({\mathbb{R}}^*,d)$.

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

Contributors and Attributions

• Template:ContribLebl