7.1Complete Metric Spaces

Updated 28 Aug 2026Chapter (PDF)

Before we proceed, we will revisit standard metrics and properties about them.

[7.1.1]Recap#

$\bR^n$ has two common metrics: the Euclidean metric
$d(x, y) = \left[ \sum_{i = 1}^n (x_i - y_i)^2 \right]^{1/2}$
and the square metric $\rho(x, y) = \max_{1 \le i \le n} |x_i - y_i|$ .
By simple geometry, note that $\rho(x, y) \le d(x, y) \le \sqrt{n} \rho(x, y)$ for $x, y \in \bR^n$ . Thus properties of $\rho$ (such as sequence convergence, sequences being Cauchy, the topology generated by the metric) are brought to $d$ immediately by this inequality. Note that the constant $\sqrt{n}$ is the only obstruction, and it is exactly what fails in infinite dimensions (since $\sqrt{n} \to \infty$ as $n \to \infty$ ).
It is easy to show the metric topology generated by $\rho$ coincides perfectly with the product topology on $\bR^n$ . By the inequality from before, the metric topology generated by $d$ also coincides with the product topology on $\bR^n$ .
Note that neither metric generalizes cleanly to infinite products, since both may take the value $\infty$ . Instead, given any metric $d$ on a space $X$ , define the standard bounded metric
$\bar{d}(x, y) = \min\{d(x, y), 1\}.$
This induces the same topology as $d$ , so boundedness is not a topological property; it depends on the choice of metric. Truncating in this way costs nothing topologically and is what makes the infinite constructions below well-defined.
We now generalize the square metric to $\bR^\omega$ . The most direct approach is the uniform metric
$\bar{\rho}(x, y) = \sup_{n \in \bZ_+} \{\bar{d}(x_n, y_n) \},$
replacing $\max$ with $\sup$ and truncating the distances above by $1$ . The topology this induces is the uniform topology. Alternatively, we may define the product metric
$D(x, y) = \sup_{n \in \bZ_+} \{\bar{d}(x_n, y_n) / n\},$
which is the same as the uniform metric, but damps the $n$ th coordinate so that later coordinates are less “important.” Fittingly, it is the metric that produces the product topology on $\bR^\omega$ .
These metrics can be placed on the space $Y^X$ , the set of functions $f: X \to Y$ . Then the product metric sets the mode of convergence to pointwise on this function space, whereas the uniform metric sets the mode to uniform convergence.
In defining metrics on function spaces, consider the space $B(X, Y)$ of bounded functions $f: X \to Y$ . We may define the sup metric
$\rho(f, g) = \sup_{x \in X} \{d(f(x), g(x)) \}.$
Note that we do not need to use the bounded metric $\bar{d}$ , since the functions are already bounded above, and so $\sup$ works out of the box. In general, note that the relationship between the uniform and sup metrics are simply $\bar{\rho}(f, g) = \min\{\rho(f, g), 1\}$ .
Note that the product topology is strictly coarser than the uniform topology, which is strictly coarser than the box topology.

[7.1.2]Definition(Cauchy Sequence, Complete Metric Space)#

Let $(X, d)$ be a metric space. A sequence $(x_n) \subseteq X$ is said to be Cauchy if, for any $\epsilon > 0$ , there is some $N \in \bZ_+$ such that $d(x_m, x_n) < \epsilon$ for every $m, n \ge N$ . We say the metric space $X$ is complete if every Cauchy sequence converges.

[7.1.3]Theorem(Equivalence of Convergent Sequences and Cauchy Sequences in Metric Spaces)#

In a complete metric space $(X, d)$ , a sequence converges if and only if it is Cauchy.

This fact is important in analysis since, in complete metric spaces, we can demonstrate a sequence converges if it is Cauchy, without needing a candidate for the limit.

[7.1.4]Theorem(Closed Subspace of Complete Metric Space is Complete)#

A closed subspace of a complete metric space is complete.

Proof.

Of course, a subspace topology derived from the metric topology is metric. Consider any Cauchy sequence contained in the closed subspace. It necessarily converges to a limit point of the closed subspace, but closed spaces contain all their limit points, and so the Cauchy sequence converges to a point in the closed subspace. Thus the closed subspace is complete.

[7.1.5]Theorem(Complete if Every Cauchy Sequence has Accumulation Point)#

A metric space $(X, d)$ is complete if every Cauchy sequence has a convergent subsequence.

Proof.

Suppose if $(x_n) \subseteq X$ is a Cauchy sequence with a subsequence $x_{n_k} \to x$ . For any $\epsilon > 0$ , note there is some $N_1 \in \bZ_+$ such that $d(x_n, x_{n_k}) < \epsilon/2$ for every $n, n_k \ge N_1$ , since the sequence is Cauchy. Moreover, there is some $N_2 \in \bZ_+$ such that $d(x_{n_k}, x) < \epsilon/2$ for every $n_k \ge N_2$ by definition of convergence. Thus, for any $n \ge \max\{N_1, N_2\}$ , we have that

d(x_n, x) \le d(x_n, x_{n_k}) + d(x_{n_k}, x) < \epsilon/2 + \epsilon/2 = \epsilon.

Since all Cauchy sequences are convergent, the space is complete.

[7.1.6]Theorem(Euclidean Space is Complete in its Standard Metrics)#

Euclidean space is complete under the standard and square metrics.

Proof.

Let $(\bR^n, \rho)$ be Euclidean space endowed with the square metric. For any Cauchy sequence $x_n$ in this metric space, there is some $N \in \bZ_+$ such that $\rho(x_m, x_n) < 1$ for every $m, n \ge N$ . Then choose a number

M = \max\{\rho(x_0, \vb{0}_n), \rho(x_1, \vb{0}_n), \dots, \rho(x_{N - 1}, \vb{0}_n), \rho(x_N, \vb{0}_n) + 1\}.

Note then that $x_n$ is fully contained in $[-M, M]^n$ . Note this box is compact, and so $x_n$ has a convergent subsequence by sequential compactness. Thus Cauchy sequences converge and so the space is complete.

Finally, for $(\bR^n, d)$ , note that a sequence converges if and only if it converges with respect to $\rho$ , and the same for the sequence being Cauchy.

[7.1.7]Theorem(Sequence in Product Space Converges iff. Components Converge)#

Let $\{X_\lambda\}_{\lambda \in \Lambda}$ be an arbitrary collection of spaces with $X = \prod_{\lambda \in \Lambda} X_\lambda$ its product. Then a sequence $\vb{x}_n \to \vb{x}$ in the product space if and only if $\pi_\lambda(\vb{x}_n) \to \pi_\lambda(\vb{x})$ for every $\lambda \in \Lambda$ .

Proof.

$(\Longrightarrow)$ : Note that $\pi_\lambda$ is continuous for every $\lambda \in \Lambda$ , so the image of $\vb{x}_n$ under each projection maps to the projection of $\vb{x}$ .

$(\Longleftarrow)$ : Let $\vb{x}_n$ be some sequence such that there exists some $\vb{x} \in X$ for which $\pi_\lambda(\vb{x}_n) \to \pi_\lambda(\vb{x})$ for every $\lambda \in \Lambda$ . Let $U$ be any basic open neighborhood of $\vb{x}$ . By definition, we have that there exist indices $\{\lambda_k\}_{k = 1}^n$ such that

U = \bigcap_{k = 1}^n \pi_\lambda^{-1}(U_\lambda),

where $U_{\lambda_k} \subseteq X_{\lambda_k}$ is open. For each $k \in \{1, \dots, n\}$ , there is some $N_k \in \bZ_+$ such that $\pi_{\lambda_k}(\vb{x}_n) \in \pi_{\lambda_k}(U_\lambda)$ for every $n \ge N_k$ . Choosing $N = \max\{N_k\}_{k = 1}^n$ , note that $\vb{x}_n \in U$ for every $n \ge N$ , completing the proof.

[7.1.8]Theorem(

(\bR^\omega, D)

is complete)#

The metric space $(\bR^\omega, D)$ , with $D(x, y) = \sup_{n \in \bZ_+} \{ \bar{d}(x, y) / n \}$ is complete.

Proof.

Suppose $\vb{x}_n$ is Cauchy in this metric space. Note that

\bar{d}(\pi_n(\vb{x}_n), \pi_n(\vb{y}_n)) \le nD(\vb{x}, \vb{y}),

and so $\pi_n(\vb{x}_n)$ is a Cauchy sequence in $(\bR, \bar{d})$ for each $n \in \bZ_+$ . This metric space is complete, and so the projections of the sequence converge. By the previous theorem, this means $\vb{x}_n$ converges.

[7.1.9]Example(Completeness is not a Topological Property)#

Note that $\bR$ is complete in its usual metric. Now consider the subspace $(-1, 1)$ with the same metric, $d(x, y) = |x - y|$ . But note the Cauchy sequence $x_n = 1 - 1/n$ converges to $1$ , which is outside of this interval. Thus this space is not complete, even though $(-1, 1) \cong \bR$ . Thus completeness is not a topological property.

[7.1.10]Definition(Uniform Metric, Revisited)#

Let $(Y, d)$ be a metric space, together with an induced bounded metric $\bar{d}(x, y) = \min\{d(x, y), 1\}$ . Let $Y^X$ be the space of functions $\{f \mid f: X \to Y\}$ . Then we define the uniform metric

\bar{\rho}(f, g) = \sup_{x \in X} \{d(f(x), g(x))\}.

[7.1.11]Remark#

Note that if $X$ is an indexing set, the space $Y^X$ is the set of sequences in $Y$ indexed by $X$ . Then the uniform metric defined for sequences coincides with the uniform metric defined in terms of functions. Of course, this is all true because a sequence is truly a function from the indexing set into $Y$ .

[7.1.12]Theorem(Criterion for Complete Metric Space Under Unifom Metric)#

Suppose $(Y, d)$ is a complete metric space. Then the metric space $(Y^X, \bar{\rho})$ is complete as well.

Proof.

Let $(f_n)_{n \in \bZ_+}$ be a Cauchy sequence in $Y^X$ , where each $f_n: X \to Y$ . For a fixed $x \in X$ , consider the sequence $(f_n(x))_{n \in \bZ_+}$ in $Y$ . Fix $1 \ge \epsilon > 0$ , and let $N \in \bZ_+$ such that $\bar{\rho}(f_m, f_n) < \epsilon$ for every $m, n \ge N$ . Then, for any $m, n \ge N$ , note that

d(f_m(x), f_n(x)) < \sup_{x \in X} \{d(f_m(x), f_n(x))\} = \bar{\rho}(f_m, f_n) < \epsilon,

and so $(f_n(x))$ is Cauchy. Since $Y$ is complete, we know this sequence converges. Thus, for every $x \in X$ , define $f(x)$ as the value $(f_n(x))$ converges to.

Finally, we show $f_n \to f$ in $Y^X$ . Fix some $\epsilon > 0$ , then note there is some $N \in \bZ_+$ such that $\bar{\rho}(f_m, f_n) < \epsilon$ for every $m, n \ge N$ . Fix $n$ and let $m \to \infty$ , then note that by the limit comparison theorem and continuity of the metric, $\bar{\rho}(f, f_n) \le \epsilon$ . Thus $f_n \to f$ .

If $\epsilon$ were chosen to be greater than $1$ in either case, the bounded metric $\bar{\rho}$ would evaluate to $1$ , making the proceeding statements trivial.

Now, when defining $Y^X$ for $Y$ a metric space, let $X$ be a topological space. The set $Y^X$ is not different, but we may now consider subsets that we were unable to do before.

[7.1.13]Definition(Space of Continuous and Bounded Functions)#

Let $X$ be a topological space and $Y$ be a metric space. The space of continuous maps, denoted $C(X, Y) \subseteq Y^X$ , is given by

C(X, Y) = \{f \mid f: X \to Y \text{ is continuous}\}.

Similarly, the space of bounded maps, denoted $B(X, Y) \subseteq Y^X$ , is given by

B(X, Y) = \{f \mid f: X \to Y \text{ is bounded} \}.

Recall that a map $f: X \to Y$ , with $Y$ a metric space, is said to be bounded if $f(X)$ is contained in a single ball.

[7.1.14]Theorem#

Let $X$ be a topological space and $(Y, d)$ a metric space. Then the spaces $C(X, Y)$ and $B(X, Y)$ are closed.

Proof.

Note these spaces are metric, so sequences probe limit points. We show that any sequences contained in these spaces converge inside the space.

Let $(f_n)$ be a sequence of functions in $C(X, Y)$ , meaning each $f_n: X \to Y$ is continuous, such that $f_n \to f$ . Note that convergence in this space is uniform convergence, so $f$ is automatically continuous and so $f \in C(X, Y)$ . Since $f_n \to f$ , for any $\epsilon > 0$ there is some $N \in \bZ_+$ such that $\bar{\rho}(f_n, f) < \epsilon$ . Then, for any $x \in X$ , note that

\bar{d}(f_n(x), f(x)) < \bar{\rho}(f_n, f) < \epsilon.

Note that $\epsilon$ is an upper bound for $\bar{d}(f_n(x), f(x))$ for every $x \in X$ , independent of $\epsilon$ . Thus $f_n \to f$ uniformly, and so $f$ is continuous, meaning $f \in C(X, Y)$ and so the space is closed.

Now let $(f_n)$ be a sequence of bounded functions in $B(X, Y)$ , meaning there is some $\epsilon_n > 0$ for each $f_n: X \to Y$ such that $f_n(X) \subseteq B_{\epsilon_n}(\vb{0})$ . Note that for $\epsilon = 1$ , there is some $N \in \bZ_+$ such that $d(f_n, f) < 1$ for every $n \ge N$ . Note then that

d(\vb{0}, f) \le d(\vb{0}, f_n) + d(f_n, f) \le \epsilon_n + 1

for every $n \ge N$ . Thus $f \in B_{\epsilon_N + 1}(\vb{0})$ , meaning $f \in B(X, Y)$ . Thus the space is closed, completing the proof.

[7.1.15]Theorem#

Let $X$ be a topological space and $(Y, d)$ a complete metric space. Then the spaces $C(X, Y)$ and $B(X, Y)$ are complete.

[7.1.16]Definition(Sup Metric)#

If $(Y, d)$ is a metric space, one can define another metric, the sup metric, on the set $B(X, Y)$ of bounded functions $X \to Y$ by the equation

d_\infty(f, g) = \sup_{x \in X} \{ d(f(x), g(x)) \}.

Note that this metric is well defined since $f(X) \cup g(X)$ is bounded above, and so $d(f(x), g(x))$ is bounded above for all $x \in X$ . In fact, the sup metric may be put on all other bounded sets. The sup metric is almost identical to the uniform metric, in which $\bar{\rho}(f, g) = \min\{d_\infty(f, g), 1\}$ .

[7.1.17]Theorem(Metric Space Completion Theorem)#

Let $(X, d)$ be a metric space. Then there is an isometric embedding of $X$ into a complete metric space.

Proof.

Let $x_0 \in X$ be fixed. For any $a \in X$ , define the map

\phi_a(x) = d(x, a) - d(x, x_0).

We show $\phi_a \in B(X, \bR)$ . Indeed, note that

d(x, a) \le d(x, x_0) + d(x_0, a),

d(x, x_0) \le d(x, a) + d(a, x_0).

Then we have that

\Big(|\phi_a(x)| = |d(x, a) - d(x, x_0)|\Big) \le |d(x, x_0) - d(x, a)| \le d(x, x_0) + d(x, a) \le d(x_0, a).

Thus $|\phi_a(x)| \le |d(x_0, a)|$ for every $x \in X$ , meaning $\phi_a$ is bounded and so $\phi_a \in B(X, \bR)$ . Finally, define the map

\Phi_a: X \to B(X, \bR) \quad \Phi_a(x) = \phi_a(x).

Note that $B(X, \bR)$ , where $\bR$ takes the Euclidean metric, is complete with the sup metric. We show $\Phi_a$ is an isometry. Note that an isometry between metric spaces automatically constitutes an embedding, completing the proof.

To show $\Phi_a$ is an isometry, we show

\rho(\phi_a, \phi_b) = d(a, b).

For any $a, b \in X$ , note that

\begin{align*} \rho(\phi_a, \phi_b) &= \sup_{x \in X} \Big\{ d(\phi_a(x), \phi_b(x)) \Big\} \\ &= \sup_{x \in X} \left\{ \abs{\phi_a(x) - \phi_b(x)} \right\} \\ &= \sup_{x \in X} \Big\{ \abs{ \left( d(x, a) - d(x, x_0) \right) - \left( d(x, b) - d(x, x_0) \right) } \Big\} \\ &= \sup_{x \in X} \Big\{ \abs{ d(x, a) - d(x, b) } \Big\} \\ &\le d(x, a) + d(x, b) \le d(a, b). \end{align*}

Finally, note that this inequality cannot be strict. Take $x = a$ , and note

|d(x, a) - d(x, b)| = |d(a, b)| \not < d(a, b).

[7.1.18]Definition(Completion of a Metric Space)#

Let $(X, d)$ be any metric space. Let $h: X \to Y$ be an isometric embedding into a complete metric space. Then the space $\bar{h(X)}$ is a complete metric space in $Y$ , which we call the completion.

[7.1.19]Example(Completion of

(\bQ, d)

A classic result is that the completion of $\bQ$ under the metric $d(x, y) = |x - y|$ is $\bR$ equipped with the standard Euclidean metric. Using the isometric embedding $\Phi$ into the complete space $B(\bQ, \bR)$ , we map any $a \in \bQ$ to the function $\phi_a \in B(\bQ, \bR)$ defined by $\phi_a(x) = |x - a| - |x|$ (where $x \in \bQ$ ).

Consider, for example, a Cauchy sequence $(a_n) \subseteq \bQ$ that approximates $\sqrt{2}$ . In $\bQ$ , this sequence has no limit. However, the sequence of their images $\Phi(a_n) = \phi_{a_n}$ is Cauchy in $B(\bQ, \bR)$ . Because $B(\bQ, \bR)$ is complete, this sequence of functions must converge uniformly. Indeed, $\phi_{a_n}(x) = |x - a_n| - |x|$ converges uniformly to the function $f(x) = |x - \sqrt{2}| - |x|$ . This limit function $f$ is a well-defined bounded function in $B(\bQ, \bR)$ , but it is not in the image $\Phi(\bQ)$ . The “hole” at $\sqrt{2}$ has thus materialized as a concrete boundary point of $\Phi(\bQ)$ in the function space.

Repeating this process for every Cauchy sequence in $\bQ$ fills in all such holes. The closure of the image of $\bQ$ is therefore exactly:

\overline{\Phi(\bQ)} = \Big\{ f_r(x) = |x - r| - |x| : r \in \bR \Big\}.

This closed subspace of $B(\bQ, \bR)$ is, by definition, the completion of $\bQ$ . Note that this abstract space of functions is structurally identical to $\bR$ . The image of $\bR$ under this isometry is precisely $\overline{\Phi(\bQ)}$ , and so there is a surjective isometry between them. Since a surjective isometry is a perfect equivalence of metric spaces, we safely identify the completion of $\bQ$ directly with $\bR$ .

[7.1.20]Remark(Intuition behind Completions)#

A completion effectively fills the holes in a metric space caused by Cauchy sequences that fail to converge. A Cauchy sequence consists of points that get arbitrarily close to one another; geometrically, it ought to converge to a limit. When a space is incomplete, it simply lacks the points that these sequences are aiming at. The completion process adjoins exactly these missing limit points. By embedding $X$ isometrically into the humongous complete function space $B(X, \bR)$ , we force the missing limit functions to exist. Thus we patch the holes of $X$ and guarantee that every Cauchy sequence in the new space converges.

7.1.1Exercises#

[7.1.21]Problem#

Let $(X, d)$ be a metric space. Suppose that, for some $\epsilon > 0$ , every $\epsilon$ -ball in $X$ has compact closure. Show that $X$ is complete.

Proof.

Let $(x_n)$ be any Cauchy sequence. There exists some $N \in \bZ_+$ such that $d(x_m, x_n) < \epsilon/2$ for every $m, n \ge N$ . Then there exists finitely many closed $\epsilon$ -balls whose union contains $x_1, \dots, x_{N - 1}$ . Adjoin to this list of balls a single ball of radius $\epsilon$ containing all $x_N, x_{N + 1}, \dots$ . Let this list be $\{U_1, \dots, U_n\}$ , and note that $\{\bar{U_1}, \dots, \bar{U_n}\}$ still covers the sequence. Since each closure of the finitely many sets is compact, the set $U = \bigcup_{k = 1}^n U_k$ is compact. Since $x_n$ is contained in the compact set $U$ , and compactness is equivalent to sequential compactness in metrizable spaces, we have that $x_n$ admits a convergent subsequence. Thus $x_n$ is a convergent sequence, making $X$ complete.

[7.1.22]Problem#

Let $(X, d_X)$ and $(Y, d_Y)$ be metric spaces, with $Y$ complete. Let $A \subseteq X$ . Show that, for any uniformly continuous $f: A \to Y$ , there exists a unique, uniformly continuous extension to $g: \bar{A} \to Y$ .

Proof.

Let $x \in \bar{A} \setminus A$ . Then there exists a sequence $(x_n)$ in $A$ such that $x_n \to x$ . By uniform continuity of $f$ , for any $\epsilon > 0$ there exists a $\delta > 0$ such that

d_X(x_1, x_2) < \delta \implies d_Y(f(x_1), f(x_2)) < \epsilon

for every $x_1, x_2 \in A$ . There is some $N \in \bZ_+$ such that $d_X(x_m, x_n) < \delta$ for every $m, n \ge N$ , and so $d_Y(f(x_m), f(x_n)) < \epsilon$ for every $m, n \ge N$ . Since $\epsilon$ is arbitrary, we have that $f(x_n)$ is Cauchy in the complete space $Y$ . Thus $f(x_n) \to y_x$ .

Now suppose $(z_n)$ is some other sequence in $A$ such that $z_n \to x$ . Fix $\epsilon > 0$ . Once again, by uniform continuity, there exists a $\delta > 0$ such that

d_X(z_n, x_n) < \delta \implies d_Y(f(z_n), f(x_n)) < \epsilon

for every $n \in \bZ_+$ . Since $z_n \to x$ and $x_n \to x$ , note that $d_X(z_n, x_n) \to 0$ , and so $d_Y(f(z_n), f(x_n)) \to 0$ . Since $f(x_n) \to y_x$ , we have that $f(z_n) \to y_x$ .

Define a function $g: \bar{A} \to Y$ such that $g|_A = f$ and $g|_{\bar{A} \setminus A}(x) = y_x$ , where $y_x$ is chosen as per before. Note that any sequence tending to the same point $x$ in $\bar{A} \setminus A$ has the same limit point $y_x$ in the image, and so $g$ is well-defined. We prove this function is uniformly continuous.

Fix $\epsilon > 0$ . By the uniform continuity of $g|_A = f$ , there exists $\delta > 0$ such that

d_X(x_1, x_2) < \delta \implies d_Y(g(x_1), g(x_2)) < \epsilon

for every $x_1, x_2 \in A$ . We show this $\delta$ works for all points in $\bar{A}$ . Indeed, suppose $u, v \in \bar{A}$ are such that $d_X(u, v) < \delta$ . There exist sequences $(u_n)$ and $(v_n)$ in $A$ such that $u_n \to u$ and $v_n \to v$ . Because the metric $d_X$ is continuous, $\lim_{n \to \infty} d_X(u_n, v_n) = d_X(u, v) < \delta$ . Thus, there exists some $N \in \bZ_+$ such that $d_X(u_n, v_n) < \delta$ for every $n \ge N$ .

Applying the uniform continuity of $g|_A$ to these sequence elements, we find that $d_Y(g(u_n), g(v_n)) < \epsilon$ for every $n \ge N$ . By the construction of $g$ , we know $g(u_n) \to g(u)$ and $g(v_n) \to g(v)$ . Taking the limit as $n \to \infty$ and using the continuity of the metric $d_Y$ , we obtain $d_Y(g(u), g(v)) \le \epsilon$ . Since $\epsilon > 0$ was arbitrary, we conclude that $g$ is uniformly continuous on $\bar{A}$ .

Recall that continuous functions preserve limits. $g$ does this, since for every sequence $x_n \to x$ , we define $g(x)$ such that $g(x_n) \to g(x)$ . If $g|_{\bar{A} \setminus A}$ were defined in any other way, the resulting function would not be continuous. Thus $g$ is the unique, uniformly continuous extension of $f$ .

[7.1.23]Problem#

Let $(X, d)$ be a metric space. Recall that a map $f: X \to X$ is said to be a contraction map if there exists some $C < 1$ such that

d(f(x), f(y)) \le Cd(x, y).

Show that if $X$ is complete, then any contraction map admits a unique fixed point.

Proof.

Fix any $x_0 \in X$ , then consider the sequence $x_n$ defined by $x_n = f^{n - 1}(x_0)$ , where $f^0 = \id$ .

We will show this sequence is Cauchy. Fix $\epsilon > 0$ . Note that $d(x_n, x_{n + 1}) \le Cd(x_{n - 1}, x_n)$ for any $n \in \bZ_+$ . Then note

\begin{align*} d(x_m, x_n) &\le d(x_m, x_{m + 1}) + d(x_{m + 1}, x_{m + 2}) + \cdots + d(x_{n - 1}, x_n) \\ &\le C^m \underbrace{d(x_0, x_1)}_{:= r} + C^{m + 1} d(x_0, x_1) + \dots + C^{n - 1} d(x_0, x_1) \\ &= r C^m \left( \frac{1 - C^{n - m}}{1 -C} \right) = r \left( \frac{C^m - C^n}{1 - C} \right). \end{align*}

Note that $d(x_m, x_n)$ can be made arbitrarily small since $C < 1$ (and so $C^m - C^n \to 0$ as $m, n \to \infty$ ). Thus there is some $N \in \bZ_+$ such that $d(x_m, x_n) < \epsilon$ for every $m, n \ge N$ . Thus the sequence is Cauchy and thus convergent, and so $x_n \to x^*$ . We show $x^*$ is our desired fixed point.

Note that $f$ is $C$ -Lipschitz and thus continuous, so $f(x_n) \to f(x^*)$ . Of course, $d(f(x_n), x_n) \to 0$ , and so they converge to the same limit. Thus $x^* = f(x^*)$ .

Finally, we show $x^*$ is the unique fixed point. Suppose $x' \in X$ is another fixed point. Then note $d(f(x^*), f(x')) = d(x^*, x') \le Cd(x^*, x')$ . Then either $C = 0$ , in which there is only one unique fixed point by inspection, or $d(x^*, x') = 0$ (and so $x^* = x'$ ), completing the proof.