
Chapter 2 - Metric Spaces

2.7 Complete Metric Spaces

 Main: Index Previous: 2.6 Convergence. Open and Closed Sets Next: 2.8 Contraction Mappings

Results

Definition 1

A sequence $\{x_n\}$ of points in a metric space $(X, \rho)$ is said to satisfy the Cauchy criterion if, given $\eps > 0$, there is an integer $N_\eps$ such that $\rho(x_n, x_{n'}) < \eps$ for all $n, n' > N_\eps$.

Definition 2

A subsequence $\{x_n\}$ of points in a metric space $(X,\rho)$ is called a Cauchy sequence (or fundamental sequence) if it satisfies the Cauchy criterion.

Theorem 1

Every convergent sequence $\{x_n\}$ is fundamental.

Proof.
If $\{x_n\}$ converges to a limit $x$, then given any $\eps > 0$, there is an integer $N_\eps$ such that $$\rho(x_n, x) < \frac{\eps}{2}$$ for all $n\geq N_\eps$. But then, $$\rho(x_n, x_{n'}) \leq \rho(x_n, x) + \rho(x_{n'}, x) < \eps$$ for all $n, n' > N_\eps$.

Definition 3

A metric space is said to be complete if every Cauchy sequence in $(X, \rho)$ converges to an element of $X$. Otherwise, $(X, \rho)$ is said to be incomplete.

Comment: A way to verify completeness.

Theorem 2 (Nested Sphere Theorem)

A metric space $(X,\rho)$ is complete if and only if every nested sequence $\{S_n\} = \{S[x_n, r_n]\}$ of closed spheres in $X$ such that $r_n\rightarrow 0$ as $n\rightarrow\infty$ has a nonempty intersection $$\bigcap_{n=1}^\infty S_n.$$

Theorem 3 (Baire)

A complete metric space $(X,\rho)$ cannot be represented as the union of a countable number of nowhere dense sets.

Corollary

A complete metric space $(X,\rho)$ without isolated points is uncountable.

Definition 4

Given a metric space $R = (X,\rho)$ with closure $[R]$, a complete metric space $R^*$ is called a completion of $R$ if $R\subset R^*$ and $[R] = R^*$, i.e. if $R$ is a subset of $R^*$ and $R$ is everywhere dense in $R^*$.

Theorem 4

Every metric space $R$ has a completion. This completion is unique to within an isometric mapping, carrying every point $x\in R$ to itself.

Problem 1

Prove that the limit $f(t)$ of a uniformly convergent sequence of functions $\{f_n(t)\}$ continuous on $[a, b]$ is itself a function continuous on $[a, b]$.

Hint: Clearly; $$|f(t) - f(t_0)| \leq |f(t) - f_n(t)| + |f_n(t) - f_n(t_0)| + |f_n(t_0) - f(t_0)|,$$ where $t,t_0\in[a,b]$. Use the uniform convergence to make the sum of the first and third terms on the right small for sufficiently large $n$. Then use the continuity of $f_n(t)$ to make the second term small for $t$ sufficiently close to $t_0$.

Proof.
To show continuity of $f$, then for any $\eps > 0$, we must find some $\delta > 0$ such that: $$|t - t_0| < \delta \;\Longrightarrow\; |f(t) - f(t_0)| < \eps.$$ From the hint, and applying the triangle inequality, we have: $$|f(t) - f(t_0)| \leq |f(t) - f_n(t)| + |f_n(t) - f_n(t_0)| + |f_n(t_0) - f(t_0)|.$$ By convergence of $f_n$, then for any $\eps/3 > 0$ there exists some $N_1, N_2\in\N$, and we set $N_3 = \max(N_1, N_2)$, such that: $$|f(t) - f_n(t)| < \frac{\eps}{3},\quad |f_n(t_0) - f(t_0)| < \frac{\eps}{3}.$$ Next, by continuity of $f_n$, then for $\eps/3 > 0$ we can find some $\delta_2 > 0$ such that $$|t - t_0| < \delta_2 \;\Longrightarrow\; |f_n(t) - f_n(t_0)| < \frac{\eps}{3}.$$ Now, for any $n\geq N_3$ and by setting $\delta = \delta_2$, we get: $$|f(t) - f(t_0)| < \frac{\eps}{3} + \frac{\eps}{3} + \frac{\eps}{3} = \eps$$ whenever $|t - t_0| < \delta$ and $n\geq N_3$, which proves that $f$ is continuous.

Problem 2

Prove that the following set is complete. The set $m$ of all bounded infinite sequences of real numbers $x = (x_1, x_2, \ldots)$ with $$\rho(x, y) = \sup_k|x_k - y_k|.$$

Proof.
We show that $m$ is complete by verifying that it contains the limit of a converging sequence. Assume the sequence of infinite sequences $\{x^{(n)}\}$ converges to some $y = (y_1,y_2,\ldots)$, so $$\lim_{n\rightarrow\infty}\rho(x^{(n)}, y) = 0 \;\Longrightarrow\; \lim_{n\rightarrow\infty}\sup_k|x^{(n)}_k - y_k| = 0.$$ By definition, $y$ is an infinite sequence. Assume for contradiction that $y$ is unbounded, but then $$\lim_{n\rightarrow\infty}\sup_k|x^{(n)}_k - y_k| = \infty$$ since $x^{(n)}$ is bounded which is a contradiction to the convergence. So $y$ must be bounded. Since it is also infinite, it follows that $y\in m$ which shows that $m$ is complete.

Problem 3

If $R = (X,\rho)$ is complete, then the intersection: $$\bigcap_{n=1}^\infty S_n$$ from Theorem 2, consists of a single point.

Proof.
From Theorem 2, we know that the intersection is nonempty, so let $x\in\bigcap_{n=1}^\infty S_n$. Now let $y\not= x$ be some other arbitrary point in $R$. Since the point is distinct from $x$, then $\rho(x, y) = k > 0$ for some $k\in\R$. Then there exists some $m\in\N$ such that $r_m < k$ so $y\not\in S[x_m, r_m]$ and $y\not\in \bigcap_{n=1}^\infty S_n$. Since this is true for any point distinct from $x$, it follows that the intersection contains a single point, i.e. $$\bigcap_{n=1}^\infty S_n = \{x\}.$$

Problem 4

By the diameter of a subset $A$ of a metric space $R$ is meant the number $$d(A) = \sup_{x,y\in A}\rho(x, y).$$ Suppose $R$ is complete, and let $\{A_n\}$ be a sequence of closed subsets of $R$ nested in the sense that $$A_1 \supset A_2 \supset \cdots \supset A_n \supset \cdots.$$ Suppose further that $$\lim_{n\rightarrow\infty} d(A_n) = 0.$$ Prove that the intersection $\bigcap_{n=1}^\infty A_n$ is nonempty.

Proof.
For each set $A_k$ let $r_k$ define the smallest possible radius around some point $x_k$ so we can define a closed ball $S_k[x_k, r_k] \subset A_k$ and $S_k[x_k, r_k]\subset S_{k-1}[x_{k-1}, r_{k-1}]$. Then we get a sequence of nested, closed spheres where $r_n\rightarrow 0$ as $n\rightarrow\infty$. Then, according to Theorem 2 and Problem 3, $$\{x\} = \bigcap_{n=1}^\infty S_n \subset \bigcap_{n=1}^\infty A_n$$ which shows that the intersection is nonempty.

Problem 5

A subset $A$ of a metric space $R$ is said to be bounded if its diameter $d(A)$ is finite. Prove that the union of a finite number of bounded sets is bounded.

Proof.
Assume $M\in\N$ is a fixed, finite number and assume $A_1,\ldots, A_M$ are $M$ bounded sets with diameters $d_1,\ldots, d_M$.

Assume for contradiction that the diameter $d(A_1\cup\ldots\cup A_M) = \infty$, then at least one of the sets must be of infinite size, but that is impossible since they are all bounded. Contradiction! Thus, the diameter of the union of the sets must be bounded.

Problem 6

Give an example of a complete metric space $R$ and a nested sequence $\{A_n\}$ of closed subsets of $R$ such that $$\bigcap_{n=1}^\infty A_n = \emptyset.$$ Reconcile this example with Problem 4.

Proof.
The real line $\R$ with $\rho(x,y) = |x-y|$ is a complete metric space. Define $A_n = \emptyset$ for each $n\in\N$ which we can do since $\emptyset$ is a closed set. Then the intersection of all $A_n$ becomes $\emptyset$.

In Problem 4 we have indirectly assumed that the sets are nonempty.

Problem 7

Prove that a subspace of a complete metric space $R$ is complete if and only if it is closed.

Proof.
Let $A\subset X$ be a subset of the complete metric space $(X, \rho)$.

$\Rightarrow$) Assume $A$ is complete. Then any convergent sequence $\{x_n\}$ in $A$ have limits $x$ also included in $A$, or in other words, $A$ contains all its limit points. Hence $[A] = A$ which means it is closed.

$\Leftarrow$) Assume $A$ is closed. Since a closed set contains all its limit points, then any sequence $\{x_n\}$ in $A$ will converge to a point in $A$, which means it must be complete.

Problem 8

Prove that the real line $\R$ equipped with the distance $$\rho(x, y) = |\arctan(x) - \arctan(y)|$$ is an incomplete metric space.

Proof.
Set $x_n = n$. Let $\eps > 0$ and $N > \tan(\frac{\pi}{2} - \eps)$. For $m,n > N$ then $$\rho(x_n, x_m) = |\arctan(x_n) - \arctan(x_m)| \leq |\frac{\pi}{2} - \arctan N| < |\frac{\pi}{2} - \frac{\pi}{2} + \eps| = \eps,$$ which shows that $\{x_n\}$ is Cauchy. As $n\rightarrow\infty$, $\arctan x_n\rightarrow \frac{\pi}{2}$ but $\{x_n\}$ does not converge to any element in $\R$ since there is no $x\in\R$ such that $\arctan x = \frac{\pi}{2}$.

Problem 9

Give an example of a complete metric space homeomorphic to an incomplete metric space.

Proof.
The function $$f(x) = \frac{2}{\pi}\arctan x$$ defines a homeomorphism between $\R$ and $(-1, 1)$. But this is an open set and from Problem 7 it is incomplete since only closed sets are complete subspaces.

Skipped...