8.1 Metrics
Definition 8.1.1 (Metric Space).label Let $X$ be a set and $d: X \times X \to [0, \infty]$, then $d$ is a metric if:
- (PM1)
For any $x \in X$, $d(x, x) = 0$.
- (M)
For any $x, y \in X$ with $x \ne y$, $d(x, y) > 0$.
- (PM2)
For any $x, y \in X$, $d(x, y) = d(y, x)$.
- (PM3)
For any $x, y, z \in X$, $d(x, z) \le d(x, y) + d(y, z)$.
The pair $(X, d)$ is a metric space, which comes with the metric uniformity induced by $d$, and the corresponding topology.
Proposition 8.1.2.label Let $(X, d)$ be a metric space, then the following are equivalent:
- (1)
$X$ is second countable.
- (2)
$X$ is Lindelöf.
- (3)
$X$ is separable.
In particular, since second countability is hereditary, separability and the Lindelöf property are both hereditary in metric spaces.
Proof. (1) $\Rightarrow$ (2): Let $\seqi{U}\subset 2^{X}$ be an open cover of $X$. Let $\seq{V_n}\subset 2^{X}$ be a countable base for the topology on $X$, and
For any $x \in X$, there exists $i \in I$ such that $x \in U_{i}$, and $n \in \natp$ such that $x \in V_{n} \in U_{i}$. Therefore $\bigcup_{k \in K}V_{k} = X$.
Let $J \subset I$ be countable such that for each $k \in K$, there exists $j \in J$ with $V_{n} \subset U_{j}$, then $X = \bigcup_{k \in K}V_{k} = \bigcup_{j \in J}V_{j}$.
(2) $\Rightarrow$ (3): Let $n \in \nat$, then $\bracs{B(x, 1/n)|x \in X}$ is an open cover of $X$, so there exists $\seq{x_{n, k}}\subset X$ such that $X = \bigcup_{k \in \natp}B(x_{k}, 1/n)$. Let $D = \bracs{x_{n, k}| n, k \in \natp}$, then for any $x \in X$ and $n \in \natp$, there exists $k \in \natp$ such that $x \in B(x_{n, k}, 1/n)$, so $x_{n, k}\in B(x, 1/n)$. Therefore $D$ is dense in $X$, and $X$ is separable.
(3) $\Rightarrow$ (1): Let $\seq{x_n}\subset X$ be a countable dense subset. Let $x \in X$ and $k \in \natp$, then there exists $x_{n} \in \natp$ such that $d(x, x_{n}) < 1/(2k)$. In which case, $x \in B(x_{n}, 1/(2k)) \subset B(x_{n}, 1/k)$. Therefore $\bracs{B(x_n, 1/k)|n, k \in \natp}$ forms a countable basis for $X$.$\square$
Proposition 8.1.3.label Let $\seq{(X_n, d_n)}$ be metrisable spaces, then $\prod_{n \in \natp}X_{n}$ is also metrisable.
Proof. For each $n \in \natp$, let
then $d_{n}'$ is a pseudometric on $X$, and $\bracsn{d_n'}_{1}^{\infty}$ induces the product uniformity on $\prod_{n \in \natp}X_{n}$. By Theorem 6.3.10, $\prod_{n \in \natp}X_{n}$ is also metrisable.$\square$
Theorem 8.1.4 (Banach’s Fixed Point Theorem).label Let $(X, d)$ be a complete metric space and $f: X \to X$. If there exists $C \in (0, 1)$ such that
then:
- (1)
There exists a unique $x \in X$ such that $f(x) = x$.
- (2)
For any $y \in X$, $\limv{n}f^{n}(y) = x$.
Proof. Let $x_{0} \in X$ be arbitrary, and $x_{n} = f^{n}(x_{0})$, then for ecah $n \in \natp$,
Thus $\seq{x_n}\subset X$ is Cauchy, and converges to a point $x \in X$.
(2): For any $y_{0} \in X$, let $y_{n} = f^{n}(y_{0})$, then $d(x_{n}, y_{n}) \to 0$ as $n \to \infty$, so $\limv{n}f^{n}(y_{0}) = x$.
(1): Since $f$ is Lipschitz continuous,
$\square$
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