Metric Spaces

Norms, distances and metrics

Definition (Norm)

A norm on $\mathbb{R}^{d}$ is a function $\|\cdot\| : \mathbb{R}^{d} \to [0, +\infty)$ that satisfies:

$\|x\| \geq 0$ for every $x \in \mathbb{R}^{d}$, with $\|x\| = 0$ if and only if $x = 0$;
$$|\lambda x| = \lambda \,|x|$for every$\lambda \in \mathbb{R}$and$x \in \mathbb{R}^{d}$$;
$\|x + y\| \leq \|x\| + \|y\|$ for all $x, y \in \mathbb{R}^{d}$.

Note (Distance induced by a norm on $\mathbb{R}^{d}$)

A norm induces a distance (or metric) $d : \mathbb{R}^{d} \times \mathbb{R}^{d} \to [0, +\infty)$ by $d(x,y) = \|x - y\|$, which satisfies:

$d(x,y) = d(y,x)$;
$d(x,y) = 0 \iff x = y$;
$d(x,z) = \|x-z\| = \|x-y+y-z\| \leq \|x-y\| + \|y-z\| = d(x,y) + d(y,z)$.

Definition (Metric space)

Let $E$ be a set. A metric is a function $d : E \times E \to [0, +\infty)$ that satisfies:

$d(x,y) = d(y,x)$;
$d(x,y) = 0$ if and only if $x = y$;
$d(x,y) \leq d(x,z) + d(z,y)$ (triangle inequality).

The pair $(E, d)$ is called a metric space.

Example (The usual metric on $\mathbb{R}^{d}$)

On $\mathbb{R}^{d}$ with the Euclidean norm, $d(x,y) = \|x - y\|$ is a metric; the triangle inequality is the usual one for the modulus.

Metric topology: balls, open sets and closed sets

Definition (Open ball in a metric space)

Let $(E, d)$ be a metric space. For $x_{0} \in E$ and $r > 0$ we define

\[B(x_{0}, r) = \{ y \in E : d(x_{0}, y) < r \}.\]

Definition (Open set in a metric space)

$D \subseteq E$ is an open set if for every $x_{0} \in D$ there exists $r > 0$ such that $B(x_{0}, r) \subseteq D$. In particular $\emptyset$ and $E$ are open.

Lemma (Balls are open sets)

For all $x_{0} \in E$ and $r > 0$, $B(x_{0}, r)$ is an open set.

Proof: Let $x_{1} \in B(x_{0}, r)$. We shall show that if $0 < r_{1} < r - d(x_{0}, x_{1})$, then $B(x_{1}, r_{1}) \subseteq B(x_{0}, r)$. Indeed, if $y \in B(x_{1}, r_{1})$, then

\[d(x_{0}, y) \leq d(x_{0}, x_{1}) + d(x_{1}, y) < d(x_{0}, x_{1}) + r_{1} < d(x_{0}, x_{1}) + r - d(x_{0}, x_{1}) = r,\]

whence $y \in B(x_{0}, r)$.

Lemma (A finite intersection of open sets is open)

If $G_{1}, G_{2}, \dots, G_{m}$ are open in $(E, d)$, then $\bigcap_{i=1}^{m} G_{i}$ is open.

Proof: It suffices to prove it for $m = 2$ and apply induction. Let $x_{0} \in G_{1} \cap G_{2}$. There exist $r_{1}, r_{2} > 0$ such that $B(x_{0}, r_{1}) \subseteq G_{1}$ and $B(x_{0}, r_{2}) \subseteq G_{2}$. Taking $r = \min(r_{1}, r_{2})$,

\[B(x_{0}, r) \subseteq B(x_{0}, r_{1}) \cap B(x_{0}, r_{2}) \subseteq G_{1} \cap G_{2}.\]

Lemma (An arbitrary union of open sets is open)

If $\{ G_{\lambda} \}_{\lambda \in \Lambda}$ is any collection of open sets in $(E, d)$, then $\bigcup_{\lambda \in \Lambda} G_{\lambda}$ is open.

Proof: If $x_{0} \in \bigcup_{\lambda \in \Lambda} G_{\lambda}$, then there exists $\lambda_{0}$ with $x_{0} \in G_{\lambda_{0}}$. Since $G_{\lambda_{0}}$ is open, there exists $r > 0$ such that $B(x_{0}, r) \subseteq G_{\lambda_{0}} \subseteq \bigcup_{\lambda \in \Lambda} G_{\lambda}$.

Definition (Closed set)

$F \subseteq E$ is closed if $E \setminus F$ is open.

Note (Basic properties of closed sets)

If $\{ F_{\lambda} \}_{\lambda \in \Lambda}$ is a family of closed sets, $\bigcap_{\lambda \in \Lambda} F_{\lambda}$ is closed, since
\[E \setminus \bigcap_{\lambda \in \Lambda} F_{\lambda} = \bigcup_{\lambda \in \Lambda} (E \setminus F_{\lambda})\]
is a union of open sets.
If $F_{1}, \dots, F_{m}$ are closed, then $\bigcup_{i=1}^{m} F_{i}$ is closed.

Example (The properties of closed/open sets are not preserved when passing to arbitrary collections)

$\bigcap_{n=1}^{\infty} \left( a - \tfrac{1}{n}, a + \tfrac{1}{n}\right) = \{ a\}$ is not open. That is, a countable intersection of open sets is not necessarily open.
$\left( a, b \right) = \bigcup_{n=1}^{\infty} \left[a + \tfrac{1}{n}, b - \tfrac{1}{n}\right]$ is not closed. That is, a countable union of closed sets is not necessarily closed.