Metric Spaces and Metrizability

# Metric Spaces and Metrizability

## Metric Spaces

 Definition: A Metric Space is a pair $(E, d)$ where $E$ is a set, and $d$ is a Metric (or Distance Function), i.e., a function $d : E \times E \to [0, \infty)$ that satisfies the following properties: (1) $d(x, y) = 0$ if and only if $x = y$. (2) $d(x, y) = d(y, x)$ for all $x, y \in E$. (3) $d(x, z) \leq d(x, y) + d(y, z)$ for all $x, y, z \in E$.
• Property (1) states that the distance between any pair of points is $0$ if and only if those points coincide.
• Property (2) states that the distance between $x$ and $y$ is the same as the distance between $y$ and $x$.
• Property (3) states that the distance between $x$ and $z$ will always be less than or equal to the sum of the distance between $x$ and an intermediary point $y$, and that intermediary point $y$ and $z$. This is sometimes referenced to as the triangle inequality property for metrics.
 Definition: Let $(E, d)$ be a metric space. Given $\epsilon > 0$, the Open Ball Centered at $x$ with Radius $\epsilon$ is the set $V(x, \epsilon) := \{ y \in E : d(x, y) < \epsilon \}$.

If $(E, d)$ is a metric space and if $x \in E$, then a set $U$ is a neighbourhood of $x$ if there exists an $\epsilon > 0$ such that $x \in V(x, \epsilon) \subseteq U$. If $\mathcal U_x$ is the collection of all neighbourhoods of $x$, then:

• (1) $x \in U$ for all $U \in \mathcal U_x$.
• (2) If $U_1, U_2 \in \mathcal U_x$, then there exists $\epsilon_1, \epsilon_2 > 0$ with $x \in V(x, \epsilon_1) \subseteq U_1$ and $x \in V(x, \epsilon_2) \subseteq U_2$. By setting $\epsilon = \min \{ \epsilon_1, \epsilon_2 \}$ we see that $x \in V(x, \epsilon) \subset U_1 \cap U_2$, so that $U_1 \cap U_2 \in \mathcal U_x$.
• (3) If $U \in \mathcal U_x$, then there exists an $\epsilon > 0$ such that $x \in V(x, \epsilon) \subseteq U$. So if $U \subseteq V$ then $x \in V(x, \epsilon) \subseteq V$ so that $V \in \mathcal U_x$.
• (4) If $U \in \mathcal U_x$ then again, there exists an $\epsilon > 0$ such that $x \in V(x, \epsilon) \subseteq U$. By setting $V := V(x, \epsilon)$, we see that $V \in \mathcal U_y$ for all $y \in V(x, \epsilon)$.

Therefore, as noted on the Topologies and Topological Spaces page, $(E, d)$ becomes a topological space where for each $x \in E$, $\mathcal U_x$ is a collection of neighbourhoods for $x$, and moreover, for each $x \in E$, $\{ V(x, \epsilon) : \epsilon > 0 \}$ is a base of neighbourhoods of $x$, and also, $\{ V \left ( x, \frac{1}{n} \right ) : n \in \mathbb{N} \}$ is a countable base of neighbourhoods of $x$.

## Metrizability

Every metric space $(E, d)$ is a topological space $(E, \tau)$, where $\tau$ consists of all unions of open balls on $E$. However, in general, there are topological spaces that are not metric spaces.

 Definition: A topological space $(E, \tau)$ is said to be Metrizable if there exists a metric $d : E \times E \to [0, \infty)$ such that the topology induced by $d$ is the same as the topology $\tau$, where the topology induced by $d$ is the topology generated by the open balls of $E$, i.e., the topology generated by $\{ V(x, \epsilon) : x \in E, \epsilon > 0 \}$.

When we say that the topology induced by $d$ is the same as the topology $\tau$, we mean that the topology on $E$ consisting of all possible unions of open balls in $E$ is the same as the topology $\tau$.

 Proposition 1: Let $(E, \tau)$ be a topological space. If $(E, \tau)$ is metrizable, then $(E, \tau)$ is Hausdorff and first countable.