> [!definition] > > Let $(X, d)$ be a [[Metric Space|metric space]]. A set $U \subseteq X$ is [[Open Set|open]] if > $ > \forall x \in U, \exists \varepsilon > 0: \bracs{y: d(x, y) < \varepsilon} \subseteq U > $ > The topology $\topo$ would be the set of all such open sets, and the pair $(X, d, \topo)$ forms the metric [[Topological Space|topology]] over $(X, d)$. > [!theorem] > > A metric topology $(X, d)$ is [[Hausdorff Space|Hausdorff]]. > > *Proof*. Take any points $x, y \in X, x \ne y$ and $\varepsilon = d(x, y)/3$. Then the sets $\bracs{z: d(x, z) < \varepsilon}$ and $\bracs{y: d(y, z) < \varepsilon}$ cannot intersect by the triangle inequality. > [!theorem] > > Let $(X, d)$ be a metric topology. The [[Basis|basis]] of $X$ is the set of all [[Open Set|open]] *balls* > $ > \bracs{B(x, r):x \in X, r > 0} > $ > [!theorem] > > Let $(X, d_1)$, $(X, d_2)$ be metric spaces. The metric topology over $(X, d_1)$ and $(X, d_2)$ are *equivalent* (have the same [[Open Set|open sets]]) if > $ > \exists c \in \real: \forall x, y \in X, \frac{1}{c} \lt \frac{d_1(x, y)}{d_2(x, y)} \lt c > $ > > *This is a sufficient but not necessary condition*. > [!theorem] > > Let $(X, d)$ be a metric space and fix $o \in X$, then $d(o, x)$ is a continuous function in the metric topology. > > *Proof*. Let $O \subset \real$ be an open set, then > $ > \forall d \in O, \exists \varepsilon > 0: (d - \varepsilon, d + \varepsilon) \subseteq O > $ > The pre-image of $(d - \varepsilon, d + \varepsilon)$ is > $ > \bracs{x \in X: d - \varepsilon < d(o, x) < d + \varepsilon} > $ > which can be expressed as > $ > B_{open}(o, d + \varepsilon) \cap (B_{closed}(o, d - \varepsilon))^c > $ > which is the intersection of two open sets. Therefore the pre-image of $(d - \varepsilon, d + \varepsilon)$ is an open set, and > $ > \forall x \in f^{-1}((d - \varepsilon, d + \varepsilon)) \subseteq f^{-1}(O) > $ > $ > \exists \delta > 0: B_{open}(x, \delta) \subseteq f^{-1}((d - \varepsilon, d + \varepsilon)) > $ > so is the pre-image of $O$. > > Since every continuous set has a continuous preimage, $d$ with one point fixed is a continuous function.