Topology Induced by Metric / Equivalence of Metrics

虽然 topology 的定义已经摆脱了 metric，但是这并不妨碍我们从 topology 的角度去研究 metric，这并不存在矛盾。

打个比方，研究 topology 就好比研究 “路程长短” 这个问题，一个具体的 topology $\tau$ 就好比 “SF to LA 比 LA to SD 远” 这么一个结论，这个结论并不关心你是用 km 还是 mile 去量的距离，但是这个结论对 km 和 mile 都成立。

Topology Induced by MetricPermalink

Let $(X,d)$ be a metric space, and let $\tau$ be the collection of all subsets of $X$ that are open in the metric space sense. $\tau$ can be called the topology induced (or generated) by $d$.

Equivalence of MetricsPermalink

Topological EquivalencePermalink

Two metrics $d_1$ and $d_2$ are said to be topologically equivalent if they generate the same topology on $X$.

• The adjective “topological” is often dropped.

This is equivalent to: a subset $A\subseteq X$ is $d_1$-open $⟺$ it is $d_2$-open.

Strong equivalence is a sufficient but not necessary condition for topological equivalence, i.e. $\text{Strong Equivalence }\Rightarrow \text{ Topological Equivalence}$

Strong EquivalencePermalink

Two metrics $d_1$ and $d_2$ are strongly equivalent (or vaguely metrically equivalent) $⟺$ there exist positive constants $\alpha$ and $\beta$ such that, for every $x,y\in X$,

ExamplePermalink

Let $x\in {\mathbb{R}}^n$, written out in componenet form as $x=(x_1,\dots ,x_n)$.

• The Euclidean norm of $x$ is $\Vert x\Vert =\sqrt{x_1^2+\cdots +x_n^2}$
• The sup norm of $x$ is $|x|=\underset{i}{max}|x_i|$

Similarly we obtain the Euclidean distance $\Vert x-y\Vert$ and sup distance $|x-y|$. It can be proved that,