---- > [!definition] Definition. ([[metric topology]]) > Let $(X,d)$ be a [[metric space]]. Then the collection of all [[epsilon-ball]]s $B_{d}(x, \varepsilon)$, for $x \in X$ and $\varepsilon>0$, is a [[basis for a topology|basis for a topology]] on $X$, called the **metric topology induced by** the [[metric]] $d$. > \ > Explicitly, we define $U \subset X$ to be open in $X$ wrt the [[topological space|topology]] induced by $d$ if for all $y \in U$ there exists $\delta > 0$ s.t. $B_{d}(y, \delta) \subset U.$ > > (We can do the same construction for [[pseudometric|pseudometrics]].) > [!justification] > We need to check that the balls indeed form a [[basis for a topology|basis]]. The first condition is trivial. To see the second condition, let $x \in B_{d}(x_{1}, \varepsilon_{1}) \cap B_{d}(x_{2}, \varepsilon_{2})$ for some $x_{1}, x_{2} \in X$ and $\varepsilon_{1}, \varepsilon_{2}>0$. Define $\varepsilon:=\min(\varepsilon_{1}-d(x,x_{1}), \varepsilon_{2} - d(x,x_{2}) )$, then we have $B_{d}(x, \varepsilon) \subset B_{d}(x_{1}, \varepsilon_{1}) \cap B_{d}(x_{2}, \varepsilon_{2}),$ > for letting $y \in B_{d}(x, \varepsilon)$ we see that >$\begin{align} d(y, x_{1}) \leq & d(y, x) + d(x, x_{1}) \\ < & \varepsilon + d(x,x_{1}) \\ = & \min(\varepsilon_{1}-d(x,x_{1}), \varepsilon_{2} - d(x,x_{2}) ) + d(x,x_{1}) \\ \leq & \varepsilon_{1} - d(x,x_{1}) + d(x,x_{1}) \\ = & \varepsilon_{1} \end{align}$ and thus $y \in B_{d}(x_{1}, \varepsilon_{1})$. Near-identical reasoning shows $y \in B_{d}(x_{2}, \varepsilon_{2})$, hence $y$ lies in the intersection as required. > [!basicexample] > For $(X,d)$ a [[metric space]], prove that the function $\sqrt{ d }: X \times X \to \mathbb{R}$, defined by $\sqrt{d }(x,y)=\sqrt{ d(x,y) }$ for all $x,y \in X$, is a [[metric]] that induces the same [[topological space|topology]] as $d$. ^b45ae9 We verify first that $\sqrt{ d }$ is indeed a metric: > **Lemma. $\sqrt{ \cdot }$ is subadditive: for $a,b \in \mathbb{R}$ we have $\sqrt{ a+b } \leq \sqrt{ a }+\sqrt{ b }$ .** > **Proof of Lemma.** We have $\begin{align} (\sqrt{ a } + \sqrt{ b })^{2} = & a + 2\sqrt{ ab } + b \geq a+b = (\sqrt{ a+b })^{2} \end{align}$ and the result follows. > >- $\sqrt{ \cdot }:[0, \infty) \to [0, \infty)$ is strictly increasing, from which it follows that $\sqrt{ x }=0$ iff $x=0$. Also, the composition of the nonnegative functions $\sqrt{ \cdot }$ and $d$ is again nonnegative. Therefore, $\sqrt{ d }$ is positive definite. >- $\sqrt{ d }(x,y)=\sqrt{ d(x,y) }=\sqrt{ d(y,x) }=\sqrt{ d }(y,x)$, so symmetry holds. >- Let $x,y,z \in X$. Using first that $\sqrt{ \cdot }$ is strictly [[increasing]] (and $d$ follows the [[triangle inequality]]) and then the subadditivity lemma, we have $\begin{align} \sqrt{ d }(x,z) = & \sqrt{ d(x,z) } \\ \leq & \sqrt{ d(x,y) + d(y,z) } \\ \leq & \sqrt{ d(x,y) } + \sqrt{ d(y,z) } \\ = & \sqrt{ d } (x,y) + \sqrt{ d }(y,z). \end{align}$ > Next, we show that the metric $\tau_{d}$ induced on $X$ by $d$ equals the metric $\tau_{\sqrt{ d }}$ induced on $X$ be $\sqrt{ d }$. Let $B_{\varepsilon}^{(d)}(x_{0})$ be the open ball in $\tau_{d}$ of arbitrary radius $\varepsilon>0$ around arbitrary $x_{0} \in X$: $B^{(d)}_{\varepsilon}(x_{0})=\{ x \in X : d(x,x_{0}) < \varepsilon \}.$ Then if we set $\varepsilon':= \sqrt{ \varepsilon }$, we get $B_{\varepsilon'}^{(\sqrt{ d })}(x_{0})=\{ x \in X : \sqrt{ d }(x,x_{0}) < \varepsilon'\}= \{ x \in X : d(x,x_{0}) < \varepsilon \}=B_{\varepsilon}^{(d)}(x_{0}),$ (And likewise, the $\varepsilon$-ball in $(X,\sqrt{ d })$ around $x_{0}$ equals the $\varepsilon^{2}$-ball in $(X,d)$ around $x_{0}$). > This implies that the open balls of radius $\varepsilon$ in $(X,d)$ are the open balls of radius $\sqrt{ \varepsilon }$ in $(X,\sqrt{ d })$. So, the bases of the topologies (and hence the topologies themselves) are the same. > > **Find a specific example of a metric space $(X,d)$ such that $d^{2}= X \times X \to \mathbb{R}$ defined by $d^{2}(x,y)=d(x,y)^{2}$ is not a metric.** Endow $\mathbb{R}^{2}$ with the [[metric]] $d$ induced by the $\ell_{\infty}$ norm: $d( (x_{1},x_{2}), (y_{1},y_{2}) ):=\max \big( |x_{1}-y_{1}|, |x_{2}-y_{2}| \big).$ Then $d^{2}$ is defined as $d^{2}\big( (x_{1},x_{2}) , (y_{1},y_{2}) \big):= \max ^{2} \big( |x_{1}-y_{1}|, |x_{2}-y_{2}| \big).$ We observe that $d^{2}$ violates the [[triangle inequality]]: consider, for example, the points $x:=(10 0, 10 0), y:= (9 9.5, 10 0), z:=(0, 10)$ in $\mathbb{R}^{2}$. We have $d^{2}(x,z)=10 0^{2}$, $d^{2}(x,y)=.5 ^{2}$, and $d^{2}(y,z)=9 9.5^{2}$. But $10 0^{2} > .5^{2} + 9 9.5 ^{2}$. ^47a408 ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```