Nonexamples:: *[[Nonexamples]]* Constructions:: *[[Constructions|Used in the construction of...]]* Specializations:: *[[Specializations]]* Generalizations:: [[Euclidean diffeomorphism]] Justifications and Intuition:: *[[Justifications and Intuition]]* Examples:: [[euclidian metric]], [[sup Metric]], [[norms induce metrics|norm-induced metric]] > [!definition] > Given a set $X$, a [[metric]] on $X$ is a function $d: X \times X \to \mathbb{R}$ such that the following properties hold for all $x,y,z\in X$ — > >1. $d(x,y) = d(y,x)$; >2. $d(x,y) \geq 0$ with equality iff $x=y$; > 3. $d(x,z)\leq d(x,y) + d(y,z)$. > [!basicexample] > > Let $\sim$ be the [[equivalence relation]] on $\mathbb{R}$ such that for all $x,y \in \mathbb{R}$, we have $x \sim y$ if and only if there exists $\ell \in \mathbb{Z}$ s.t. $x=y+\ell$. Let $d$ be the [[Euclidean metric]] on $\mathbb{R}$. Then $\sim$ induces a [[metric]] $\overline{d}$ on $\mathbb{R} / \sim$ given by $\overline{d}(\overline{x}, \overline{y})= \inf \{ d(x, y+k) : k \in \mathbb{Z} \}$ for $\overline{x}=[x], \overline{y}=[y]$ in the [[quotient set]] $\mathbb{R} /\sim$. For example, $\overline{d}([1.1], [5.4])=d(1.1, 1.4)=0.3$. > We verify: > To ensure $\overline{d}$ is [[well-defined]], we should verify that $\overline{d}(\overline{x}, \overline{y})$ does not depend on the representatives chosen for $\overline{x}$ and $\overline{y}$, i.e., that $ \inf \{ d(x, y+k) : k \in \mathbb{Z}\}=\inf \{ d(x', y' + k) : k \in \mathbb{Z} \}$ whenever $x' \sim x$ and $y' \sim y$. But since $x'=x+a$ for some $a \in \mathbb{Z}$ and $y'=y+b$ for some $b \in \mathbb{Z}$, the above can be rewritten as $ \inf \{ d(x, y+k) : k \in \mathbb{Z}\}=\inf \{ d(x + a, y + b + k) : k \in \mathbb{Z} \}$ and the [[infimum]] on the RHS is unchanged by our integer-shifting of $x$ and $y$ (all we are doing is reindexing $\mathbb{Z}$ with $k-b+a$). Now, > >1. $\overline{d}(\overline{x}, \overline{y}) \geq 0$ because the infimum of a nonnegative set of real numbers is nonnegative. $\overline{d}(\overline{x}, \overline{x})=\inf\{ d(x, x+k) : k \in \mathbb{Z}\}=0$ as witnessed by $k=0$, and conversely if $\inf\{ d (x,y+k) : k \in \mathbb{Z} \}=0$ then because $k$ is an integer we have that the infimum is achieved for some $\ell \in \mathbb{Z}$: $x=y+\ell$. Then $x \sim y$ and thus $\overline{x}=\overline{y}$. >2. That $\overline{d}(\overline{x}, \overline{y})=\overline{d}(\overline{y}, \overline{x})$ is a direct consequence of the symmetry of $d$: $\begin{align} \overline{d}(\overline{x}, \overline{y}) = & \inf \{ d(x, y+k) : k \in \mathbb{Z}\} \\ = & \inf\{ d (y+k, x) : k \in \mathbb{Z}\} \\ = & \inf \{ d(y, x+k) : (-k) \in \mathbb{Z} \} \\ = & \overline{d}(\overline{y}, \overline{x}). \end{align}$ >3. Let $z \in \mathbb{R}$. We'll use translation-invariance of the [[Euclidean metric]]. We have $\begin{align} \overline{d}(\overline{x}, \overline{z}) = & \inf_{k \in \mathbb{Z}} \{ d (x, z+k)\} \\ \leq & \inf_{m,k \in \mathbb{Z}} \{ d(x, y+m) + d(y+m, z+k) \} \\ & \overbrace{\leq}^{k':=k-m} \inf_{m \in \mathbb{Z}} \{ d (x, y+m) \} + \inf_{k' \in \mathbb{Z}} \{ y, z+k' \} \\ = & \overline{d}(\overline{x}, \overline{y}) + \overline{d}(\overline{y}, \overline{z}). \end{align}$ ^99fe49 > [!basicnonexample] > It is not always the case that we can obtain a 'quotient metric' in such a way. As an counterexample, endow $X:=\left\{ \frac{1}{n} \right\}_{n \in \mathbb{N}} \cup \{ 0 \}$ with the [[metric]] $d(x,y):=|x-y|$. '' Next, define an [[equivalence relation]] $\sim$ on $\mathbb{N}$ via the [[partition]] $\left\{ \frac{1}{n} \right\}_{n \in \mathbb{N}} \sqcup \{ 0 \}$. The space $X / \sim$ has two elements, say, $[1]$ and $[0]$. Now, $\overline{d}([1], [0])= \inf \left\{ | \frac{1}{n} - 0 |: n \in \mathbb{N} \right\}=0$ which violates positive definiteness. ^8fd8aa