A pseudometric keeps most metric axioms but permits distinct points to have zero distance. This occurs naturally in analysis when two objects differ only on a negligible set or when a formula ignores part of the data. The quotient construction repairs this by identifying all zero-distance points and then defining a genuine metric on equivalence classes. :::definition[Pseudometric] Let $X$ be a nonempty set. A function $d:X\times X\to\mathbb{R}$ is called a $pseudometric$ on $X$ if for all $x,y,z\in X$: (i) $d(x,y)\geq 0$. (ii) If $x=y$, then $d(x,y)=0$. (iii) $d(x,y)=d(y,x)$. (iv) $d(x,y)\leq d(x,z)+d(z,y)$. A pseudometric may have $d(x,y)=0$ even when $x\neq y$. ::: :::example Let $X=\mathbb{R}^2$ and define $d(x,y)=|x_2-y_2|$ for $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Then $d$ is a pseudometric but not a metric. ::: :::proof Given that $d(x,y)=|x_2-y_2|$ on $\mathbb{R}^2$. To prove that $d$ is a pseudometric but not a metric. The first, third, and fourth conditions follow from the absolute value on the second coordinate. If $x=y$, then $x_2=y_2$, so $d(x,y)=0$. However, for $x=(0,1)$ and $y=(2,1)$, we have $x\neq y$ and $d(x,y)=0$. Hence $d$ is not a metric. $\square$ ::: :::example Let $X$ be the set of all Riemann integrable functions on $[a,b]$. Define $$\begin{equation*} d(f,g)=\int_a^b |f(x)-g(x)|\,dx. \end{equation*}$$ Then $d$ is a pseudometric on $X$. It is not a metric on the full class of Riemann integrable functions because functions that differ at finitely many points may have zero integral distance. ::: :::definition[Equivalence Relation Induced by a Pseudometric] Let $(X,d)$ be a pseudometric space. Define a relation $\sim$ on $X$ by $$\begin{equation*} x\sim y \iff d(x,y)=0. \end{equation*}$$ This is the $equivalence relation induced by the pseudometric$. ::: :::proof Given that $x\sim y$ if and only if $d(x,y)=0$. To prove that $\sim$ is an equivalence relation. Reflexivity follows from $d(x,x)=0$. Symmetry follows from $d(x,y)=d(y,x)$. If $x\sim y$ and $y\sim z$, then $d(x,y)=0$ and $d(y,z)=0$. By the triangle inequality, $$\begin{equation*} d(x,z)\leq d(x,y)+d(y,z)=0. \end{equation*}$$ Since $d(x,z)\geq 0$, we get $d(x,z)=0$, so $x\sim z$. Hence $\sim$ is an equivalence relation. $\square$ ::: :::theorem[Quotient Metric Induced by a Pseudometric] Let $(X,d)$ be a pseudometric space and define $x\sim y$ if and only if $d(x,y)=0$. On $X/\sim$, define $$\begin{equation*} \widetilde d([x],[y])=d(x,y). \end{equation*}$$ Then $\widetilde d$ is a metric on $X/\sim$. ::: :::proof Given that $\widetilde d([x],[y])=d(x,y)$. To prove that $\widetilde d$ is a metric. If $[x]=[x']$ and $[y]=[y']$, then $d(x,x')=0$ and $d(y,y')=0$. The triangle inequality gives $d(x,y)\leq d(x',y')$, and the reverse inequality follows similarly. Thus $\widetilde d$ is well-defined. The metric axioms then follow from the pseudometric axioms, and identity of indiscernibles follows from $$\begin{equation*} \widetilde d([x],[y])=0\iff x\sim y\iff [x]=[y]. \end{equation*}$$ Hence $\widetilde d$ is a metric. $\square$ ::: :::exercise [1] State the difference between a metric and a pseudometric. [2] For $d(x,y)=|x_2-y_2|$, give distinct points at distance zero. [3] Define the quotient metric induced by a pseudometric. ::: :::answer [1] A metric requires zero distance only between equal points; a pseudometric may assign zero distance to distinct points. [2] $(0,1)$ and $(2,1)$ have distance zero. [3] It is $\widetilde d([x],[y])=d(x,y)$ on the equivalence classes determined by $d(x,y)=0$. ::: :::call-to-action[Continue to Extended Spaces] subtitle: Next, study metrics on spaces obtained by adding points at infinity. button: Next Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/metrics-on-extended-spaces button-ghost: Previous Lesson | /functional-analysis/metric-spaces/introduction-to-metric-spaces/bounded-metrics-from-existing-metrics :::