_For this proof, both [partitions](Equivalence%20class.md#Partitions%20of%20sets) and [equivalence classes](Equivalence%20class.md) are denoted with capital letters in "Euler font," (e.g. $\mathscr{E}$), equivalence relations with lowercase Euler font (e.g. $\mathscr{e}$) and sets and elemnts of sets they apply to are denoted in the default manner (e.g. $a\in A$). We then show that every equivalence relation is a partition on a set and vice versa._ ^a81c48 Consider a [partition,](Equivalence%20class.md#Partitions%20of%20sets) $\mathscr{P}$ on a [set,](Sets.md) $A.$ We show below that it gives rise to exactly one [equivalence relation,](Equivalence%20relations.md) $\mathscr{E}.$ ^578037 * Define a [relation](Relations%20on%20sets.md) $\mathscr{c}$ on $A,$ $x\mathscr{c}y,$ where $x$ and $y$ are related solely by the fact that they belong to the same element, $\mathscr{p}$, in the partition $\mathscr{P}.$ ^31ffba * Since this is the case, it is also true that $y\mathscr{c}x.$ Thus, this relation is [symmetric.](Equivalence%20relations.md#^81607a) ^a97740 * Since $A$ is also the [union](Union%20of%20sets.md) of elements of the [partition,](Equivalence%20class.md#Partitions%20of%20sets) the $\mathscr{P},$ the [relation](Relations%20on%20sets.md) still holds for $y\mathscr{c}y$ and $x\mathscr{c}x$ on $A$ and thus [reflexivity](Equivalence%20relations.md#^f197c0) holds. * No $x$ or $y$ belongs to more than one element of $\mathscr{P},$ since [partitions](Equivalence%20class.md#Partitions%20of%20sets) are [disjoint,](Intersection%20of%20sets.md#Disjoint%20sets) thus if $x,y,z\in\mathscr{P}$ where $x\mathscr{c}y$ and $y\mathscr{c}z$ implies that $x\mathscr{c}z$ holds and the [relations](Relations%20on%20sets.md) is [transitive.](Equivalence%20relations.md#^0d57a3) * Thus $\mathscr{c}$ is an [equivalence relation.](Equivalence%20relations.md) * Here we show that $\mathscr{c}$ is the _only_ [equivalence relation](Equivalence%20relations.md) that arises from the partition $\mathscr{P}.$ To do this we show that if for $x\in A$ we define equivalence relations $y\mathscr{c}_1 x$ and $y\mathscr{c}_2 x$ then it must be the case that $\mathscr{c}_1=\mathscr{c}_2.$ * Define corressponding [equivalence classes](Equivalence%20class.md) as $\mathscr{E}_1$ and $\mathscr{E}_2,$ * $\mathscr{E}_1$ is an element of $\mathscr{P}$ and thus must be equal to a unique element of $\mathscr{P},$ $\mathscr{p},$ that also contains $x.$ $\mathscr{E}_2$ must then also equal $\mathscr{p}.$ * $\mathscr{E}_1$ must also consist of $y$ such that $y\mathscr{c}_1 x$ and $\mathscr{E}_2$ must also consist of $y$ such that $y\mathscr{c}_2 x.$ * Since $\mathscr{E}_1=\mathscr{p}=\mathscr{E}_2$ then the equivalence relation must be unique. Therefore [equivalence classes](Equivalence%20class.md) and [partitions of sets](Equivalence%20class.md#Partitions%20of%20sets) are equivalent to each other. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\blacksquare$ %%This proof is on pgs 23 and 24 of Munkres topology and also answers the question posed by exercise 1.2.%% ^ad566a #MathematicalFoundations/SetTheory #Proofs