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  "body": "\n## Proof\n\nWe check the three properties of a [[partition|09CCE86A6D]].\n\n1. Let $\\sim$ be an [[equivalence relation|0SGTMBEWRF]] on a set $A$. For $a \\in A$, the [[equivalence class|082RMRSY26]] of $a$ is $[a] = \\set{x \\in A \\mid x \\sim a}$. Let $P = \\set{[a] \\mid a \\in A}$.\n2. Each $[a]$ is a subset of $A$ by its definition. So $P$ is a set of subsets of $A$.\n3. Let $a \\in A$. By reflexivity, $a \\sim a$, so $a \\in [a]$. Hence each block $[a]$ is nonempty.\n4. By step 2, the union of the blocks is contained in $A$. By step 3, each $a \\in A$ lies in the block $[a]$. So the union of the blocks is $A$.\n5. Let $a, b \\in A$ and suppose that some $c$ lies in $[a] \\cap [b]$. Then $c \\sim a$ and $c \\sim b$.\n6. In step 5, let $x \\in [a]$, so $x \\sim a$. By symmetry, $a \\sim c$. By transitivity, $x \\sim a$ and $a \\sim c$ give $x \\sim c$.\n7. By transitivity again, $x \\sim c$ and $c \\sim b$ give $x \\sim b$, so $x \\in [b]$. Hence $[a] \\subseteq [b]$.\n8. Exchanging the roles of $a$ and $b$ in steps 6 and 7 gives $[b] \\subseteq [a]$. So $[a]$ and $[b]$ have the same elements, and $[a] = [b]$ by the [[axiom of extensionality|0TB16T2C97]].\n9. By steps 5 to 8, two blocks with a common element are equal. So two different blocks have no common element. Thus the blocks are pairwise [[disjoint|0W4RPA8QH3]].\n10. By steps 3, 4 and 9, the blocks are nonempty, pairwise disjoint, and their union is $A$. So $P$ is a partition of $A$.\n\n## Idea\n\nReflexivity puts each element in its own class, so the classes are nonempty and cover $A$. Symmetry and transitivity show that two classes sharing one element have exactly the same elements. So distinct classes cannot overlap.\n",
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