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  "body": "\n## Proof\n\nWe apply Zorn's lemma to the well-ordered subsets of $A$, ordered by extension at the top.\n\n1. Let $A$ be a set. Let $W$ be the set of pairs $(B, R)$ with $B \\subseteq A$ and $R \\subseteq B \\times B$ a [[well-order|0F1D6PK9RZ]] on $B$. Write $x \\mathrel{R} y$ for $(x, y) \\in R$.\n2. Define $(B, R) \\preceq (B', R')$ when three conditions hold: $B \\subseteq B'$; $R = R' \\cap (B \\times B)$; and $b \\mathrel{R'} b'$ for all $b \\in B$ and $b' \\in B' \\setminus B$.\n3. The relation $\\preceq$ is a [[partial order|011C29Q1S1]]. It is reflexive. If $(B, R) \\preceq (B', R')$ and $(B', R') \\preceq (B, R)$, then $B = B'$ and $R = R' \\cap (B \\times B) = R'$.\n4. Transitivity: let $(B, R) \\preceq (B', R') \\preceq (B'', R'')$. Then $B \\subseteq B''$ and $R = R' \\cap (B \\times B) = R'' \\cap (B \\times B)$, since $B \\subseteq B'$.\n5. Let $b \\in B$ and $b'' \\in B'' \\setminus B$. If $b'' \\in B'$, then $b \\mathrel{R'} b''$, so $b \\mathrel{R''} b''$. If $b'' \\notin B'$, then $b \\mathrel{R''} b''$, since $b \\in B'$. So $(B, R) \\preceq (B'', R'')$.\n6. The set $W$ is nonempty, since $(\\emptyset, \\emptyset) \\in W$. So the empty [[chain|0GP9ASVQ1X]] has an [[upper bound|0H5J02P8S9]].\n7. Let $K$ be a nonempty chain of $(W, \\preceq)$. Put $B^* = \\bigcup_{(B, R) \\in K} B$ and $R^* = \\bigcup_{(B, R) \\in K} R$, two [[unions|05GK32H72H]].\n8. For $(B, R) \\in K$, we have $R^* \\cap (B \\times B) = R$. One inclusion is clear. If $x, y \\in B$ and $x \\mathrel{R'} y$ with $(B', R') \\in K$, then $(B', R') \\preceq (B, R)$ gives $R' \\subseteq R$, and $(B, R) \\preceq (B', R')$ gives $R = R' \\cap (B \\times B)$. Either way $x \\mathrel{R} y$.\n9. Finitely many elements of $B^*$ lie in finitely many members of $K$. These are comparable, so all the elements lie in the largest of these members, $B$ say.\n10. By steps 8 and 9, any three elements of $B^*$ lie in some $B$ where $R^*$ agrees with the total order $R$. So $R^*$ is a [[total order|0316AYH1SC]] on $B^*$.\n11. Let $(B, R) \\in K$, $b \\in B$ and $y \\in B^* \\setminus B$. Pick $(B', R') \\in K$ with $y \\in B'$. Then $(B', R') \\preceq (B, R)$ fails, as $y \\notin B$. So $(B, R) \\preceq (B', R')$, and $b \\mathrel{R'} y$, hence $b \\mathrel{R^*} y$.\n12. The order $R^*$ is a well-order. Let $S \\subseteq B^*$ be nonempty, and pick $(B, R) \\in K$ with $S \\cap B \\neq \\emptyset$. Let $m$ be the [[least element|0T2NTJCV83]] of $S \\cap B$ for $R$.\n13. Let $t \\in S$. If $t \\in B$, then $m \\mathrel{R} t$, so $m \\mathrel{R^*} t$. If $t \\notin B$, then $m \\mathrel{R^*} t$ by step 11. So $m$ is the least element of $S$.\n14. So $(B^*, R^*) \\in W$. By steps 8 and 11, $(B, R) \\preceq (B^*, R^*)$ for each $(B, R) \\in K$. So every chain of $W$ has an upper bound.\n15. By [[Zorn's lemma|04RXTQA5ZA]], $W$ has a [[maximal element|0TZEEFTNCC]] $(M, R_M)$.\n16. Suppose that $M \\neq A$, and pick $a \\in A \\setminus M$. Put $M' = M \\cup \\{a\\}$ and $R' = R_M \\cup \\{(x, a) : x \\in M'\\}$.\n17. Then $R'$ is a total order on $M'$ with $a$ as largest element. A nonempty $S \\subseteq M'$ has a least element: the least element of $S \\cap M$ if this set is nonempty, and $a$ otherwise.\n18. So $(M', R') \\in W$, and $(M, R_M) \\preceq (M', R')$ with $M \\neq M'$. This contradicts maximality. Hence $M = A$, and $R_M$ is a well-order on $A$.\n\n## Idea\n\nLook at all ways to well-order part of $A$, where one is below another when the bigger one only adds new elements on top. A chain of such orders glues into one well-order, because new elements never slip in below old ones. Zorn's lemma gives a maximal one, and it must cover all of $A$, since otherwise we could add a new element on top.\n",
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