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  "body": "\n## Proof\n\nWe apply Zorn's lemma to a set of chains ordered by inclusion.\n\n1. Let $(P, \\le)$ be a [[partially ordered set|011C29Q1S1]] and let $C_0$ be a [[chain|0GP9ASVQ1X]] of $P$.\n2. Let $Q$ be the set of chains $C$ of $P$ with $C_0 \\subseteq C$. Order $Q$ by inclusion $\\subseteq$, which is a partial order. The set $Q$ is nonempty, since $C_0 \\in Q$.\n3. Let $K$ be a chain of $(Q, \\subseteq)$. If $K$ is empty, then $C_0$ is an [[upper bound|0H5J02P8S9]] of $K$ in $Q$.\n4. Let $K$ be nonempty, and let $W = \\bigcup_{C \\in K} C$ be the [[union|05GK32H72H]] of its members. Then $C_0 \\subseteq W$.\n5. The set $W$ is a chain of $P$. Let $x, y \\in W$, with $x \\in C$ and $y \\in D$ for some $C, D \\in K$. Since $K$ is a chain, $C \\subseteq D$ or $D \\subseteq C$.\n6. So $x$ and $y$ lie in the larger of $C$ and $D$, which is a chain of $P$. Hence $x \\le y$ or $y \\le x$.\n7. So $W \\in Q$, and $C \\subseteq W$ for each $C \\in K$. Thus $W$ is an upper bound of $K$ in $Q$.\n8. By steps 2 to 7, [[Zorn's lemma|04RXTQA5ZA]] applies to $(Q, \\subseteq)$. It gives a [[maximal element|0TZEEFTNCC]] $M$ of $Q$.\n9. Then $M$ is a chain with $C_0 \\subseteq M$. Suppose that $D$ is a chain of $P$ with $M \\subsetneq D$.\n10. Then $C_0 \\subseteq D$, so $D \\in Q$, and $M \\subsetneq D$ contradicts the maximality of $M$ in $Q$.\n11. So $M$ is a [[maximal chain|WZKPDWBW2Z]] of $P$ that contains $C_0$.\n\n## Idea\n\nChains that contain $C_0$, ordered by inclusion, satisfy the hypothesis of Zorn's lemma: a nested family of chains has a union that is again a chain. A maximal element of this family is a chain that cannot be enlarged, which is exactly a maximal chain.\n",
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