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  "body": "\n## Statement\n\nAssume the [[axiom of choice|0HTGB13FE2]].\nLet $I$ be a [[countable|0ZG61TFBES]] set and let $(A_i)_{i \\in I}$ be a [[family of sets|05Q6G4EH2G]] such that each $A_i$ is countable.\nThen the [[union|05GK32H72H]] $\\bigcup_{i \\in I} A_i$ is countable.\n\n## Remarks\n\nProof: choose for each $i$ a surjection from $\\mathbb{N}$ onto $A_i$, when $A_i$ is nonempty. Together they give a surjection from a subset of $\\mathbb{N} \\times \\mathbb{N}$ onto the union, and $\\mathbb{N} \\times \\mathbb{N}$ is countable.\nThe proof uses the axiom of countable choice to choose the surjections. Without any choice, the statement cannot be proved in ZF.\nWhen the enumerations of the $A_i$ are given, no choice is needed.\n\n## Explanations\n\n### L1 human\n\nList each set $A_i$ as a row of a grid.\nThen list the grid diagonal by diagonal.\nEach element of the union appears in the list.\n",
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