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  "body": "\n## Proof\n\n1. Let $A$ and $B$ be [[countable sets|0ZG61TFBES]]. By definition, there are [[injective functions|0PVG1CRAJN]] $f \\colon A \\to \\N$ and $g \\colon B \\to \\N$.\n2. Define $h \\colon A \\times B \\to \\N \\times \\N$ on the [[Cartesian product|0C81YBVTWH]] by $h(a, b) = (f(a), g(b))$.\n3. The map $h$ is injective. If $h(a, b) = h(a', b')$, then $f(a) = f(a')$ and $g(b) = g(b')$. So $a = a'$ and $b = b'$, since $f$ and $g$ are injective.\n4. Let $\\pi \\colon \\N \\times \\N \\to \\N$ be the [[Cantor pairing function|0NP5N5PYW5]], $\\pi(x, y) = (x + y)(x + y + 1)/2 + y$. It is a bijection, so it is injective.\n5. The composite $\\pi \\circ h \\colon A \\times B \\to \\N$ is [[injective|01FPXMTBX1]], as a composition of injections.\n6. Hence $A \\times B$ is countable.\n\n## Idea\n\nCountable means \"can be labelled by natural numbers without repeats\". A pair gets a pair of labels, and the Cantor pairing function packs a pair of natural numbers into one natural number without collisions.\n",
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