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  "body": "\n## Proof\n\nWe argue by contradiction, with nested closed intervals.\n\n1. Suppose that $\\R$ is [[countable|0ZG61TFBES]]. Then there is an [[injective function|0PVG1CRAJN]] $j \\colon \\R \\to \\N$.\n2. Fix $n \\in \\N$. Since $j$ is injective, at most one real number $x$ satisfies $j(x) = n$.\n3. Let $[a, b]$ be a closed [[interval|0TK8Y3J5RW]] with $a \u003c b$, and put $d = (b - a)/3$. Let $L = [a, a + d]$ and $R = [b - d, b]$.\n4. Since $a + d \u003c b - d$, the intervals $L$ and $R$ are disjoint. By step 2, at least one of them contains no $x$ with $j(x) = n$.\n5. Define $T(n, [a, b])$ to be $L$ if $L$ contains no $x$ with $j(x) = n$, and $R$ otherwise. By step 4, $T(n, [a, b])$ is a closed subinterval of $[a, b]$ of length $d \u003e 0$ that contains no $x$ with $j(x) = n$.\n6. By [[definition by recursion|0DSHAVH45F]], there are closed intervals $I_n = [a_n, b_n]$ with $I_0 = [0, 1]$ and $I_{n+1} = T(n, I_n)$ for each $n \\in \\N$. Each has $a_n \u003c b_n$.\n7. By step 5, $I_{n+1} \\subseteq I_n$ for each $n$, and $I_{n+1}$ contains no real $x$ with $j(x) = n$.\n8. By the [[nested intervals theorem|MF56GK7QY2]], applied to $I_1 \\supseteq I_2 \\supseteq \\cdots$, there is a real number $y$ with $y \\in I_n$ for all $n \\ge 1$.\n9. Let $m = j(y)$. Then $y \\in I_{m+1}$ by step 8. But $I_{m+1}$ contains no $x$ with $j(x) = m$ by step 7. This is a contradiction.\n10. Hence there is no injection $\\R \\to \\N$, and $\\R$ is [[uncountable|0X55M9XET4]].\n\n## Idea\n\nEach natural number $n$ labels at most one real number. At stage $n$ we shrink the interval to a third that avoids that real. The completeness of $\\R$ gives a point in all the intervals, and this point has avoided its own label, which is absurd.\n",
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