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  "body": "\n## Counterexample\n\nLet $f_n(x) = x^n$ on $[0, 1]$. Let $f(x) = 0$ for $0 \\le x \u003c 1$ and $f(1) = 1$.\nFor $0 \\le x \u003c 1$, $x^n \\to 0$ by the [[geometric sequence limit|0MNM54WB99]]. So $f_n \\to f$ [[pointwise|056FSKG32J]] on $[0, 1]$.\nBut $\\sup_{0 \\le x \u003c 1} |x^n - 0| = 1$ for every $n$. So the convergence is not [[uniform|08FPABDQ4C]].\n\n## Remarks\n\nEach $f_n$ is continuous, but $f$ is not. This also shows the failure, by the [[uniform limit theorem|0X9XT05JVC]].\nOn $[0, r]$ with $0 \u003c r \u003c 1$, the convergence is uniform, since $\\sup_{[0, r]} x^n = r^n \\to 0$.\n\n## Explanations\n\n### L1 human\n\nNear $x = 1$, the values $x^n$ stay close to $1$ for a long time before they drop to $0$.\nSo no single $n$ makes all the values small at once.\n",
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