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  "body": "\n## Example\n\nView $\\R$ as a vector space over the field $\\Q$ of rational numbers.\nThe numbers $1$ and $\\sqrt{2}$ are linearly independent over $\\Q$, because $\\sqrt{2}$ is irrational.\nBy the [[axiom of choice|0HTGB13FE2]], this independent set extends to a [[basis|0KCQ9Y3QKA]] $B$ of $\\R$ over $\\Q$, called a Hamel basis.\nEvery real $x$ is then a unique finite sum $x = \\sum_{b \\in B} q_b(x)\\, b$ with rational coefficients $q_b(x)$, almost all zero.\nDefine $f \\colon \\R \\to \\R$ by $f(x) = q_1(x)$, the coefficient of the basis vector $1$.\n\nCheck of additivity: the coefficients of $x + y$ are the sums of the coefficients of $x$ and $y$, so $f(x + y) = f(x) + f(y)$.\nSo $f$ is an [[additive function|KF719SZD5T]].\nIt is not of the form $f(x) = cx$: $f(1) = 1$ would force $c = 1$, but $f(\\sqrt{2}) = 0 \\ne \\sqrt{2}$.\nHence $f$ is not [[continuous|0X3WGX95FE]], since every continuous additive function is linear.\nOne can also see this directly: $f$ takes only rational values, while a continuous function on $\\R$ with only rational values is constant, and $f(0) = 0 \\ne 1 = f(1)$.\n\n## Explanations\n\n### L1 human\n\nImagine writing every real number in a strange coordinate system with infinitely many rational coordinates.\nThe function $f$ just reads off one coordinate.\nReading a coordinate respects addition, so $f$ is additive.\nYet $f$ jumps around wildly and cannot be drawn, because the coordinate system itself comes from the axiom of choice.\n",
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