= Discontinuous additive function
Assuming the <axiom of choice>, <Cauchy's functional equation> has solutions that are not <linear maps> over the <real numbers>. Regard $\mathbb R$ as a <vector space over the rational numbers>. Because $\sqrt2$ is an <irrational number>, $1$ and $\sqrt2$ are <linearly independent> over $\mathbb Q$. <Zorn's lemma> extends them to a <Hamel basis> $B$; assigning arbitrary real outputs to its <basis vectors> and extending by finite rational <linear combinations> gives an <additive function>.<Zorn’s Lemma and Some Applications by Keith Conrad>[§4, printed pp. 19–20, Corollaries 4.2 and 4.5: basis extension and an additive map not of the form cx]{ref}
For a particular choice, assign $g(1)=3$ and $g(b)=0$ for every $b\in B\setminus\{1\}$. Every <real number> has a unique finite <linear combination> $x=\sum_{b\in F}q_bb$ with $F\subset B$ finite and $q_b\in\mathbb Q$. Define
$$
g(x)=3q_1,
$$
where the coefficient $q_1$ is zero if $1\notin F$. The uniqueness of the <Hamel basis> representation makes this definition unambiguous, and adding the coefficients proves $g(x+y)=g(x)+g(y)$. On the rational two-dimensional <vector subspace> spanned by $1$ and $\sqrt2$, the rule is particularly simple:
$$
g(p+q\sqrt2)=3p\qquad(p,q\in\mathbb Q).
$$
For instance, $u=2/3+\sqrt2$ and $v=5/6-2\sqrt2$ give $g(u)=2$, $g(v)=5/2$, and $g(u+v)=9/2$. Thus the map still preserves <addition>, even though $g(\sqrt2)=0\ne3\sqrt2$ rules out being a <linear map> over the <real numbers>.
The rational approximants $q_n$ from <additive function continuous at one point is linear> expose the failure of <continuity>: $q_n\to\sqrt2$ but $g(q_n)=3q_n\to3\sqrt2\ne g(\sqrt2)$. Moreover $h_n=\sqrt2-q_n\to0$ while $g(h_n)=-3q_n\to-3\sqrt2$. The map is <continuous> nowhere, since <continuity> at any point would imply <additive function continuous at one point is linear>. It is unbounded on every <interval> of positive length by <additive function bounded on an interval is linear>.
The <axiom of choice> is used to obtain the <Hamel basis>; the coefficient rule does not provide an explicit way to find the representation of an arbitrary <real number>. A plot sampled only at <rational numbers> would show exactly the same values as $f(x)=3x$, so that plot cannot reveal this <discontinuous additive function>.
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