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In the previous post, we were looking for a monster—a nonlinear additive function. We found that such a function is extremely pathological: it is nowhere locally monotone, nowhere continuous and nowhere locally bounded. Worse than that, it's easy to prove that the graph of a monster is dense in \(\R \times \R\), that is, for every \(x\) and \(y\), an arbitrary neighborhood of \(x\) contains a point that monster sends arbitrarily close to \(y\).
Recall our attempt to construct a monster. Any additive function is linear on any \(\Q\)-set and is fully determined on this set by a value it has in any single of its points. Our Direchlet function derived monster failed (or rather fell) because the slopes an additive function has on different \(\Q\)-sets are not independent. Indeed, given that \(f\) has a slope \(k_1\) on a \(\Q\)-set \(\Q\cdot\alpha_1\) and a slope \(k_2\) on a \(\Q\)-set \(\Q\cdot\alpha_2\), it has to have a slope \(k_1 + k_2\) on a \(\Q\)-set \(\Q\cdot(\alpha_1 + \alpha_2)\). This shows a way to construct a monster: one has to find a collection \(B\) of real numbers \(r_1, r_2, \ldots\) such that (i) every real number can be represented as a sum \(q_1\cdot r_1 + q_2\cdot r_2 + \ldots\), with rational coefficients \(q_1, q_2, \ldots\) of which only finite number is non-zero (so that the sum is defined) and (ii) that such representation is unique. Then one can select arbitrary values on elements of \(B\) and take moster's value on \(q_1\cdot r_1 + q_2\cdot r_2 + \ldots\) to be \(q_1\cdot f(r_1) + q_2\cdot f(r_2) + \ldots\), which is well-defined thanks to (ii).
Looks familiar? It should be: the definition of \(B\) is exactly the definition of a basis of a vector space. Real numbers can be added to each other and multiplied by rationals and, therefore, form a vector space over \(\Q\). This space is very different from a usual one-dimensional vector space real numbers form over \(\R\) (i.e., over themselves).
After a streak of bad and unlikely properties that a monster has, we now got something positive: a monster exists if and only if \(\R\) as a vector space over \(\Q\) has a basis. Does it?
But of course. Any vector space has a basis—this is a general theorem almost immediately following from the Zorn's lemma. The basis we are looking for even got a name of its own: Hamel basis.
At last we stumbled across the whole family on monsters. Specifically, there exists a set \(B \subset \mathbb{R}\) and a function \(q : \mathbb{R}\times B \rightarrow \Q\) such that every real number r can be uniquely represented as
$$r = \displaystyle\sum_{b\in B}q(r, b)\cdot b$$where only finite number of \(q(r, b)\) are non-zero for a given \(r\). From this it immediately follows that \(q(r_1 + r_2, b) = q(r_1, b) + q(r_2, b)\).
Take an arbitrary function \(f_0 : B \rightarrow \mathbb{R}\), and define
$$f(r) = \displaystyle\sum_{b\in B} f_0(b)\cdot q(r, b)\cdot b$$ $$\begin{array}{r@{\;}c@{\;}l@{\quad}} f(r_1) + f(r_2) &\;=\;& \\ &\;=\;& \displaystyle\sum_{b\in B} f_0(b)\cdot q(r_1, b)\cdot b + \displaystyle\sum_{b\in B} f_0(b)\cdot q(r_2, b)\cdot b \\ &\;=\;& \displaystyle\sum_{b\in B} f_0(b)\cdot\left(q(r_1, b) + q(r_2, b)\right)\cdot b \\ &\;=\;& \displaystyle\sum_{b\in B} f_0(b)\cdot q(r_1 + r_2, b) \cdot b \\ &\;=\;& f(r_1 + r_2) \end{array}$$that is, \(f\) is additive. Intuitively, \(f_0(b)\) is a slope \(f\) has at the \(\Q\)-set \(\Q\cdot b\). \(f\) is linear if and only if \(f_0\) is a constant function, in all other cases \(f\) is a monster. If one takes \(f_0 : b \mapsto 1/b\), then
$$f(r) = \displaystyle\sum_{b\in B} q(r, b)$$is an especially weird monster function: it takes only rational values!
Note that almost all additive functions are, after all, monsters—only very small sub-set of them is linear.