Ukrainian Mathematical Competitions

Suppose that the set $A$ is more than countable set. Divide half-plane $\mathbb{R}\times (0,+\infty )$ to the counted number of rectangles: $$D_{n,m}=[n,n+1)\times \left[\frac{1}{m+1},\frac{1}{m}\right),n\in \mathbb{Z},m\in \mathbb{N},D_{n,0}=[n,n+1)\times [1,+\infty ).$$ By construction, ${\bigcup }_{n\in \mathbb{Z}}{\bigcup }_{m\in {\mathbb{Z}}^{+}}{D}_{n,m}=\mathbb{R}\times (0,+\infty )$, in addition, any two of determined earlier rectangles do not intersect. Consider the graph $G=\{(x,f(x))\mid x\in A\}$ of the function $f$. Because $A$ is more than countable, the set $G$ is also more than countable ($A\sim G$). Suppose that $\forall n\in \mathbb{Z},m\in {\mathbb{Z}}^{+}$ (i.e. the set $G\cap D_{n,m}$) is finite, but then the whole set $G$ is at most countable, that contradicts its construction. Therefore, $\exists n_0\in \mathbb{Z},m_0\in {\mathbb{Z}}^{+}:G\cap D_{n_0,m_0}$ is infinite set. This means that there is an infinite set $X\subset [n_0,n_0+1)\cap A$ such that $\forall x\in X$ $f(x)>\frac{1}{m_0+1}$. Then due to Bolzano-Weierstrass theorem there is a sequence of different elements that converges to some number. As a result, this sequence is fundamental and $\forall k\in \mathbb{N}$ $f(x_k)>\frac{1}{m_0+1}$. But then we have the following relationship: $0={\lim }_{k\to \infty }|x_{k+1}-x_k|$ and $\min \{f(x_k),f(x_{k+1})\}\ge \frac{1}{m_0+1}$, which contradicts the problem's condition.

Thereby, $A$ should be no more than countable. Now determine the required function. Let $A=\{x_1,x_2,...\}$. Define the function by induction: $f(x_1)=1$. Now let us know $f(x_1),...,f(x_n)$. Let us make the following notation: $r={\min }_{k=1,n}|x_{n+1}-x_k|>0$. Let's set $f(x_{n+1})=\frac{r}{2}$. We should verify that the function which is defined for $A$ satisfies the conditions of the problem.

Example: for $A=\mathbb{Q}$: if $x=\frac{p}{q}$, where $(p,q)=1$, $p\in \mathbb{Z}$, $q\in \mathbb{N}$ ($0=\frac{0}{1}$), we will set $f\left(\frac{p}{q}\right)=\frac{1}{q^2}$. Then if $\frac{p_1}{q_1}\ne \frac{p_2}{q_2}$ we have: $$|x_1-x_2|=\frac{|p_1{q}_2-q_1{p}_2|}{q_1{q}_2}\ge \frac{1}{q_1{q}_2}\ge \frac{1}{\max \{q_1^2,q_2^2\}}=\min \{f(x_1),f(x_2)\}.$$