Continuous self-maps in the Golomb space that are neither increasing nor decreasing

Question

Let $\mathbb{N}$ denote the set of the positive integers. The Golomb space is a space ${\bf G} =(\mathbb{N},\tau)$ where a basis of $\tau$ is generated by $$\big\{\{a+bn: n\in \mathbb{N}\cup\{0\}\}: a,b\in\mathbb{N} \text{ and } a,b \text{ relatively prime}\big\}.$$ Is there a continous map $f: {\bf G}\to {\bf G}$ that is neither increasing ($n<m \in \mathbb{N}$ implies $f(n) \leq f(m)$) nor decreasing ($n<m \in \mathbb{N}$ implies $f(n) \geq f(m)$)?

Answer 1

For polynomials with non-negative integer coefficients and no constant term, the following simple (but not obvious) fact was observed by Paulina Szczuka.

Theorem. Each polynomial $f:\mathbb N\to\mathbb N$, $f:x\mapsto a_1x+a_2x^2+\dots+a_nx^n$, with integer coefficients and no constant term is continuous in the Golomb topology on $\mathbb N$.

Proof. Take any number $x\in\mathbb N$ and a basic neighborhood $f(x)+b\mathbb N_0$ of its image in the Golomb topology.

Since $f(x)$ is divisible by $x$, the number $b$ is relatively prime with $x$, so $x+b\mathbb N_0$ is a well-defined basic neighborhood of $x$ in $\mathbb G$. Observe that $$f(x+n\mathbb N_0)\subset f(x+b\mathbb Z)\subset f(x)+b\mathbb Z.$$

Since $f(\mathbb N)\subset\mathbb N$ and $f(0)=0$, the polynomial $f$ is not constant, so, for any $y\in Y$ in the finite set $Y:= (f(x)+b\mathbb Z)\setminus (f(x)+b\mathbb N_0)$ the set $f^{-1}(y)$ is finite. The neighborhood $O_x=(x+b\mathbb N_0)\setminus \bigcup_{y\in Y}f^{-1}(y)$ of $x$ in the Golomb topology has the required property: $f(O_x)\subset f(x)+b\mathbb N_0$.

Corollary, The polynomial $f:\mathbb G\to\mathbb G$, $f:x\mapsto x^3-12x^2+45x$, on the Golomb space is continuous but it is neither increasing nor decreasing (since $f(1)=34$, $f(3)=54$ and $f(5)=50$).

Remark. It can be shown (see Theorem 5 in this paper) that the semigroup $S(\mathbb G)$ of all continuous self-maps of the Golomb space has cardinality $|S(\mathbb G)|=\mathfrak c$.