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

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 $f(n) \geq f(m)$)?

• Related to a question in mathoverflow.net/a/285890/14094 (if there's no such map, then the homeomorphism group of $\mathbf{G}$ is trivial). – YCor Nov 13 '17 at 10:10
• The Golomb space admits many continuous increasing maps, in particular, for any $a\in\mathbb N$ the multiplication map $\mathbb G\to\mathbb G$, $x\mapsto ax$, is continuous. But I do not know other maps (shifts are no continuous). – Taras Banakh Nov 13 '17 at 10:33
• Constant maps are also increasing in the sense indicated by Dominic – YCor Nov 13 '17 at 11:13
• Each (continuous) decreasing map $f:\mathbb G\to\mathbb G$ has finite image (and hence is constant) since the set $\mathbb G=\mathbb N$ is well-ordered. So, the question is equivalent to the existing a non-constant continuous map which is not increasing. – Taras Banakh Nov 13 '17 at 19:42
• By the way, the Golomb space is a topological semigroup with respect to the multiplication of natural numbers. So, at least the maps $f(x)=ax^n$ are continuous. But all such maps are increasing. – Taras Banakh Nov 16 '17 at 16:22

## 1 Answer

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 $|S(\mathbb G)|=\mathfrak c$.

• This argument works for every $f:\mathbb{N}\to\mathbb{N}$ such that $n|f(n)$ for all $n$ and $b|f(n+b)-f(n)$ for all $n,b$. How many such maps are there? – YCor Nov 21 '17 at 8:00
• A smaller basis for the topology consists of the $a+b\mathbb{N}_0$ for $a,b$ coprime and $1\le a<b$ (i.e. I discard those with "holes" such as $3+2\mathbb{N}_0$). Using this basis makes life simpler and avoids an unnecessary argument involving finite fibers. (btw you probably mean $Y=(f(x)+b\mathbb{Z})\cap (\mathbb{N}\smallsetminus (f(x)+b\mathbb{N}_0))$) – YCor Nov 21 '17 at 8:12
• @YCor Thank for your comment. Indeed, the set $Y$ had wrong definition. Now it is fixed. Concerning the cardinality of the set of self-maps of the Golomb space I am sure that it is continuum (the number of branches of some recursively defined tree of partial finite functions) but the proof of this fact still escapes from me (I hope not for long time). – Taras Banakh Nov 21 '17 at 8:22
• Can you elaborate about your new remark (that there are $2^{\aleph_0}$ self-maps)? – YCor Nov 24 '17 at 18:19
• @YCor This is a bit long. I will add the proof of this fact to the update of the corresponding paper in arXiv. – Taras Banakh Nov 25 '17 at 8:46