Is the divisor topology from Steen-Seebach contractible?

Question

The divisor topology appears as Counterexample 57 in [Steen-Seebach]. It is a topological space $X := \{x \in \mathbb{N} : x \ge 2\}$ with the topology generated by open sets $U_n := \{x \in X : x\ \mathrm{divides}\ n\}$, $n \in X$. Is the divisor topology contractible?

A topological space is ultraconnected if it has no disjoint nonempty closed sets. The closure of a singleton $\{x\} \subset X$ is equal to the set of multiples of $x$ ([Steen-Seebach, 57.2]), so it follows that $X$ is ultraconnected. Hence if the divisor topology is not contractible, then it will provide a counterexample to 'ultraconnected $\Rightarrow$ contractible', which is why I am thinking about this. It is known that an ultraconnected space is path-connected, which is a necessary condition for contractible. A stronger condition than ultraconnected is the existence of a point $p \in X$ such that the only neighborhood of $p$ is $X$; this is equivalent to $p$ being an element of every nonempty closed set. It is known that this stronger condition does imply that $X$ is contractible. However, the previous remark about closures of singletons shows that $X$ does not satisfy this stronger condition.

Here are some other remarks I thought might be worth sharing, just because they helped me understand the space a bit better.

• Every neighborhood of a point $x \in X$ contains $U_x$ as a subset, so it follows that $X$ is an Alexandrov space (arbitrary intersections of open sets are open).
• ([Steen-Seebach, 57.1]) $X$ is $T_0$ (topological distinguishability) but not $T_1$ (points are closed), since $x < y$ implies $y \notin U_x$, but every neighborhood of $6$ contains $3$.
• ([Steen-Seebach, 57.4]) The set of primes is dense in $X$, and each singleton containing a prime is open.

Steen, Lynn Arthur; Seebach, J. Arthur jun., Counterexamples in topology. 2nd ed, New York - Heidelberg - Berlin: Springer-Verlag. XI, 244 p. (1978). ZBL0386.54001.

Answer 1

Define $G : X \to X$, $$G(x) := \begin{cases} x & x \mathrm{\ is\ even} \\ 2x & x \mathrm{\ is\ odd,} \end{cases}$$ and $F : X \times [0, 1] \to X$, $$F(x, t) := \begin{cases} x & t < 1 \\ G(x) & t = 1.\end{cases}$$

Let $x \in X$ and $t \in [0, 1)$. Since $U_x$ is an open subset of every neighborhood of $F(x, t) = x$, and $U_x \times [0, 1)$ is an open neighborhood of $(x, t)$ in $X \times [0, 1]$, and lastly because $F(U_x \times [0, 1)) = U_x$, it follows that $F$ is continuous at $(x, t)$. Since $U_{G(x)}$ is an open subset of every neighborhood of $F(x, 1) = G(x)$, and $$F(U_x \times [0, 1]) = U_x \cup G(U_x) \subset U_{G(x)},$$ it follows that $F$ is continuous at $(x, 1)$. Together these imply that $F$ is continuous. For $x \in \overline{\{2\}} = \{\mathrm{evens}\}$, $F(x, 1) = G(x) = x$. And likewise for odd $x$, $F(x, 1) = 2x \in \overline{\{2\}}$. So $F$ is a deformation retraction of $X$ onto $\overline{\{2\}}$, and it follows that $X$ is homotopy equivalent to $\overline{\{2\}}$. Since $\overline{\{2\}}$ has a generic point (a point $p$ for which $\{p\}$ is a dense subset), it follows that $\overline{\{2\}}$ is contractible; see the end of David Gao's answer. Therefore $X$ is contractible.