Is the interlocking interval topology contractible?

Question

This is intended to be a self-answering question, which is allowed on StackExchange sites (see here).

The interlocking interval topology is the topology on $X = (0, \infty) \setminus \mathbb{Z}$ generated by the subbasis consisting of the following sets $S_n$, where $n$ ranges over all positive integers:

$$S_n = (0, \frac{1}{n}) \cup (n, n + 1)$$

It is defined as Counterexample #54 in the classic text Counterexamples in Topology.

It is known that this space is path-connected. See here or this answer on MSE. A natural further question is whether this space is actually contractible.

It should also be noted that the question of whether certain classes of non-$T_1$ spaces are contractible or not has received some attention on MSE recently. See, for example, this question, this question, and this question. The current question can be seen as an effort to further complete the project of classifying when non-$T_1$ but highly connected spaces are contractible. (As the interlocking interval topology is hyperconnected, see here.) Moreover, if this space is not contractible, it would provide another example of nonempty hyperconnected path-connected space that is not contractible, other than the space constructed in this answer of mine.

Answer 1

The space is contractible. We shall first show that there is a strong deformation retract from $X$ to the subspace $(0, 1)$ of $X$. Indeed, the following map $F: X \times [0, 1] \to X$ is such a strong deformation retract:

$$F(x, t) = \begin{cases} x &, \text{ if } x \in (0, 1) \text{ or } t = 0\\ \frac{1}{n + 1} &, \text{ if }x \in (n, n + 1) \text{ for some positive integer } n \text{ and }t > 0 \end{cases}$$

That $F(x, 0) = x$, $F(x, 1) \in (0, 1)$, and $F(a, t) = a$ for all $x \in X$, $a \in (0, 1)$, and $t \in [0, 1]$ is obvious, so it suffices to show $F$ is continuous. For this, we note that,

$$\begin{split} F^{-1}(S_n) &= S_n \times \{0\} \cup ((S_n \cap (0, 1)) \times (0, 1]) \cup \left(\left(\bigcup_{m = n}^\infty (S_m \cap (1, \infty))\right) \times (0, 1]\right)\\ &= S_n \times [0, 1] \cup \left(\left(\bigcup_{m = n}^\infty S_m\right) \times (0, 1]\right) \end{split}\tag{1}$$

where in the second equality we have used that $S_n = (S_n \cap (0, 1)) \cup (S_n \cap (1, \infty))$ and $S_m \cap (0, 1) \subset S_n \cap (0, 1)$ for $m \geq n$. The RHS of $(1)$ is open in $X \times [0, 1]$, so $F$ is indeed continuous.

Now, to conclude the proof, we note that the subspace $(0, 1)$ of $X$ has a focal point (i.e., a point whose only open neighborhood is the entire space, or equivalently it is contained in all nonempty closed sets), for example, $\frac{3}{4}$, as $\frac{3}{4} \in S_n$ iff $n = 1$ and $S_1 \cap (0, 1) = (0, 1)$. Thus, the subspace $(0, 1)$ is contractible, see, for example, the Theorem in this answer of mine. This shows $X$ is contractible as well.