Let $p_n$ denote the $n$th prime throughout and define $p_n\# := p_n p_{n-1} \cdots p_1$ to be the $n$th "primorial".
Define the topology $\tau$ on $\Bbb{Z}/p_n\#$ to be that generated by the basis $B = \{ X_d : d \mid p_n\#\}$ and $X_d = \{ x \in \Bbb{Z}/p_n\#: x^2 \neq 1 \pmod q, \forall \text{ primes } q \mid d\}$. Then clearly $X_c \cap X_d = X_{\text{lcm}(c,d)}$.
The full space $\Bbb{Z}/p_n\#$ is given vacuously by $X_1 = \{ x\in \Bbb{Z} : x^2 \neq 1 \pmod q, \forall q \mid 1\}$ but there are no $q \mid 1$ t check, so $X_1 =$ the whole space.
Clearly a sub-basis for this topology is given by $B = \{X_q$ such that $q$ is a prime and $q \mid p_n\#\}$.
The monoids $M_{d }= \{ x \in \Bbb{Z}/p_n\# : x^2 = 1 \pmod {d}\}$ for $d \mid p_n\#$ are the complements:
$$ M_d = \bigcap_{q \mid d} X\setminus X_q\\ X := \Bbb{Z}/p_n\# $$ And are thus closed submonoids of $(\Bbb{Z}/p_n\#)^{\times}$ in this topology. $M_{p_n\#}$ is a group.
Now let $I = [0, \dots, p_{n+1}^2 - 2]$ for sufficiently large $n$ that all these residues occur distinctly within $\Bbb{Z}/p_n\#$ as $\overline{0}, \overline{1}, \dots, \overline{p_{n+1}^2 - 2}$, etc. It doesn't take a large $n$ because primorial grows so fast.
Conjecture. The set $\{0\}$ is not open in the subspace topology $I \cap \tau = \{ I \cap G : G = \bigcup_{d} X_d \text{ is open in } \tau\}$.
This is true because $X_1 \cap I = I$ and $X_d \cap I$ contains ?
That is my question, is this provable?
Further Attempt. Clearly there exist primes $q \in I$ that are coprime to $p_n\#$, in which case some power $q^k \in M_{p_n\#} =X \setminus \bigcup_{q \mid p_n\#}X_q$.
This Might Help.
Point of accumulation of a subspace $S$
They say $x \Bbb{Z}/p_n\# := S$ is a point of accumulation for subspace $I$ if and only if for all open balls $X_d$, $X_d \cap I$ contains a distinct other point $y \neq x$. This is almost exactly what is sought!
Namely, I want to prove that $0$ is a point of accumulation for $I$.