GRE topology question Consider $\mathbb{Z}$ and let $\mathcal{U}$ be the elements of $\mathcal{P}(\mathbb{Z})$ in arithmetic progressions. So $U \in \mathcal{U}$ if there exists $a \in \mathbb{Z}, b \in \mathbb{Z}^+$ such that $$U = \{a+bn : n \in \mathbb{Z}\}.$$ Let $X$ be the integers with $\mathcal{U}$ as a base. Which of the following are true? also if the sequence $n!$ just $\{1,2,6,24,...\}$
(I) $X$ is metrizable.
(II) The sequence $n!$ converges in $X$.
(III) Addition, multiplication, negation are continuous in $X$. (i.e., $X$ is a topological ring).
Thanks in advance. just a tip or hint would be good. My thoughts are that this is a discrete space? even more so than the integers ? So do We get (I) for free right off the bat? I am a bit unsure about the other two.
 A: Just the continuity question.
Write $U_{a,b}=\{a+bn\}$ for a basis element.
That negation is continuous is easy. If $n(x)=-x,$ then $n^{-1}(U_{a,b})=U_{-a,b}.$
Partial proof just of continuity of addition
Let $s(x,y)=x+y.$ Then we need $$s^{-1}(U_{a,b})=\{(x,y)\in\mathbb Z^2\mid x+y\in U\}$$
to be open in the product topology on $\mathbb Z^{2}.$
Show: $$s^{-1}(U_{a,b})=\bigcup_{c\in \mathbb Z}U_{c,b}\times U_{a-c,b}$$
Multiplication is not much harder
Let $p(x,y)=xy.$
Define $$p^{-1}(U_{a,b})=\{(u,v)\mid uv\equiv a\pmod b\}$$
Then show: $$p^{-1}(U_{a,b})=\bigcup_{(u,v)\in p^{-1}(U_{a,b})} U_{u,b}\times U_{v,b}$$
A: To show (I), define $d(x, y) = 1 / ($the largest $n$ such that $n!$ is a factor of $x - y$). Note that when $x = y$, we suggestively say that the largest $n$ such that $n!$ is a factor of $x - y$ is $\infty$, so the distance will be $\frac{1}{\infty} = 0$. To formalise this, we would define $d$ by doing case analysis on whether $x = y$.
Note that $d(x, y) = 0$ iff $x = y$. Note that $d(x, y) \geq 0$. And note that $d(x, y) = d(x, y)$.
Now we must show $d(x, y) + d(y, z) \geq d(x, z)$. WLOG, assume $x, y, z$ are all distinct (since in any other case, the proof is immediate). Take the largest $n$ such that $n!$ is a factor of $x - y$. Take the largest $m$ such that $m!$ is a factor of $y - z$. WLOG, suppose that $n \geq m$. Then we see that $m!$ is a factor of $x - y$ and $y - z$, hence of $x - z$. Therefore, $d(x, z) \leq \frac{1}{m} = d(x, y) \leq d(x, y) + d(y, z)$.
Thus, we see that $d$ is a metric.
Note that $d(x, y) < \delta$ if and only if there is some $n > 1/\delta$ such that $n!$ is a factor of $x - y$.
It remains to show that the metric space topology induced by $d$ is the topology above. We first show that open balls are open under the given topology. Consider an open ball $B_\delta(m)$ where $\delta > 0$. Then we can write $B_\delta(m) = \bigcup\limits_{n \in \mathbb{N}_+, n > 1/m} \{k \in \mathbb{Z} \mid n!$ is a factor of $k - m\}$. And the set $\{k \in \mathbb{Z} \mid n!$ is a factor of $k - m\}$ is exactly the set $\{m + n! j \mid j \in \mathbb{Z}\}$, which is open. Thus, we see that $B_\delta(m)$ is open.
Conversely, we show all basis sets of the given topology are open with respect to the metric topology.
Consider $S = \{a + bn \mid n \in \mathbb{Z}\}$, where $a \in \mathbb{Z}$ and $b \in \mathbb{Z}_+$. Then consider some $s \in S$. Consider that the set $J_s := \{k \in \mathbb{Z} \mid b!$ is a factor of $k - s\}$ is exactly the set $B_\delta(s)$ where $\delta = \frac{1}{b - 1/2}$, so $J_s$ is open. Then we can write $S = \bigcup\limits_{s \in S} J_s$, and therefore $S$ is open with respect to the metric topology.
Thus, the metric topology is exactly the given topology.
To show (II), we will use the above metric. Note that $d(n!, 0) = \frac{1}{n}$, so we have $\lim\limits_{n \to \infty} d(n!, 0) = \lim\limits_{n \to \infty} \frac{1}{n} = 0$ and hence $\lim\limits_{n \to \infty} n! = 0$.
Now, I'll show (III).
Consider the set $U = \{a + bn \mid n \in \mathbb{Z}\}$ where $b > 0$. Let $\pi : \mathbb{Z} \to \mathbb{Z} / (b)$ be the quotient ring homomorphism. Give $\mathbb{Z} / (b)$ the discrete topology. Note that $\pi$ is a continuous map, since $\pi^{-1}(\{\pi(x)\}) = \{x + bn \mid n \in \mathbb{Z}\}$ is open and the basis sets of the topology of $\mathbb{Z} / (b)$ are of the form $\pi^{-1}(\{\pi(x)\})$.
Note that because $\mathbb{Z} / (n)$ has the discrete topology, all maps out of it are continuous. And also, it must be the case that $(\mathbb{Z} / (n))^2$ has the discrete topology, and hence maps out of this space are also continuous. So the maps $- : \mathbb{Z} / (n) \to \mathbb{Z} / (n)$, as well as $+, \cdot : (\mathbb{Z} / (n))^2 \to \mathbb{Z} / (n)$, are all continuous.
Now consider the function $+ : \mathbb{Z}^2 \to \mathbb{Z}$. Note that $U = \pi^{-1}(\{\pi(a)\})$, and hence we have $+^{-1}(U) = +^{-1}(\pi^{-1}(\{\pi(a)\})) = (\pi \circ +)^{-1}(\{\pi(a)\}) = (+ \circ (\pi, \pi))^{-1})(\{\pi(a)\})$. Since $+ \circ (\pi, \pi)$ is continuous and $\{\pi(a)\}$ is open, we see that $+^{-1}(U)$ is open.
Note that the analogous argument applies for $-$ and $\cdot$.
