# Limit in a topology.

Let $$\mathbb{Z}$$ denote the set of integers. For $$c$$ and $$r$$ in $$\mathbb{Z}$$, define: $$B(c, r):=\{c+kr\ |\ k\in \mathbb{Z}\}.$$ As $$c$$ varies over all integers and $$r$$ over all positive integers, the sets $$B(c, r)$$ form a basis for a topology on $$\mathbb{Z}$$.

Does the following limit exist with respect to this topology? $$\lim_{n\to\infty}(n!-2)^2.$$

The limit is $$4$$. For each $$n \in \Bbb N$$, the sequence element is $$4$$ plus a multiple of $$n!$$. Any basis element that contains $$4$$ can be expressed in the form $$\{ 4+ kr \mid k \in \Bbb Z \}$$, which will contain all elements of the sequence with $$n \geq r$$.
• How can you say $4$ is only limit? Commented Jan 26, 2022 at 17:35
• @data I think you can show that the space is $T_2$ hence the limit is unique. For distinct $x$ and $y$ there exists a prime power $p^n$ such that $p^n|x$ and $p^n\nmid y$. Then $\{x+kp^n|k\in \mathbb Z\}$ and $\{y+k'p^n|k'\in \mathbb Z\}$ are disjoint open nbds of $x$ and $y$ since $x+kp^n=y+k'p^n \implies y=x+(k-k')p^n$ but only the R.H.S. is divisible by $p^n$. Perhaps Mr. Shore could confirm.
• Using the same logic by showing that the limit of $n!-2$ is $2$, can we conclude that the limit of the given sequence is $4$ ? Commented Mar 14, 2023 at 12:05