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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? $$\lim_{n\to\infty}(n!-2)^2.$$

I have no idea how to think about this problem. Any hint or help.

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1 Answer 1

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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 $n \geq r$.

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  • $\begingroup$ How can you say $4$ is only limit? $\endgroup$
    – prashant dattatrey
    Jan 26, 2022 at 17:35
  • $\begingroup$ @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. $\endgroup$
    – ZSMJ
    Mar 10, 2022 at 10:32
  • $\begingroup$ @AkashGaur I agree with your analysis. $\endgroup$
    – Robert Shore
    Mar 10, 2022 at 16:21
  • $\begingroup$ 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 $4$ ? $\endgroup$
    – nkh99
    Mar 14 at 12:05

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