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Let $\tau$ be the cofinite topology on $\mathbb{Z}$. Let $(x_n)_n = \{1,2,3,\dots\}$ be a sequence. How can we show that this sequence converges to each point of $\mathbb{Z}$?

My attempt:

Let $k$ be an integer. Then $x_n \rightarrow k$ if any open set containing $k$ contains all $x_n$ for all $n \ge N_k$ for some $N_k \in \mathbb{N}$. Set $M$ to be the maximum of elements that is not contained in an open set except $X$. Then chosing $N_k$ as $M$ gives the result.

I can not find any other idea rather than the one above and it looks somehow wrong to me. My first question is the proof of the theorem and second question is that how can we describe convergent sequences on $(\tau, \mathbb{Z})$ in general?

Ninja
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  • Any non-empty open set in $\tau$ must contain a set of the form ${n\in \mathbb{Z}:n\ge m}$ for some $m\ge 1$. Prove this is equivalent to being a cofinite set (i.e. complement is finite) then you can see that your sequence is automatically convergent to any member of $\mathbb{Z}$. Then it's easy to see that any convergent sequence must be convergent to $\infty$ in standard topology induced from $\mathbb{R}$ – user160738 Feb 14 '17 at 00:48
  • Can I say the following? Let $U$ be open so $U^c$ must be finite. Then, $U^c = {a_1, \dots, a_n}$. Since $U \cup U^c = \mathbb{Z}$, any integer greater than max$(U^c)$ must be in $U$. – Ninja Feb 14 '17 at 00:52
  • Yes, as long as $U\neq \emptyset$ – user160738 Feb 14 '17 at 00:54
  • Thank you! Lastly, why a sequence must be convergent to $\infty$ in standard topology induced from $\mathbb{R}$? – Ninja Feb 14 '17 at 01:21
  • Actually, that wasn't entirely correct. If sequence $x_n$ converges to $a$, then either $x_n$ is eventually $a$ (i.e. for some $k$, $x_n=a$ for all $n\ge k$) or $x_n\to \infty$. This can be easily seen from the definition of convergence: $x_n\to a$ means for all open $U$ containing $a$ there is $k$ such that $n\ge k$ implies $x_n\in U$ so either $x_n\to\infty$ or if that's not the case then $x_n=a$ eventually – user160738 Feb 14 '17 at 01:58
  • Check out http://math.stackexchange.com/a/2117586/4280 which has a characterization of the convergent sequences. – Henno Brandsma Feb 14 '17 at 12:35

1 Answers1

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Rememeber that the open sets in the cofinite topology are the complements of the finite sets. Now if $U\subseteq \mathbb{Z}$ is open then it is of the following form $$U=\mathbb{Z}\setminus \lbrace n_1,\ldots , n_k\rbrace.$$ Now consider a sequence $(x_n)_{n\in\mathbb{N}}$ such that $\sup x_n=\infty$, we claim that every point of $\mathbb{Z}$ is a limit of $(x_n)$. Chosen a neightborhood of $x\in\mathbb{Z}$ we have $(x_n)\cap U=\lbrace x\in (x_n): x\geq \max n_i\rbrace$, i.e. the sequence $(x_n)$ is finally contained in every open neightborhood of $x$.

Vincenzo Zaccaro
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