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Consider the topology on $\mathbb{R}$ in which a set is open if $U = V \setminus C$, where $V$ is open in the usual topology and $C$ is a countable set.

Prove that in this space a sequence converges iff it is eventually constant.

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    Mh, I got that in this space $\overline{(0,1)} = [0,1]$ but no sequence in $(0,1)$ converges to either 0 or 1. That was a sort of previous exercise, but I cannot link it with the claim I wrote in the question.2012-11-29
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    If $\{x_n\}$ is a sequence in $(0,1)$ then I can consider $(0,2) \setminus \cup_{n \in \mathbb{N}} \{x_n\}$, which is a neighborhood of 1 containing no point of $\{x_n\}$. So the sequence does not converge to 1 as it would do with the usual topology. I think (and hope) it's correct.2012-11-29
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    Your last comment is *almost* correct: it’s correct if none of the $x_n$’s is $1$, but if some $x_n=1$, then your open set isn’t a nbhd of $1$, because $1$ isn’t in it. However, $\{1\}\cup\left((0,2)\setminus\{x_n:n\in\Bbb N\}\right)$ is an open nbhd of $1$, and it contains a tail of the sequence if and only if there is an $m$ such that $x_n=1$ for all $n\ge m$. This shows that the sequence converges to $1$ iff it’s eventually constant at $1$. Now generalize this to all points of $\Bbb R$, not just $1$.2012-11-29
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    Ok, I try an answer, thanks for the hint.2012-11-30

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