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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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What have you tried? –  akkkk Nov 29 '12 at 17:04
    
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. –  FDP Nov 29 '12 at 17:12
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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. –  FDP Nov 29 '12 at 17:29
    
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 $1$. –  Brian M. Scott Nov 29 '12 at 19:52
    
Ok, I try an answer, thanks for the hint. –  FDP Nov 30 '12 at 15:12
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2 Answers

We know that a sequence $\{x_n\}$ converges iff there exists a neighborhood base containing eventually the sequence. So, we can try to consider a neighborhood base $\mathcal{N}_x$ formed by $V_\epsilon = B_\epsilon (x) \setminus C$, where $B_\epsilon (x)$ is the ball with the usual topology and $C$ a countable set such that $x \notin C$.

Then, if a sequence is eventually equal to $x$, it eventually belongs to any $V_\epsilon$; conversely, if a sequence isn't eventually equal to $x$, choosing $C = \cup_{n \in \mathbb{N}} \{x_n\} \setminus \{x\}$ we have that $\{x_n\}$ doesn't belong to any $V_\epsilon$.

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Recall the definition of convergence of a sequence in a general topological space:

$\{x_n\}_{n\in\mathbb N}$ converges to $x$ if and only if for every open set $U$ such that $x\in U$ there is some $k\in\mathbb N$ such that for all $n>k$ we have $x_n\in U$. That is, $\{x_n\}$ is eventually in $U$ for every open neighborhood $U$ of $x$.

One implication is trivial, of course that eventually constant sequences converge. (You should still write that, though.)

Now suppose that $\{x_n\}$ is not eventually constant. Suppose that $x$ was the limit of $\{x_n\}$, let $V$ be an open interval around $x$, then $U=V\setminus\{x_n\neq x\mid n\in\mathbb N\}$ is an open set in our topology. Because the sequence is not eventually constant we can find arbitrarily large $n\in\mathbb N$ such that $x_n\notin U$. However $x\in U$ and $U$ is open so $x_n$ is not eventually in $U$, which is a contradiction to the definition of convergence.

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