= Sequence convergence in the cofinite topology
By <convergence of a sequence in a topological space>, $x_n\to a$ means that every <open> <neighbourhood> of $a$ contains all sufficiently late terms. In the <cofinite topology>, this is equivalent to the condition that, for every $b\ne a$, the value $b$ occurs only finitely often. Necessity uses the <open> <neighbourhood> $X\setminus\{b\}$; sufficiency chooses an index beyond every last occurrence of the finitely many points excluded by a given <open> <neighbourhood>.
Consequently a <sequence> with distinct terms converges to every point of $X$. For the original choice $x_n=n$ on $X=\mathbb N_{>0}$, take the two <open> <neighbourhoods>
$$
U_1=X\setminus\{2,5,8\}\ni1,\qquad U_2=X\setminus\{1,3,7\}\ni2.
$$
All terms with $n\geq9$ lie in $U_1$ and all terms with $n\geq8$ lie in $U_2$. More generally, each finite excluded <set> has a last visit, so $x_n\to1$ and $x_n\to2$, and in fact $x_n\to a$ for every $a\in X$. No numerical distance is tending to zero here; the <open sets> specify the meaning of convergence.
\Image[/cofinite-sequence-multiple-neighbourhood-limits.png]
{title=The same <sequence> eventually stays in neighbourhoods of two different limits}
{description=Original sequence $x_n=n$ in the <cofinite topology>. Two sample <neighbourhoods> have different finite exclusions and different last-visit indices. Their memberships illustrate the general finite-exclusion argument; testing just these two neighbourhoods would not by itself prove convergence.}
Distinct terms are sufficient, not necessary. The <sequence> $1,1,2,2,3,3,\ldots$ also converges to every point, since each value occurs only twice. By contrast, the <constant sequence> $x_n=1$ converges only to $1$: an <open> <neighbourhood> of any $a\ne1$ can exclude $1$. The alternating <sequence> $1,2,1,2,\ldots$ converges to no point, since some value different from any proposed limit occurs infinitely often.
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