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A constant sequence has the same value $a$ at every index. It converges to $a$ by convergence of a sequence in a topological space.

Choose an uncountable set $X$ and a point $a\in X$. Every point other than $a$ is isolated, and a set containing $a$ is open precisely when its complement is countable. Equivalently, the closed sets are all subsets containing $a$ together with all countable subsets. Arbitrary intersections and finite unions preserve this latter condition, so it defines a topology.

Every convergent sequence in this space is eventually equal to its limit. Indeed the set of its values different from its limit is countable and closed; its complement is an open neighbourhood of the limit. Eventual membership in this complement forces eventual equality. Nevertheless $a$ is not isolated, because its singleton has uncountable complement. In particular sequence convergence alone does not determine this topology.

A real-valued function on this space is continuous exactly when $\{x:f(x)\ne f(a)\}$ is countable. Sufficiency follows by considering the preimages of closed subsets of the real line: each either contains $a$ or is a subset of that countable exceptional set. For necessity, each set $\{x:|f(x)-f(a)|\ge 1/n\}$ is closed and excludes $a$, hence is countable; their countable union is the exceptional set.

Convergence of a sequence to $\ell$ means that every open neighbourhood $U$ of $\ell$ contains all sufficiently late terms: there is $N$ such that $x_n\in U$ for all $n\ge N$. The threshold may depend on $U$.

Set $S=\{x_n:n\in \mathbb{N},x_n\ne \ell \}$. This is a countable set, hence a closed set in the stated topology, and $\ell \notin S$. Consequently $X\setminus S$ is an open neighbourhood of $\ell$. By convergence of a sequence, all sufficiently late terms lie outside $S$. A sequence term outside $S$ must equal $\ell$. Thus the sequence is eventually constant, and only finitely many indices have $x_n\ne \ell$.

The definition of convergence of a sequence in a topological space to $\ell$ is that every open neighborhood $U$ of $\ell$ contains all sufficiently late terms: there exists $N$ such that $x_n\in U$ for every $n\ge N$.

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 neighbourhoodsAll terms with $n\ge 9$ lie in $U_1$ and all terms with $n\ge 8$ lie in $U_2$. More generally, each finite excluded set has a last visit, so $x_n\to 1$ and $x_n\to 2$, 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.

Distinct terms are sufficient, not necessary. The sequence $1,1,2,2,3,3,\dots$ 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 $a\ne 1$ can exclude $1$. The alternating sequence $1,2,1,2,\dots$ converges to no point, since some value different from any proposed limit occurs infinitely often.