Uniqueness of limit of convergent sequence I know that limit of convergent sequence is unique for some spaces like metric spaces, Hausdorff spaces, etc. Is there any space the limit of the convergence of sequence is not unique?
-Thanks.
 A: Here is another informative example. 
Consider $X=\mathbb R$ with the open sets $\tau=\{ \mathbb R,\emptyset \}$. Then a given sequence $(x_n)_{n\in \mathbb N}$ converges to every real number!
A: Let $X$ be an infinite space, endowed with the cofinite topology (the closed sets are the finite sets).  A sequence $(x_n)$ converges in this space under one of three conditions:

*

*The sequence is eventually constant, meaning that there is some $N$ such that $x_n = x_N$ for all $n \ge N$.  In this case, the sequence converges to $x$ (this sequence has a unique limit).


*The sequence does not attain any value more than a finite number of times.  In this case, the situation is a little more subtle
Claim:  Let $X$ be an infinite set endowed with the cofinite topology.  Suppose that $(x_n)$ is a sequence in $X$ in which each term appears only finitely many times, and let $x \in X$ be arbitrary.  Then $x_n \to x$.
Proof:  Let $U$ be a neighborhood of $x$.  As $U$ is open in the cofinite topology, its complement,
$$X \setminus U = \{u_1, u_2, \dots, u_m\},$$ is finite.  For each $j = 1, 2, \dots, m$, let $n_j$ denote the largest natural number such that
$$ x_{n_j} = u_j,$$
(set $n_j = 0$ if $u_j$ does not appear anywhere in the sequence), and choose $N > \max\{n_j\}$.  If $n > N$, then $x_n \in U$, which implies that $x_n \to x$, as claimed. $\square$
Note that this demonstrates that such a sequence converges, but that the limit of the sequence is not unique.


*The sequence attains one value, $x$, infinitely many times, but no other value is attained infinitely many times.  In this case, the sequence will converge to $x$ (uniquely).  The argument is similar to that in (2), but needs to be modified to take the infinite constant subsequence into account.

A sequence in such a space which attains two (or more) different values an infinite number of times will fail to converge.  To see this, let $x, x' \in X$ and suppose that $(x_n)$ is a sequence such that both $x$ and $x'$ appear infinitely often.  Note that this implies that for any $N$, there exists some $m > N$ such that $x_m = x$, and $n > N$ such that $x_n = x'$.  Let $y \in X$ and suppose that $y \ne x$.  Let $U = X \setminus \{x\}$.  Notice that $x_n$ cannot converge to $y$:  for any $N$, there exists some $n > N$ such that $x_n = x \not\in U$.  If $y = x$, then take $U = X \setminus \{x'\}$ and apply the same reasoning.
Thus the three categories of sequences listed above are the only convergent sequences in an infinite space with the cofinite topology.
A: Suppose $X$ is infinite and has the co-finite topology. That is, $S \subset X$ is closed if and only if $|S|$ is finite or $S=X$. Then, if $s_n$ is a sequence in $X$ with this topology, then $s_n\rightarrow s$ for every $s \in X$.
So, limits are not unique.
NB: If $X$ is finite, then the cofinite topology is just the discrete topology.
A: To make this thing clear. As shown in this post, a sequence converges in cofinite topology if and only if there is at most one value in the sequence that is repeated infinitely many times. In the case that there is no term repeated infinitely many times, the sequence converges to every point of the space.
A: Let X be a non empty set of cardinality greater than 1. Then in indiscrete topological space (X,I) every sequence converges to every point of X and hence limit of convergence sequence is not unique.
