Two sequences, same set, different limit I recently saw a user write a sequence as $\{x_n\}_{n=1}^{\infty}$. This is generally bad notation, since it could lead people to think of the sequence as a set rather than as a sequence, although we (hopefully) all know what they mean.
However, this raises the question: when does the set $\{x_n: n \in \mathbb{N}\}$ determine the limit $\lim_{n \rightarrow \infty} x_n$? Clearly, when the set is finite, it can't, e.g. $(1, 0, 0, 0, ...)$ and $(0, 1, 1, 1, ...)$. But if not? More precisely:
Let $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ be real sequences such that:


*

*$\lim \limits_{n \rightarrow \infty} x_n$ and $\lim \limits_{n \rightarrow \infty} y_n$ both exist 

*$\{x_n: n \in \mathbb{N}\} = \{y_n: n \in \mathbb{N}\} =: \mathcal{Z}$

*for all $z \in \mathcal{Z}$, the sets $\{n: x_n = z\}$ and $\{n: y_n = z\}$ are both finite. 
Then do we always have $\lim \limits_{n \rightarrow \infty} x_n = \lim \limits_{n \rightarrow \infty} y_n$?
 A: If $\lim_{n\to\infty}x_n$ exists, then it is a limit point of $\mathcal{Z}$, because of your finiteness condition. Conversely, if $\mathcal{Z}$ has more than one limit point, then so does $(x_n)$, hence $\lim_{n\to\infty}x_n$ doesn't exist. So $\mathcal{Z}$ has exactly one limit point. By symmetry, this is also the limit of the sequence $(y_n)$.
A: Two points:


*

*To me, writing $\{x_n\}_{n=1}^\infty$ is a clear and unambiguous sign that you are talking about a sequence and not a set. It's not bad notation, it's just different from yours.

*The answer to your question is, I think, yes:
Let $x$ be the limit of $\{x_n\}_{n=1}^\infty$ and let $\epsilon > 0$. Then, we know there exists such an $N$ that for $n>N$, we have $|x_n-x|<\epsilon$.
We also know that only a finite number of elements of $\{y_n\}_{n=1}^\infty$ are equal to $x_1,x_2,\dots, x_N$. This means that there exists some $M$, such that for all $m>M$, we know that $y_m\notin \{x_1,\dots, x_N\}$. But that means that for all $m>M$, $y_m=x_{n'}$ for some $n'>N$, meaning that $$|x-y_m| = |x-x_{n'}|<\epsilon$$
so $x$ is also the limit of $\{y_n\}_{n=1}^\infty$
A: Let us denote $x=\lim_n x_n$ and $y=\lim_n y_n.$ By definition, given $\epsilon>0$ there exists $N\in \mathbb{N}$ such that
$$n\ge N \implies |x_n-x|< \epsilon \quad \mathrm{and} \quad |y_n-y|< \epsilon.$$ We have that, if $n,m\ge N$ then
$$|x-y|\le |x_n-x|+|y_m-y|+|x_n-y_m|<|x_n-y_m|+2\epsilon.$$ Now, since $\{x_n:n\in\mathbb{N}\}=\{y_n:n\in\mathbb{N}\}$ and $\{n: x_n = z\}$ and $\{n: y_n = z\}$ are finite, there exist $x_n$ and $y_m$ with $n,m\ge N$ such that $y_m=x_n.$ Thus,
$$|x-y|<2\epsilon.$$ Since $\epsilon$ is arbitrary we can conclude that $x=y.$
A: I think your criterion work (at least when the topology is Hausdorff). Because if you take $x$ (resp. $y$) to be the limit of $(x_n)$ (resp. $(y_n)$) then you have that $x\in Adh(\{x_n|n\in\mathbb{N}\})$ the adherence of the set of the sequence $(x_n)$ (you have the same for $(y_n)$.
Now you know (here you use that the topology is Hausdorff and also that the topology has a countable system of neighborhood in each point) that :
$$Adh(\{x_n|n\in\mathbb{N}\})=\{x_n,n\in\mathbb{N}\}\cup \{\text{limits of converging subsequences}\}=\{x_n,n\in\mathbb{N}\}\cup \{x\} $$
Because of what you suppose $x$ cannot be in $\{x_n,n\in\mathbb{N}\}$ so the union is disjoint. But now, for the same reason :
$$Adh(\{y_n|n\in\mathbb{N}\})=\{y_n,n\in\mathbb{N}\}\cup \{y\} $$
And because :
$$Adh(\{x_n|n\in\mathbb{N}\})=Adh(\{y_n|n\in\mathbb{N}\})$$
and :
$$\{x_n,n\in\mathbb{N}\}=\{y_n,n\in\mathbb{N}\}$$
You get that the extra point  $x$ from $Adh(\{x_n|n\in\mathbb{N}\})$ to  $\{x_n|n\in\mathbb{N}\}$ and the extra point $y$ from $Adh(\{y_n|n\in\mathbb{N}\})$ to $\{y_n|n\in\mathbb{N}\}$ must be equal.
I have made strong assumption about the topology (Hausdorff and countable system of neighborhoods) this is verified if your topological space is a metric space. I think it doesn't work if you leave out the countable system of neighborhood but leaving Hausdorff for a $T_1$-topology should work as well (a $T_1$-topology is a topology for which every points are closed).
