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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 $\lim \limits_{n \rightarrow \infty} x_n = \lim \limits_{n \rightarrow \infty} y_n$?

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    $\begingroup$ The notation $\{x_n\}_{n=1}^{\infty}$ is not to be interpreted as $\{x_n: n \in \mathbb{N}\}$. That would be disgusting. $\endgroup$ – Git Gud Mar 23 '15 at 12:16
  • $\begingroup$ If $x_n\to x$ and $y_n\to y$ in $\Bbb R$ and $\{x_n:n\in \Bbb N\}=\{y_n:n\in\Bbb N\}$, then $\{x_n\}$ can be treated as a subsequence of $\{y_n\}$ (after rearrangement of terms, if necessary) and so $x_n\to y$. Thus $x=y$. I do not understand what is the role of the last condition. $\endgroup$ – user149418 Mar 23 '15 at 12:27
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    $\begingroup$ @user149418 Be careful. Since we are here anyway talking about convergence of sequences, you might as well state that this is only true under $\Bbb R$ with the usual (standard) topology. If $\Bbb R$ is equipped with a different topology, we could have $x \neq y$. I'm 100% sure people won't like that I'm bringing this up since "it's so obvious we are only talking about the standard topology here" but it's worth mentioning that $\Bbb R$ can be equipped with topologies where a sequence can converge to multiple different points. $\endgroup$ – layman Mar 23 '15 at 12:29
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    $\begingroup$ @user149418 It's to try to exclude things like $(0, 1, 1, 1, ...)$ and $(1, 0, 0, 0, ...)$, which are obviously counterexamples. $\endgroup$ – Christopher Mar 23 '15 at 12:37
  • $\begingroup$ Suppose we regarded instead the objects $\{x_n : n\in\mathbb{N} \}$ and $\{y_n : n\in\mathbb{N} \}$ as multisets, i.e. for each member we kept track of its multiplicity. Then I guess we could drop the criterion that each sequence attains a value $z$ at most finitely many times. Now suppose these two multisets were equal (as multisets). It would be true that $(x_n)_{n=1}^\infty$ was convergent if and only $(y_n)_{n=1}^\infty$ was, and their limits would agree in that case. Unless I overlook something? $\endgroup$ – Jeppe Stig Nielsen Mar 23 '15 at 14:44
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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 $(y_n)$.

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  • $\begingroup$ You might want to consider pathological topologies to do this. For instance if your topology is not $T_0$ then your sequence can converge toward more than $1$ point. Furthermore the link between the limit points of the set given by $(x_n)$ with its limit is "good" if and only if you have a countable system of neighborhoods in each point of your topological space. $\endgroup$ – Clément Guérin Mar 23 '15 at 12:34
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    $\begingroup$ @ClémentGuérin: "Let $(x_n)_{n=1}^{\infty}$ and $(y_n)_{n=1}^{\infty}$ be real sequences..." And unless otherwise stated, the reader is entitled to assume the usual topology on $\mathbb R$. $\endgroup$ – TonyK Mar 23 '15 at 12:37
  • $\begingroup$ Of course, my mistake I just read the beginning. $\endgroup$ – Clément Guérin Mar 23 '15 at 12:40
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Two points:

  1. 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.

  2. 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 $\{y_n\}_{n=1}^\infty$

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  • $\begingroup$ I've definitely seen undergraduates write "$x \in \{x_n\}$". Although, maybe even that's not as terrible a thing to do as my first reaction to it thinks it is. $\endgroup$ – Christopher Mar 23 '15 at 12:33
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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 $x=y.$

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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).

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