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This is related to a previous post of mine (link) regarding how to show that for any sequence $\{x_{n}\}$, the limit superior of the sequence, which is defined as $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}$, is equal to the to supremum of limit points of the sequence. Below I think I have found a counter-example to this (although I know I am wrong but just don't know where!).

Define the sequence $x_{n}=\sin(n)$. We have $\text{sup }_{k\geq n} x_{k}=1$ for any $k\geq 1$ (i.e. for any subsequence). Thus $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}=1$. Now the sequence has no limit points since it does not converge to anything so the supremum of all the subsequence limits is the supremum of the empty set which presumably is not equal to 1 (I actually do not know what it is). So we have a sequence where the supremum of limit points does not equal $\text{inf}_{n\geq 1}\text{sup }_{k\geq n} x_{k}=1$?

Any help with showing where this is wrong would be much appreciated.

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A limit point of a sequence is a number that some subsequence of the sequence converges to. This is a different notion than the limit of a sequence. (how do you know the sup is 1?) –  David Mitra Nov 10 '11 at 18:51
Every point in $[-1,1]$ is a limit point of the sequence $(\sin(n))$. –  TonyK Nov 10 '11 at 18:52
Thank-you David and Tony for these comments. I have made a very lazy error in assumig a limit point is the limit of a sequence. As you point out since [-1,1] is the set of limit points of $x_{n}=sin(n)$ then the supremum of the set is 1 which agrees with the limit superior. Also David I have looked at a graph of the function and made a gues that the supremum of any subsequence is 1 - I have not tried a formal proof. –  dandar Nov 10 '11 at 20:56

2 Answers 2

up vote 2 down vote accepted

The problem is that you’ve misunderstood the notion of limit point of a sequence. I prefer the term cluster point in this context, but by either name it’s a point $x$ such for any nbhd $U$ of $x$ and any $n\in\mathbb{N}$ there is an $m\ge n$ such that $x_n \in U$. Thus $1$ and $-1$ are both cluster points of the sequence $\langle (-1)^n:n\in\mathbb{N}\rangle$, though this sequence doesn’t converge to anything.

Equivalently $x$ is a cluster point of $\langle x_n:n\in\mathbb{N}\rangle$ iff for each nbhd $U$ of $x$, $\{n\in\mathbb{N}:x_n \in U\}$ is infinite. If the space is first countable, $x$ is a cluster point of $\langle x_n:n\in\mathbb{N}\rangle$ iff some subsequence of $\langle x_n:n\in\mathbb{N}\rangle$ converges to $x$.

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Thanks Brian for this. Yes indeed I have misunderstood what a limit point is - I was fooled by the word limit! The example you give is a good illustration of the behaviour of the function $x_{n}=sin(n)$ which has no limit but does have limit points. –  dandar Nov 10 '11 at 20:59

The sequence does not converge to anything, but a subsequence might converge. In fact there exists a subsequence whose limit is 1.

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Thanks Florian for this answer. I am sure you are correct but could you give an example? –  dandar Nov 10 '11 at 21:03

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