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Let $(x_n)$ be a sequence in $\mathbb R$. We call $y$ a cluster point of $(x_n)$ iff for every neighborhood $N$ of $y$ there are infinitely many $n$ such that $x_n \in N$. Let $C$ denote the set of cluster points of $(x_n)$.

By definition, $\limsup_{n \to \infty} x_n = \inf_n \sup_{k \geq n} x_k$.

Can you show me how to prove $\limsup_n x_n = \sup C$? Thanks.

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In fact, you can show that 1) $\lim \sup x_n$ is a cluster point of $(x_n)$, and that 2) it's greater than any other cluster point of $(x_n)$. That is, $\sup C$ is actually $\max C$. You could try to construct a sequence that converges to $\lim \sup x_n$ (then it's easy to see that $\lim \sup x_n$ is a cluster point of $(x_n)$). –  talmid Jul 3 '12 at 14:23
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3 Answers 3

up vote 1 down vote accepted

Put $L = \limsup_n x_n$ and $s_n = \sup_{k \ge n} x_k$. Notice that $s_n$ is a decreasing sequence in $n$ and that $s_n \downarrow L$.

Let $\lambda > L$. then for some $n$, $\sup_{k\ge n} x_k <\lambda$. Hence there can be no limit point of the sequence in $(\lambda, \infty)$. Since $\lambda > L$ was chosen arbitrarily, all limit points of the sequence must lie in $(-\infty, L]$.

We will be done if we can show that the limsup is a limit point of the $x_n$. We choose a subsequence as follows. Choose $n$ so that $\sup_{k\ge n} x_k < L - 1/2$; now fix $n_1$ so $\sup_{k\ge n_1} x_k > L - 1/2$. Now suppose $n_1 < n_2 < \cdots < n_k$ are chosen. Since $s_n\downarrow L$ there is $N$ so that $s_N > L - 1/2^n$. Since $S$ is decreasing, we can choose $N > n_k$. Put $N = n_{k+1}$.

The sequence $\{x_{n_k}\}$ converges to $L$, so $L$ is a limit point of the $x_n$.

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Thank you very much. This is exactly what I was looking for. Since I'm not sure that I understand it all I have attempted to write it in my own words. Would you mind having a look and telling me whether it's right? Of course I will not accept my own answer. –  Matt N. Jul 3 '12 at 16:13
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It follows quite easily from the following characterization of $\sup C$ (see W. Rudin, Principles of mathematical analysis, Theorem 3.17):

  1. $\sup C$ is the limit of some subsequence of $\{x_n\}$;
  2. If $x>\sup C$, then there is an integer $N$ such that $n \geq N$ implies $x_n < x$.
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Isn't this preempting the whole question? –  Did Jul 3 '12 at 16:34
    
@did I do not understand the expression. I believe that this alternative definition of $\limsup$ is helpful to answer the question. I'm sorry if my answer is unfair... –  Siminore Jul 3 '12 at 16:49
    
This, by the way, is a very nice and useful result. It has a clear analog for $\liminf x_n$ too. –  ncmathsadist Jul 3 '12 at 21:00
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We need two show $\limsup_n x_n \leq \sup C$ and $\limsup_n x_n \geq \sup C$. To get $\limsup_n x_n \leq \sup C$, we show that $\limsup_n x_n \in \sup C$:

(i) We show that $\limsup_n x_n \in \sup C$: The idea is to construct a subsequence $(x_{n_k})$ of $(x_n)$ that converges to $L = \limsup_n x_n$. We construct $(x_{n_k})$ as follows: Let $s_n$ denote $\sup_{k \geq n} x_k$. Then $\lim_{n \to \infty} s_n = L$. Hence for $\varepsilon_1 = \frac12$ there is $n_1$ such that $n \geq n_1$ implies that $s_n \in B(L, \varepsilon_1)$. In particular, $L - \varepsilon_1 < \sup_{k \geq n} x_n < L + \varepsilon_1$ and hence by the definition of $\sup$ we can find an $x_{n_1}$ such that $L - \varepsilon_1 < x_{n_1} < L + \varepsilon_1$. We repeat this process for $\varepsilon_i = \frac{1}{2^i}$. Then $(x_{n_i})$ converges to $L$. Hence $L$ is a cluster (limit) point of $(x_n)$.

(ii) $\limsup_n x_n \geq \sup C$: By contradiction assume that $\limsup_n x_n < \sup C$. Then $\limsup_n x_n < \sup C - \varepsilon$ for some $\varepsilon > 0$. by the definition of $\sup$ there exists $c \in C$ such that $\limsup_n x_n < c$. Then there exists $\varepsilon^\prime > 0$ such that $\limsup_n x_n < c - \varepsilon^\prime$. Hence in particular, there exists an $n$ such that $\sup_{k \geq n} x_k < c - \varepsilon^\prime$ and hence for all $k \geq n$, $x_k < c - \varepsilon^\prime$. But then $c$ cannot be a cluster point. Therefore $\limsup_n x_n \geq \sup C$.

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Line 2: say " we show that $\limsup_n x_n \in C$. –  ncmathsadist Jul 3 '12 at 20:57
    
In (ii) if you are a little careful (look at my argument), no contradiction is needed. That will make the style a little nicer. –  ncmathsadist Jul 3 '12 at 20:59
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