$\limsup$ and cluster points 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.
 A: 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 $x_n$.
A: It follows quite easily from the following characterization of $\sup C$ (see W. Rudin, Principles of mathematical analysis, Theorem 3.17):


*

*$\sup C$ is the limit of some subsequence of $\{x_n\}$;

*If $x>\sup C$, then there is an integer $N$ such that $n \geq N$ implies $x_n < x$.

A: 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 $\limsup_n x_n \geq \sup C$.
