Need every bounded sequence contain a cluster point? Let $x_{n}$ be a sequence of real numbers. Then $x$ is a cluster point of $x_{n}$ if there are infinitely many points from the sequence contained in any neighborhood centered on $x$.
My question:
If a sequence is bounded, does it necessarily have a cluster point? I can't think of a counterexample, but then again, my textbook defines limit inferior and limit superior so that the cases of having no cluster points and being unbounded are distinct.
In other words, what is an example of a bounded sequence having no cluster points?
 A: Yes, as a consequence of the Bolzano-Weierstrass theorem, which is equivalent to the axiom of completeness that is fundamental to the definition of $\mathbb{R}$.
A: For any real $a,b$ with $a<b$ and $J=[a,b]$, for convenience let $J^-=[a,(a+b)/2]$ and $J^+=[(a+b)/2,b]....$ Let $(x_n)_{n\in N}$ be a bounded sequence where $\{x_n :n\in N\}\subset [L,M]$ (...with $-\infty <L< M<\infty$).... Let $J_1=[L,M]$. Recursively, if $\{n: x_n\in (J_m)^-\}$ is infinite then let $J_{m+1}=(J_m)^-$ ; otherwise let $J_{m+1}=(J_m)^+$. Observe that for every  $m$, the set $\{n:x_n\in J_m\}$ is infinite....Now, with $J_m=[a_m,b_m]$, we have $a_m\leq a_{m+1}<b_{m+1}\leq b_m$ and $b_m-a_m=2^{1-m}(M-L).$ So there is a unique $x$ such that $\{x\}=\cap_{m\in N}J_m$. If $U$ is any nbhd of $x$ then $\{n:x_n\in U\}$ is infinite, because $U\supset (x-d,x+d)$ for some $d>0$,and if $m$ is large enough that $2^{1-m}(M-L)< d$ then $(x\in J_m=[a_m,b_m]\land b_m-a_m<d)\implies J_m\subset (x-d,x+d)\subset U $ $\implies \{n:x_n\in J_m\}\subset \{n:x_n\in U\}$.
