TITLE: A question about sequence and subsequences
QUESTION [5 upvotes]: Give example of Sequence and Subsequences which satisfy following 

$\{x_n\}$ is not increasing , but $\{x_n\}$ has increasing subsequence.
$\{x_n\}$ is unbounded , but $\{x_n\}$ has a bounded subsequence.
A sequence of integers $\{x_n\}$ which diverse , but which has infinitely many distinct subsequnetial limits 

for (1) $x_n=\begin{cases} n& n=2k\\\frac1n &n=2k+1\end{cases}$ this not increasing it has increasing subsequence 
(2)  consider $x_n = n$ if $n$ is even and $x_n=0$  if $n$ is odd here $\{x_n\}$ is unbounded but $\{x_n\}$ has subsequence 
is i am correct for (1) and (2) if i am please can any give example of (3)

REPLY [8 votes]: A fun way of doing (3) is to use primes. (In fact it works for (1) and (2) as well.)
Consider the sequence $(a_n)$ defined by
$$a_n = \cases{p,\quad n = p^k\ \text{for some $p$ prime and $k\in\mathbb{N}_{\geq 1}$}\\0,\quad \text{otherwise}}$$
Then $(a_n) = \{0,2,3,2,5,0,7,2,3,0,11,0,13,0,0,2,17,0,19,0,0,0,23,0,5,...\}$
The sequence has subsequences converging to every prime, and we know there are infinitely many primes.
Addendum
It might look excessive, like it's a lot more complicated than it has to be, but it has the benefit of subsequences being very easy to describe. e.g.
$(a_p)_{p\ \text{prime}} = \{2,3,5,7,11,13,...\}$ is an increasing subsequence
$(a_{2k})_{k\in\mathbb{N}} = \{2,2,0,2,0,0,0,2,0,...\}$ is a bounded subsequence
for each $p$ prime, $(a_{p^k})_{k\in\mathbb{N}} = \{p,p,p,...\}$ is a convergent (constant) subsequence