Sequence can be rerranged into a monotone decreasing sequence Assume the sequence $a_{n}$ satisfy $\lim a_{n} = a$ and $a_{n} > a\ \forall n$ . Prove that this sequence can be rearranged into a monotone decreasing sequence.
I prove as process : 
First ( can let $a = 0$ ) because $\lim a_{n} = 0$ so  $\forall \epsilon > 0$ there exists $N$ such that $\forall n > N ,a_{n} < \epsilon$ . The set of all $a_{k}$ such that each element is bigger than $\epsilon$ is finite, so we can arrange this set and continue with the remainder of the sequence. 
I'm not sure with this solution because I don't see why we need limit and bounded? Anybody help me? 
I just solve this problem second way and write down again here :
I let an equivalence in this sequence , two term is equivalent if only if they're equal . So this equivalence divide the sequence in classes . If I can't find a maximum value of classes . Let $b$ is supermum of sequence , it's different from $a$ and exist $i \in N$ such that :
$$a < b_{i} < b_{i+1} < ... < b$$
But now $\lim b_{n} = a$ is different from $b$ , contradiction with supermum theorem of monotone decreasing sequence . 
 A: Note that $\{a_n:n\in\mathbb{N}\}$ is a bounded set of positive numbers. Since $a_n\rightarrow 0$ the maximum $\max \{a_n:n\in\mathbb{N}\}$ is attained at some $a_N$, and the number of elements $a_n$ with this property is finite.
So 
$$\phi(1):=\min\{ {n\in\mathbb{N}: a_n =\max\{a_m: m\in\mathbb{N}\}\}}$$
is well defined and by definition $a_{\phi(1)} \ge a_k \,\forall k \in \mathbb{N}$.
If you assume that $\phi$ is defined on $\{1,\dots, n-1\}$ with the properties that 


*

*$a_{\phi(i)} \ge a_{\phi(k)}$ whenever $k,i \in \{1,\dots, n-1\}$ such that $i<k$ and

*$a_{\phi(i)} \ge a_m$ whenever $i \in \{1,\dots, n-1\}$ and $m\notin  \phi(\{1,\dots, n-1\})$


define, recursively,
$$\phi(n):= \min\{ {n\in\mathbb{N}: a_n =\max\{a_m: m\in\mathbb{N}\backslash 
\phi(\{1,\dots,n-1 \})\}\}}$$
($\min$ and $\max$ in this definiation are attained for the same reasons as before).
It is then easy to see that 1. and 2. hold with $n-1$ replaced by $n$. By the induction principle you can now conclude that $\phi$ is defined on all of
$\mathbb{N}$ and $a_{\phi(n)}$ is a monotonically decreasing sequence.
In order to show it is actually a rearrangement of the original sequence you need to prove that $\phi :\mathbb{N}\rightarrow \mathbb{N}$ is onto. This follows from the fact that, by construction, each $a_m$ is picked as maximum in the above process once every larger $a_k$ (there are only finitely many) or equal $a_k$ (there are also at most finitely many) has been picked. 
(This is a replacement of an answer I gave earlier. I wrote a new one since I think it's a bit easier to read. If you want to see the previous version version control will allow you to do that)
A: The sequence of positive numbers converges to $0$, so there can only be finitely many elements $\geq 1$; since these are finite (perhaps even zero) in number, list them first: in particular, list them in monotonically decreasing order. The same observation holds for the finitely many elements between $1/2$ and $1$; list these, second, in monotonically decreasing order. Next, tackle the finitely many elements between $1/3$ and $1/2$. Continuing "in this way" leads to the desired outcome. 
A: An algorithm for sorting the numbers.
There is a function over the integers $f(i) = N$ where $N$ is the least value for which $\forall j \ge N: a_j < a_i$.
The existence of this function is a consequence of the given conditions: 
$\forall i: a_i > a = \lim_{j\to \infty} a_j$.
We also know that $\forall i: f(i) > i$ (this is obviously true because $a_i = a_i$).
Lemma: $\forall i, j: a_j \ge a_i \implies j < f(i)$.
Algorithm:
Pick any $i > 0$.
Determine $f(i)$.  Sort all the elements from $1$ up to $f(i)-1$.  Since we know $f(i)$ is finite, this sorting is possible.  All the entries up to the new location of $a_i$ are now sorted to their correct places.  (For optimization, we could use a heap-sort with $a_i$ as pivot, but stop after the left half is sorted, since the right half may need re-sorting).
Repeat the process with the elements after $a_i$, leaving the elements up to $a_i$ in place.
You can thus sort the entire list in finite chunks.  After each pass a non-empty set of elements will be fixed and will not need re-examination.
