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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 .

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    "... and continue with the remain of the sequence" is a bit imprecise and looks like handwaving. If you don't see why the limit is needed try to find such a rearrangement for $a_n = n$ or $a_n = 1 - \frac{1}{n}$. – 2017-02-21
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    Ok , Can you solve this and explain me why it isn't true ? – 2017-02-21
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    I think the first I should let $\epsilon=a_{1}$ so exist $N_{0},a_{n} < a_{1}$ and we arrange the set from $a_{N_{0}} \to a_{1}$ – 2017-02-21

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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.

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    No , It's what I can't explain , I think this proceed thus true with finite – 2017-02-21
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    Without a limit, the sequence could be, e.g., $1, 2, 3, 4, \ldots$, which cannot be arranged as monotonically decreasing. Without specifying the elements are positive, you could have $1, -1, 1/2, -1/2, 1/3, -1/3, \ldots$. This sequence converges to $0$, but cannot be listed in a monotonically decreasing order. **So:** Both assumptions are necessary. (And the method outlined above will work!) – 2017-02-21
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    Yes , I see how limit is necessary but the hypothesis $a_{n} > a$ isn't used in your proof , and what about second my prove ? – 2017-02-21
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    @NguyễnTràMy You edited the hypothesis $a_n > a$; when I answered your question, it said: $a_n > 0$ for all $n$. This is visible in the [**revisions**](http://math.stackexchange.com/posts/2154822/revisions). I have answered the question that you posted. – 2017-02-21
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    Using Benjamin's answer, you can define a sequence of permutations $\sigma_n$ defined on the terms $a_j$ such that $1/(n+1)$\sigma_n$ are disjoint $\cup_n \sigma_n$ defines the required rearrangement. For this to work, you do need the $a_n$ to be positve and the limit to be zero. – 2017-02-21
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    "Continuing this way" is nothing but handwaving and is exactly what should be made precise in such an exercise. – 2017-02-22
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    @Thomas If you would like to post an answer with additional details, then perhaps it would be appreciated. – 2017-02-22
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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

  1. $a_{\phi(i)} \ge a_{\phi(k)}$ whenever $k,i \in \{1,\dots, n-1\}$ such that $i
  2. $a_{\phi(i)} \ge a_m$ 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)

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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.