Infinite indices zero sequence Let $(a_n)_{n\in\mathbb N}$ be a sequence of positive real numbers with $\lim_{x\to \infty}a_n=0$ . 
Show that there is an infinite number of indices $n$, that $a_m\le a_n, \forall  m\ge n$ .
My idea is to proof this via contradiction and assume, that there is a finite number of indices $n$, so that $a_m\le a_n ,\forall  m\ge n$. 
This would mean that there is a point $a_x$ in the sequence from where $a_m\gt a_n ,\forall  m\ge n$.  Because of that, $\lim_{n\to \infty}a_n=0$ can't be fulfilled .
If I'm correct with this draft, I am having trouble formalising this idea, maybe there's someone who can help me with that.
Thanks !
 A: Your basic idea is correct, but you have to be careful!
You consider $A=\{n\in\mathbb N ~:~a_m\leq a_n \forall m\geq n\}$ and assume $A$ is finite. Then you can choose $M:=\max A$. For $n>M$ you now $n\notin A$ and therefore not $a_m\leq a_n$ for all $m\geq n$. But this doesn't mean $a_m>a_n$ for all $m\geq n$ but there exists $m\geq n$ such that $a_m>a_n$.
Hence you can construct a monoton increasing subsequence. Since $(a_n)_n$ is a sequence of positive real numbers, you can deduce a contradition to $a_n\to 0$.
How to construct the subsequence:

Consider $M+1\notin A$, so there exists $m>M+1$ such that $a_m>a_{M+1}$. Define $n_1:=m>M+1$.



Since $n_1>M$ we know $n_1\notin A$ hence there exists $m>n_1$ such that $a_m>a_{n_1}$ and define $n_2:= m$.



Since $n_2>n_1>M$ we know $n_2\notin A$ hence there exists $m>n_2$ such that $a_m>a_{n_2}$ and define $n_3:=m$.



Iterate the argument and you get a subsequence $(a_{n_k})_k$ of $(a_n)_n$ which is striktly monotonic increasing and $a_{n_1}>0$. Especially $a_{n_k}\not\to 0$.



Since $a_n\to 0$ you get a contradiction.

A: Let an element $a_n$ such that $a_n \ge a_m$ for all $m \ge n$ be called dominant. We'll construct a subsequence of dominant elements inductively. 
Take $\varepsilon = a_1 > 0$. By definition of $\lim_n a_n = 0$, there exists $n_1 \in \mathbb{N}$ such that $n \ge n_1 \implies a_n < a_1$.
Now consider $a_{k_1} = \max\{a_1, \ldots, a_{n_1}\}$.
Notice that $a_{k_1}$ is dominant:
We have that $a_{k_1} \ge a_{k_1 + 1}, a_{k_1 + 2}, \ldots, a_{n_1}$, and also for any $n \ge n_1$ we have $a_{k_1} \ge a_{1} > a_n$ by definition of $n_1$.
Now, if all elements after $a_{k_1}$ are dominant, we're finished. Otherwise, let $a_{r_2 - 1}$ be the first element after $a_{k_1}$ where $a_{r_2 - 1} < a_{r_2}$.
Take $\varepsilon = a_{r_2} > 0$. By definition of $\lim_n a_n = 0$, there exists $n_2 \in \mathbb{N}$ such that $n \ge n_2 \implies a_n < a_{r_2}$. Now define $a_{k_2} = \max\{a_{r_2}, \ldots, a_{n_2}\}$. Notice that $a_{k_2}$ is dominant:
$a_{k_2} \ge a_{k_2 + 1}, a_{k_2 + 2}, \ldots, a_{n_2}$ and also for all $n \ge n_2$ we have $a_{k_2} \ge a_{r_2} > a_n$ by definition of $n_2$.
Now continue this inductively, and you will obtain a subsequence $(a_{k_n})_{n=1}^\infty$ of dominant elements.
