Exercise over sequences of real numbers Let $(a_n)$ be a sequence of nonnegative real numbers such that $$\lim_{n\rightarrow \infty}a_n=0$$ How to prove that exists an infinite number of indices $ n $ such that $a_n\geq a_m$ for all $m\geq n$
 A: We'll show your claim by contraposition.
Assume there are only finitely many $n$ such that $a_n\geq a_m$ for all $m\geq n$ and let $N$ be the maximum of these $n$'s.
Then there is an $m_0\geq N$ such that $a_N<a_{m_0}$ and for every $m_i$ there is an $m_{i+1}\geq m_i$ such that $a_{m_i}<a_{m_{i+1}}$.
Thus we have a strictly increasing subsequence $a_{m_i}$ of non-negative integers and therefore $\lim_{i\to\infty} a_{m_i}\neq 0$ and thus $\lim_{n\to\infty} a_n\neq 0$.
A: Using the definition of $\lim_{n\rightarrow\infty}a_n=0$, for any $\varepsilon>0$ there is and $n_{\varepsilon}>0$ such that $m>n_{\varepsilon}\implies a_m<\varepsilon$. 
Define $a_{n_1}=\max\{a_m|a_m<\varepsilon\}$ for some given $\varepsilon>0$. This $a_{n_1}$ satisfy the problem assumptions. Now repeat this argument for $\varepsilon/2$ to get the element $a_{n_2}$, and inductively you construct $a_{n_1}\geq a_{n_2}\geq a_{n_3}\geq\ldots$, each one of them satisfying the problem assumptions.
The set $\{a_{n_1},a_{n_2},a_{n_3},\ldots\}$ is a set of infinite elements with the desired property.
A: Assume that a sequence, $(a_n^1)$, that converges to 0 has at least one such point (say, $n_0$) - if it does, consider the subsequence $(a^2_n) =(a_{n+n_0+1}^1)$. This sequence converges to 0, so would have such a point of its own (and the two points would be distinct). By continuing this process, we can generate an infinite sequence of such points.
Hence it suffices to show that the sequence has at least one such point. Assume the opposite - that for some $(x_n$), there are no points for which $x_n \geq x_m$ for all $n \geq m$. 
This means that each point in the sequence has a point greater than it further in the sequence. 
So take $y_1 = x_1$. Then there must exist $n_1$, with $x_{n_1} \geq y_1$. Define $y_2$ to be $x_{n_1}$. Repeat this process to create $(y_n)$, which is an increasing subsequence of $(x_n)$
But this is a contradiction, as $(y_n)$ would not tend to 0, but it is a subsequence of $(x_n)$
