Show that monotone subsequences have the same limit Let be $f:\mathbb{R}\to \mathbb{R}$ a monotone function and $\left(a_n\right)_{n\in\mathbb{N}}$ a sequence with $\lim\limits_{n\to\infty}a_n=\infty$.
Is it possible that two monotonically increasing subsequences of $\left(a_n\right)_{n\in\mathbb{N}}$ produce two different (finite) limits when plugged into $f$?

My approach:
Without loss of generality we assume $f$ to be increasing (the case where $f$ is decreasing is handled in the same way).
Let $\left(a_{n_j}\right)_{j\in\mathbb{N}}$ and $\left(a_{m_j}\right)_{j\in\mathbb{N}}$ be two monotonically increasing subsequences of $\left(a_n\right)_{n\in\mathbb{N}}$ which we plug into $f$. Then both subsequences $\left(f(a_{n_j})\right)_{j\in\mathbb{N}}$ and $\left(f(a_{m_j})\right)_{j\in\mathbb{N}}$ are monotonically increasing. We assume $\lim\limits_{j\to\infty}f(a_{n_j})=c$, $\lim\limits_{j\to\infty}f(a_{m_j})=d$ and $\epsilon:c-d>0$. Due to convergence I find an index $j_0$ such that members with $j>j_0$ satisfy $|f(a_{n_j})-c|<\frac{\epsilon}{2}$ and $|f(a_{m_j})-d|<\frac{\epsilon}{2}$, respectively. We can conclude:
$$f(a_{m_j})<d+\frac{\epsilon}{2}=c-\frac{\epsilon}{2}<f(a_{n_j})\implies f(a_{m_j})<f(a_{n_j})\text{, for all } j>j_0.$$
Both sequences $\left(a_{n_j}\right)_{j\in\mathbb{N}}$ and $\left(a_{m_j}\right)_{j\in\mathbb{N}}$ go to $\infty$, so we can find two members $a_{n_j}$ and $a_{m_j}$ such that $j_0<j$ and $a_{n_j}\leq a_{m_j}$. As $f$ is monotonically increasing it implies $f(a_{n_j})\leq f(a_{m_j})$ which is a contradiction because we had assumed that for all $j>j_0$ it must hold $f(a_{n_j})> f(a_{m_j})$. The case where we assume $d>c$ is handled in the same way and also comes to a contradiction. Hence, it must be $c=d$.
Is this correct? Although the statement is intuitively clear, the way I have proved it feels a little bit complicated. May be there is a more elegant way in which one can avoid the tedious use of indices? Any suggestions are welcome :)
 A: What you’re trying to do is correct, but there’s one serious notational error at the beginning: $c_{n_i}$ makes sense only if you already have a sequence $\langle c_n:n\in\Bbb N\rangle$ and are defining a subsequence of it. To get a subsequence of $\langle a_n:n\in\Bbb N\rangle$, you need to write $\langle a_{n_i}:i\in\Bbb N\rangle$ or the like, so that $\langle n_i:i\in\Bbb N\rangle$ is a subsequence of the indices. A second subsequence of $\langle a_n:n\in\Bbb N\rangle$ will be defined by a different subsequence of $\Bbb N$, so you can’t index it by subscripts $n_i$ or $n_j$: the $n$ implies that you’re talking about the indices picking out the first subsequence. You can instead make it $\langle a_{m_i}:i\in\Bbb N\rangle$, for instance. Now you can suppose that $\langle a_{n_i}:i\in\Bbb N\rangle$ converges to $c$ and $\langle a_{m_i}:i\in\Bbb N\rangle$ converges to $d$ and proceed just as you did, with a few obvious changes in the notation.
Given subsequences $\langle a_{n_i}:i\in\Bbb N\rangle$ and $\langle a_{m_i}:i\in\Bbb N\rangle$, you can make an alternative argument that uses the same basic intuition in a slightly different way by defining a sequence $\langle k_i:i\in\Bbb N\rangle$ as follows.
Let $N=\{n_i:i\in\Bbb N\}$ and $M=\{m_i:i\in\Bbb N\}$, and let $k_0=n_0$. Given $k_{2\ell}$ for some $\ell\in\Bbb N$, let $k_{2\ell+1}$ be the smallest member of $M$ that is larger than $k_{2\ell}$, and let $k_{2(\ell+1)}$ be the smallest member of $N$ that is larger than $k_{2\ell+1}$. Then $\langle a_{k_i}:i\in\Bbb N\rangle$ alternates terms of $\langle a_{n_i}:i\in\Bbb N\rangle$ and $\langle a_{m_i}:i\in\Bbb N\rangle$, so it’s a subsequence of both, and therefore both must have the same limit:
$$\lim_{i\to\infty}a_{n_i}=\lim_{i\to\infty}a_{k_i}=\lim_{i\to\infty}a_{m_i}\,.$$
