Suppose $\{a_n\}$ is monotonically increasing sequence. This means that for all $n, m \in \mathbb{N}$ \begin{align*} n < m \implies a_n \leq a_m \tag{1} \end{align*} If $\{a_{n_k}\}$ is an arbitrary subsequence of $\{a_n\}$, then $\{a_{n_k}\}$ is monotonically increasing, as well.
Since the sequence $\{n_k\}$ consisting of the indices of $\{a_{n_k}\}$ is strictly increasing, we know that for all $k \in \mathbb{N}: k \leq n_k$.
Question
Since $k \leq n_k$ I am wondering if I can immediately infer from $(1)$ that $a_k \leq a_{n_k}$ since both sequences are monotonically increasing.
This seems reasonable to me: If $k = n_k$, then $a_k = a_{n_k}$. If $k < n_k$, then $a_k \leq a_{n_k}$ seems to follow from $(1)$.
Is this inference valid?