All Subsequences Monotone Except The Sequence Itself Just had my Calculus 1 exam and wanted a clarification.
Say An is a sequence, the claim was:
If all of An's Subsequences are monotone except for An itself, An is monotone.
It didn't specifically state that An isn't monotone, so I couldn't find a counterexample.
If it's true, would like to know how to prove it.
Thanks a lot in advance!
 A: I guess you mean that you know that all subsequences of $a_n$ are monotone except for the sequence itself, i.e., you can assume that all proper subsequences are monotone and need to show that $a_n$ is monotone.
Additionally, I guess the statement is that all subsequences share the same monotonicity. Otherwise, it's pretty clear that whole sequence cannot be monotone. For example, if $\{a_{n_k}\}_k$ is increasing and $\{a_{m_j}\}_j$ is decreasing, we have that $a_{n_2}\geq a_{n_1}$ and $a_{m_1}\leq a_{m_2}$. Now you need to split into cases, I'll discuss here just one, the others are similar. Assume $m_1\leq n_1 < n_2\leq m_2$. If $a_{m_1}\leq a_{n_1}$, you have $a_{m_2}\leq a_{m_1}\leq a_{n_1}\leq a_{n_2}$ so you can only have monotonicity if you have equality everywhere. It's easy to see the other cases if you just try to plot $2$ pairs of values from two subsequences of different monotonicity.
Therefore I will assume that what you want to prove is this. Let $a_n$ be a sequence such that for each strictly increasing sequence of naturals $\{n_k\}_k$ that is not the entire sequence of natural numbers, the subsequence $\{a_{n_k}\}_k$ is (wlog) increasing. Show that $\{a_n\}_n$ is increasing.
Assume that $\{a_n\}_n$ is not increasing. Then there exist some indices $1\leq i <j$ such that $a_i>a_j$. Define the sequence of naturals $\{n_k\}_k$ by setting $n_1=i$ and $n_2=j$ and $n_k=j+10k$ for $k\geq 3$. Clearly this sequence is (strictly) increasing and it is not the full set of naturals (in particular, the average gap between consecutive terms is $10$). Therefore $\{a_{n_k}\}_k$ is a proper subsequence of $\{a_n\}_n$ and must be, by assumption, increasing, meaning that $a_i=a_{n_1}\leq a_{n_2}=a_j$, contradiction.
