Firstly, i'm sorry for getting stuck three times in one day! I'm pretty much hogging this board which isn't that nice.

Anyway, i'm stuck on this:

Kindly (?), the homework sheet offers some hints in the form of more questions!Prove that every sequence has a monotonic subsequence.

My answer to 1).1). If there are an infinite number of floor terms, show they form a monotonic increasing subsequence.

2).If there are a finite number of floor terms and the last one is $\displaystyle a_F$, construct a monotonic decreasing subsequence with $\displaystyle a_{F+1}$ as it's first term.

3). If there are no floor terms, construct a monotonic decreasing subsequence with $\displaystyle a_1$ as it's first term.

Since $\displaystyle a_n \geq a_f \ \forall \ n\geq f$ (ie. the definition of a floor term) then let $\displaystyle n=f+1 \Rightarrow \ a_{f+1} \geq \ a_f$

Therefore there are an infinite number of floor terms and the next floor term is greater than the previous one.

My answer to 2).

Starting once more from the definition $\displaystyle a_n \geq a_f \ \forall \ n \geq f$. $\displaystyle a_{F+1}$ is the first term and $\displaystyle a_F$ is the last term.

Let $\displaystyle n=F$ and $\displaystyle f=F+1$.

Therefore $\displaystyle a_{F} \geq a_{F+1} \ \forall \ F \ \geq F+1$

Therefore there is a monotonic decreasing subsequence as required.

(I think so anyway. Could someone double check this?)

My answer to 3).

I decided to choose $\displaystyle a_n=n$ since this doesn't have a lower bound. Would floor terms be in the sequence $\displaystyle a_f=f-1$?

Therefore making f=1:

$\displaystyle a_1=1-1=0$

$\displaystyle a_2=2-1=1$

$\displaystyle a_3=3-1=2$ etc

However, this doesn't create a monotonic decreasing subsequence and I can't think of a sequence that does!

Could I have some help with these "hints" so I can use them to solve the question?

Thank you kindly to anyone who posts!