For an increasing sequence $(a_n)_{n=m}^\infty$ of real number, can the $\sup(a_n)_{n=m}^\infty$ be an element of the sequence?
My guess is that is not the case as there are an infinitely many terms in the sequence and it is an increasing sequence. So if for some natural number $n$ we had that $a_n=\sup(a_n)_{n=m}^\infty$, then $a_{n+1}$ would have to be less than or equal to $\sup(a_n)_{n=m}^\infty$, but since it is an increasing sequence, we have that $a_{n+1}>a_n$ and thus $a_n\ne \sup(a_n)_{n=m}^\infty$.
Am I correct in saying that an infinite increasing sequence does not contain it's supremum?
Background: This came up while I was attempting another exercise in which I had set $a_n=\sup(a_n)_{n=m}^\infty$. But then I thought if this can even be the case or not. So I just wanted to make sure that I wasn't derailing.
Perhaps it would help to distinguish between increasing and strictly increasing. If a sequence is just increasing, the supremum of the sequence can certainly be an element of the sequence. For instance, a trivial example is the constant sequence $$a_n = 1$$ for all $n\ge 1$. On the other hand, for a strictly increasing sequence, your argument is correct.