Let $X$ be a set of sequences of natural numbers $s \in \mathbb N^{\mathbb N} $ such that at least two elements in $s$ appear infinite times. We define $\leqslant$ order on set $X$, $s \leqslant t$ if and only if $s(n) \leqslant t(n)$ for any n. Does $(X, \leqslant)$ have the smallest element?
My attempts: I think that $(X, \leqslant)$ doesn't have the smallest element - at least two elements in every $s$ appear infinite times, so there is no way to find the smallest element. Am I right? If not, could you please explain me how to solve this task?
You're correct. Just observe that for any $s \in \mathbb{N}^{\mathbb{N}}$ if $m, n$ appear in $s$ infinitely many times and $n < m$, replacing one occurence of $m$ with $n$ gives a sequence which is smaller with respect to this order, yet still in $X$.