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?

  • $\begingroup$ Is the constant sequence of $0$ in $X$? $\endgroup$ – Kenny Lau Aug 10 '17 at 17:35
  • $\begingroup$ I don't think so, because that would mean that only one element appears infinite times, and I think that at least two different elements should be present in a sequence. $\endgroup$ – zu_m Aug 11 '17 at 10:04

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 $X$.

  • $\begingroup$ An explicit construction would be $(0,1,0,1,0,1,0,1,\cdots) \ge (0,0,0,1,0,1,0,1,\cdots) \ge (0,0,0,0,0,1,0,1,\cdots) \ge (0,0,0,0,0,0,0,1,\cdots) \ge \cdots$ $\endgroup$ – Kenny Lau Aug 10 '17 at 17:42
  • 1
    $\begingroup$ @KennyLau But the existence of an infinite strictly decreasing sequence in $X$ doesn't imply there is no least element in $X$, right? $\endgroup$ – Adayah Aug 10 '17 at 17:45
  • $\begingroup$ Sorry, that was a brain fart. $\endgroup$ – Kenny Lau Aug 10 '17 at 17:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.