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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?

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  • $\begingroup$ 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
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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 $X$.

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  • $\begingroup$ 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
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    $\begingroup$ @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

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