Let $ X = \{ i | i \in \mathbb{N} \} = \mathbb{N}$. Then is there an infinite sequence $ \{t_i \} $ with terms from $X$ such that $\forall i \ t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?
My thought process was follows: If $X_m = \{ i | i \in \mathbb{N}, i \leq m \}$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ \{t_i \} $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < \infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = \infty$) - a process I am clueless about.
*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.
Assume that such a sequence $(t_i)_i$ exists.
$T:=\{t_i\mid i\in\mathbb N\}\subseteq\mathbb N$ is a non-empty subset of well-ordered $\mathbb N$ hence has a smallest element $k$.
Some $i\in\mathbb N$ exists such that $t_i=k$.
Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.