Can I always order a countable set of numbers? Suppose I have a countable subset of the real numbers $A$. Then since it is countable there is a function $f: \mathbb N \to A$ which is bijective. This function is of course a sequence, so we can write every element $x$ of $A$ as $x_i = f(i)$. So $A= \{ x_i \mid i \in \mathbb N \}$. Of course no two $x_i$ and $x_j$ are the same.
I can now come and reorder the $x_i$'s (instead of naming the $x \in A$ as $x_i$ I can name it $x_j$) such that they are decreasing $$x_1 >x_2>\cdots. $$
(Trivial??) Question: I CAN do that right? It seems right, since any $x_i$ and $x_j$ are different, then one is bigger than the other, thus I can do this comparison with all of them and order them.
If that is the case then take as $A= \Bbb Q \cap (0,1)$ for example. What we said above is that since $A$ is countable I can write $A= \Bbb Q \cap (0,1)=\{ x_i \mid i \in \mathbb N \}$ with $x_1 >x_2>\cdots$.
But that is absurd, there is no biggest rational in $(0,1)$. It's late, what am I missing?
EDIT. Also suppose a case where there IS a biggest number such as $A=\Bbb Q \cap (0,0.9]$ where the biggest rational is $0.9$. Then we can do $ 0.9=x_1 <x_2 <x_3<\cdots$. We have no idea what $x_2$ is, but there is a rational closest to $0.9$ than all others??

The answer is "no" and here is why. If you could always do that ordering that would mean that any countable set has a largest element. This is clearly not the case; take for example
$$
A=\{x_i={i\over 1+i}, i\in \mathbb{N}\}
$$
which clearly does not have a largest element.
(Of course there are cases where this is possible, such as $B=\{x_i={1\over 1+i}, i\in \mathbb{N}\}$ where $x_1=1$).
Maybe what you are missing is that the number of permutations of a finite set is itself finite but the permutations of an infinite set are infinite as well.
