Sequence of increasing rationals in $[0,1]$ diverges and converges

Question

Consider the set $U$=$\mathbb{Q}$${\displaystyle \cap }$ $[0,1]$.

$U$ is in bijection with $\mathbb{N}$, so we could construct the sequence $({q_n})_n$ of all rationals in $[0,1]$.

I think we can also order this sequence so that it is incresing. So the sequence $({q_n})_n$ is bounded and increasing, therefore converges.

But there are many subsequences that converge to different numbers ( like {$\frac{1}{n}$} and {$1-\frac{1}{n}$} ), so it should diverge.

I'm confused on what I got wrong but I would guess it's in the 3rd line...

Answer 1

I think we can also order this sequence so that it is incresing

Incorrect. You cannot order it. In fact, it is quite simple to show that every strictly increasing sequence of elements from $U$ cannot cover all elements from $U$.

Proof:

Take $q_1$ to be the first element in the sequence, and $q_2$ to be the second. Then, because the sequence is increasing, we know that $q_n>q_2$ for all $n\geq 3$. However, if you define $$q=\frac{q_1 + q_2}{2}$$ then you have $$0\leq q_1 < q < q_2 \leq 1$$ which means

1. $q$ is in $U$ and
2. $q$ is not in the sequence.