# A problem about the cover of the interval

We list all the rational numbers in $\mathbb{R}\bigcap\lbrack0,3\rbrack$ here, $$\{x_1,x_2,x_3,...\},$$ with the order from small to large. Then we define the ball $B(x_i,r_i)$, $$B(x_i,r_i)=\{x\arrowvert x_i-r_i\le x\le x_i+r_i\}$$ We take $r_i=1/2^i(1\le i<\infty)$.

Now there exists one question here.

Do these balls $(\bigcup_{1\le i<\infty}B(x_i,r_i))$ cover the interval $[0,3]$?

If they can cover the interval,then $$m(\bigcup_{1\le i<\infty}B(x_i,r_i))\le\sum_{1\le i<\infty} B(x_i,r_i)=2< m([0,3])$$ it leads to a contradiction.

If they cannot cover the interval,then there must exist a positive integer $N$,such that $$x_{N+1}-x_N>\frac{1}{2^N}+\frac{1}{2^{N+1}}>\frac{1}{2^N}$$ I just cannot imagine there exist the two adjacent rational numbers that are so far apart (their distance is larger than one rational number $\frac{1}{2^N}$). That's impossible! We can insert another rational number $x_N+\frac{1}{2^N}$ between $x_N$ and $x_{N+1}$.This leads to a contradiction.

I just cannot find any mistakes from above words.Who can find an answer?

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The rationals in your interval are densely ordered. That is, between any two rationals $r \lt s$ in the interval, there is a rational $x$ such that $r \lt x \lt s$. So although the rationals in your interval can be listed (they are a countably infinite set), they cannot be listed "from small to large."
Added: More concretely, as pointed out by David Mitra, let $x_1$ be the first element in the supposed list, presumably $0$, and let $x_2$ be the second element in the list. Then $(x_1+x_2)/2$ is a rational in our interval, and it lies between $x_1$ and $x_2$ in the usual ordering of the rationals. So if the list is "from small to large" it cannot include the rational number $(x_1+x_2)/2$.