Open covering for rationals in [0, 1]. The set of all rational numbers in the closed interval $[0,1]$ is not closed, and therefore not compact. Hence, there must exist an open covering for this set that has no finite sub-covering. Explicitly construct such an open covering.
My attempt at an answer is as follows. The rationals in the closed unit interval are countably infinite. So if we enumerate them, center the $i^{th}$ rational in an open segment of width $1/2^{i+1}$. This forms an open covering of all the rationals in the closed unit interval. Then any finite sub covering formed using this open covering must have length strictly less than 1, and must therefore miss some rationals. I feel uncomfortable with this proof. But am not sure what I am missing.
 A: Assume that $\{{r_1},{r_2},\cdots \}$ is a countably infinite enumeration of the rational numbers in ${\bf Q}\cap [0,1].$ Fix ${\epsilon}\in (0,1)$, and define $${I_j}=\bigg({{r_i}-{\frac{\epsilon}{2^{(i+1)}}}},{{r_i}+{\frac{\epsilon}{2^{(i+1)}}\bigg)}}.$$ Then the length of $I_j$ is ${\epsilon}/2^i.$ And all the rationals in ${\bf Q}\cap [0,1]$ must then be contained in a countable union of open intervals  whose measure is  $\le\epsilon<1.$ (We are summing a geometric series with common ratio $1/2$ starting at the index $1$, the sum being multiplied by a constant factor of $\epsilon$). Thus this forms an open covering of ${\bf Q}\cap [0,1].$ If ${\bf Q}\cap [0,1]$ were compact, it would have a finite sub-covering. We can label this finite sub-covering  $$\{I_{s_1},\cdots, {I_{s_m}}\},$$ these being open intervals centered at rationals ${r_{s_1}},\cdots ,{r_{s_m}}.$ We now observe that the measure of this subcovering cannot exceed $\epsilon.$ On the other hand as Mr. Wainfleet (Dr. Wainfleet?) observes, as this finite subcovering by open intervals must contain $0$ and $1$, it must have length at least $1$, as the rationals are dense in $[0,1]$, which is a contradiction. This proof was inspired by Mr. Wainfleet's sketch and so he deserves my sincere thanks. It seems to be the case that even the infinite open covering has length less than 1 in this particular choice of an open cover-so this choice of $\epsilon$ seems to make the finiteness of the covering less significant than it might seem-not surprising-after all the rationals have measure $0$ even in $\bf R$.
