I'm stuck on this one.
We know that $[0,1]\cap\mathbb{Q}=\{\frac{a}{b}, a,b \in \mathbb{N}:\frac{a}{b}\in[0,1]\}$
Thus, they look something like $\{0,\frac{a_1}{b_1},\frac{a_2}{b_2},…,1\}$
Pick $x\in[0,1]\cap\mathbb{Q}$
Since the irrationals are dense in $\mathbb{R},$ $x \in[0,1]\cap\mathbb{Q}, \forall \epsilon >0 ,(x-\epsilon,x+\epsilon)$ contains $i$ irrationals
Therefore, $[0,1]\cap\mathbb{Q}$ does not contain all of its accumulation points
Therefore, $[0,1]\cap\mathbb{Q}$ is not closed
This is where I get stuck though because $[0,1]\cap\mathbb{Q}$ has infinitely many holes in it (since the irrationals are dense in $\mathbb{R})$
So naturally I would want to construct an open cover that has no finite subcover, but I can't find it.
Best Answer
Take an irrational number $a\in[0,1]$ and consider the set$$\left\{(\Bbb Q\cap[0,1])\setminus\left[a-\frac1n,a+\frac1n\right]\,\middle|\,n\in\Bbb N\right\}.$$It is an open cover of $\Bbb Q\cap[0,1]$ without a finite subcover: if $F$ is a finite subset of $\Bbb N$, if $N=\max F$, and if $q\in\left[a-\frac1N,a+\frac1N\right]\cap\Bbb Q\cap[0,1]$, then$$q\notin\bigcup_{n\in F}(\Bbb Q\cap[0,1])\setminus\left[a-\frac1n,a+\frac1n\right].$$