Show that $\overline{\mathbb{Q}\cap (0,1)}=[0,1]$.

Question

Show that $\overline{\mathbb{Q}\cap (0,1)}=[0,1]$

My attemp: If $E=\mathbb{Q}\cap (0,1)$, then $$\overline{\mathbb{Q}\cap (0,1)}=\overline{\mathbb{Q}}\cap \overline{(0,1)}=\mathbb{R}\cap [0,1]=[0,1]$$

This is right? Thanks!

Answer 1

Your attempt doesn't work. $\overline{X\cap Y}=\overline{X}\cap \overline{Y}$ is not true for all $X,Y$. Try $X=(0,1)$ and $Y=(1,2)$, then $X\cap Y=\{\}$, so $\overline{X\cap Y}=\{\}$, but $\overline{X}\cap \overline{Y}=[0,1]\cap[1,2]=\{1\}$.

I would do it like this. Since $X=\Bbb Q\cap (0,1)$ is contained in $Y=(0,1)$, we have $\overline{X}\subseteq \overline{Y}=[0,1]$. To do the other way around, we note that for any $a\in [0,1]$, there exists a sequence $(x_1,x_2,\ldots)$ in $X$ with limit $a$. If $a=0$, take $x_k=1/2^k$. If $a=1$, take $x_k=1-1/2^k$. If $0<a<1$, we start with $x_1\in \Bbb Q\cap (a,1)$. For $k=2,3,\ldots$, pick $x_k\in \Bbb{Q}$ in $\left(a,\frac{a+x_{k-1}}{2}\right)$. So, $|x_k-a|\leq \frac{|x_1-a|}{2^{k-1}}$, so $x_k\to a$.