Show the closure of $\mathbb{Q}$ in $\mathbb{R}$ is $\mathbb{R}$. I know this question has been asked on here before, but the answers did not contain a formal proof which I am trying to do. 
To show $\operatorname{cl}\mathbb{Q}$ =  $\mathbb{R}$, I want to do it by showing two inclusions. Clearly we know $\operatorname{cl}\mathbb{Q} \subset \mathbb{R}$ but I am having trouble showing  $\mathbb{R}$  $\subset\operatorname{cl}\mathbb{Q}$. We can write $\operatorname{cl}\mathbb{Q} = \bigcap_{I \in J}A_I$ where each $A_I$ is a closed set in $\mathbb{R}$ containing $\mathbb{Q}$. However I don't know how this helps for me to show the inclusion I am stuck on. Also limit points have not been introduced yet and shouldn't be needed for this problem. 
 A: Let $A$ be a closed set containing $\mathbb{Q}$. Then $A^c$ is open. Suppose there is an $a\in A^c$. Then there is an open interval $(a-\epsilon,a+\epsilon)\in A^c$. So there is an open interval with no elements in $A$ and this implies there is an open interval with no elements in $\mathbb{Q}$, which is impossible since the rationals are dense. Thus $A^c=\emptyset$, so $A=\mathbb{R}$. So the only closed set containing $\mathbb{Q}$ is $\mathbb{R}$, so cl$(\mathbb{Q})=\mathbb{R}$.
A: Let $x\in \mathbb{R}$ and $q_n = \frac{\lfloor n x \rfloor}{n}$.
Then $q_n \in \mathbb{Q}$, and $q_n \to x$. Since $\mathbb{Q}\subset \mathbb{R} $ it follows that $\overline{\mathbb{Q}} = \mathbb{R}$.
A: You need to show that every point in $\mathbb R$ is in $\operatorname{cl}\mathbb(Q)$.  So let $x\in\mathbb R$.  To conclude that $x\in\operatorname{cl}\mathbb(Q)$ it is enough to show that for all $\varepsilon>0$, $(x-\varepsilon,x+\varepsilon)\cap\mathbb Q\ne \varnothing$.  If that set were empty, there would be no rational numbers between $x-\varepsilon$ and $x+\varepsilon$.  Thus the numbers $k/n,\  k\in\mathbb Z$ would never intersect $(x-\varepsilon,x+\varepsilon)$, no matter how big $n\in\mathbb N$ is.  That would imply $0<2\varepsilon<1/n$, so that $1/(2\varepsilon)>\sup\mathbb N$.  In particular $\sup\mathbb N$ would exist.  Some integer $m$ would be $\le\sup\mathbb N$ and $>(\sup\mathbb N)-1$.  Then think about $m+1$ and you get a contradiction.
To think about the collection of all closed subsets of $\mathbb R$ that contain $\mathbb Q$ is to make the argument more complicated than it needs to be.
A: The $\Bbb{R}$eals, by definition, or by proof satisfy the criteria to be a linear continuum $L$: $(1) \ \ $ L has the least upper bound property : for any subset of $L$ bounded above, there exists an l.u.b..   $(2) \ \ $  For any $x \lt y \in L$, there exists $z$ such that $x \lt z \lt y$.
Use these to prove that every point in $\Bbb{R}$ is a limit point of a rational sequence.
