Smallest dense subset of $\mathbb{R}$ I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion.
The set of real numbers $\mathbb{R}$ is obtained as completion of $\mathbb{Q}$. However, $\mathbb{Q}$ is not the only set which is dense in $\mathbb{R}$. $\mathbb{Q} \backslash \mathbb{Z}$ is also a dense subset of $\mathbb{R}$. I am wondering if it makes sense to talk of the "smallest" dense subset of $\mathbb{R}$. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set $A$ of $\mathbb{R}$ such that if $B$ is a proper subset of $A$ then $B$ is not dense in $\mathbb{R}$. I am able to "see" that the set $A$, I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why $A$ doesn't exist. Could some one throw more light on this?
 A: If $A$ is a dense subset of $\mathbb{R}$, then $A\neq \emptyset$. In particular, there exists an element $a\in A$. If $B=A\setminus \{a\}$, then is $B$ dense in $\mathbb{R}$? (Hint: if there is an open subset $U$ of $\mathbb{R}$ such that $U\cap A=\{a\}$, then $A$ is not dense in $\mathbb{R}$.)
Exercise 1: Prove or give a counterexample: a finite intersection of dense subsets of $\mathbb{R}$ is dense in $\mathbb{R}$.
Exercise 2: Prove or give a counterexample: if $\{A_i\}_{i\in I}$ ($I$ is an index set) is an infinite collection of dense subsets of $\mathbb{R}$ such that the intersection of any finite number of $A_i$'s is again dense in $\mathbb{R}$, then the intersection of all the $A_i$'s dense in $\mathbb{R}$. (If you wish to view a hint, hover your cursor over the grey region directly below:

(Hint: if $x\in \mathbb{Q}$, let $A_x=\mathbb{Q}\setminus \{x\}$; prove that the intersection of any finite number of $A_x$'s is dense in $\mathbb{R}$ and determine the intersection $\bigcap_{x\in\mathbb{Q}} A_x$.)

Exercise 3: If $A$ is dense in $B$ and if $B$ is dense in $\mathbb{R}$, $A\subseteq B\subseteq \mathbb{R}$, then is $A$ dense in $\mathbb{R}$?
Exercise 4 (if you are familiar with measure theory): Let $\epsilon>0$. Prove that there exists an open dense subset $U$ of $\mathbb{R}$ such that the Lebesgue measure of $U$ is at most $\epsilon$. (If you wish to view a hint, hover your cursor over the grey region directly below:

(Hint: enumerate $\mathbb{Q}$ as $q_1,q_2,\dots$. If $n\in\mathbb{N}$, let $I_n=(q_n-\frac{\epsilon}{2^{n+1}},q_n+\frac{\epsilon}{2^{n+1}})$; prove that the union $\bigcup_{n\in\mathbb{N}} I_n$ is an open dense subset of $\mathbb{R}$ and determine the Lebesgue measure of $\bigcup_{n\in\mathbb{N}} I_n$.)

Exercise 5 (if you are familiar with path-connected topological spaces): If $U$ is an open path-connected subspace of $\mathbb{R}$ and if $U$ is dense in $\mathbb{R}$, then prove that $U=\mathbb{R}$.
I hope this helps!
A: Let $X$ be a topological space which is separated and without isolated points: that is, for all $x \in X$, the singleton set $\{x\}$ is closed and not open.  Let $Y$ be a dense subset of $X$.  
CLAIM: For every nonempty open subset $U$ of $X$, $U \cap Y$ is infinite.  
Proof: First observe that $U$ is infinite: if not, removal of all but one of its points would make a singleton open set.  Then similarly, if $U \cap Y$ were finite, removal of all of its points would leave a nonempty open subset of $X$ which is disjoint from $Y$, contradicting denseness of $Y$.
It follows immediately from the claim that removing any finite number of points from $Y$ leaves us with a subset which still meets every nonempty open subset of $X$ so is still dense.  In particular, there is no minimal dense subset of $X$.  
In particular the argument applies to any metric space without isolated points, like 
$\mathbb{R}$.  It is a good exercise to check that both of the hypotheses imposed on $X$ are necessary.  Especially, in a non-separated space one may well have singleton dense subsets: these are called generic points and are ubiquitous (and useful) in algebraic geometry.
