A set of all rational numbers in $[0, 1]$? I have a question that is giving me some minor grief:
If $A$ is a closed set containing all rational numbers $r \in [0, 1]$, then show that $[0, 1] \subset A$.
I don't really understand this question - surely that set of all values $[0, 1]$ contains infinitely more points that the set of rational numbers over the same interval? Am I missing something large?
Thanks in advance.
 A: Since $A$ is closed, it contains all limit points to all sequences as well.
A: A is closed, so it contains the closure of the set of all rational points in [0,1], because the closure of a set S is the smallest closed set that contains S. By contrast, the interior of a set S is the greatest open set that is contained in S.  
A: All can be derived from properties of closure operation $\bar{X}$: 
$\bar{A} = A$, since A is closed.
Also, if $B~\subset~A$, then $\bar{B}~\subset\bar{A}$.
$\bar{\mathbb{Q}} = \mathbb{R}$, since $\mathbb{Q}$ is dense in $\mathbb{R}$.
Altogether gives you:
. $\mathbb{Q}\cap[0,1] \subset A$ (hypothesis)
. $\overline{\mathbb{Q}\cap [0,1]}~\subset~\bar{A}$
. $[0,1]~\subset~A$
A: A closed set contains every point near it. So the question is asking you to show that every point in $[0,1]$ is near the subset of rational numbers between $0$ and $1$.
Maybe it will help to consider the complementary problem: if $A$ is an open set that doesn't contain any rational number between $0$ and $1$, then prove that it is disjoint from $[0,1]$
Or contrapositively, if an open set contains an element in $[0,1]$, then it contains a rational number in $[0,1]$.
