Question about irrational numbers How  to show :

If $A \subseteq \mathbb{I}$ is not
  countable, then $A$ is not closed set
  in $\mathbb{R}$.

I think that if $A$ is closed, then $A^{c}$ is open thus: $A^{c}$ is an union of open intervals, but...
 A: The statement is not true.  There exist perfect1 (hence closed) subsets of $\mathbb R$ that doesn't intersect $\mathbb Q$, and perfect sets are not countable.  See: Perfect set without rationals.

1  A set $S$ is perfect if it is closed and every point of $S$ is an accumulation point of $S$. 
A: As another counterexample, enumerate the rationals as $\{q_k\}_{k=1}^\infty$, and let $U = \bigcup_{k=1}^\infty (q_k - 2^{-(k+2)}, q_k + 2^{-(k+2)})$.  $U$ is open and contains all the rationals, so its complement $U^c$ is closed and contains none of them.  But $U$ is a union of intervals of lengths $2^{-(k+1)}$, so its Lebesgue measure (i.e. length) is at most $\sum_{k=1}^\infty 2^{-(k+1)} = 1/2$.  So $U^c$ has measure at least $1/2$ and in particular must be uncountable.  You can modify this construction to make the  measure of $U^c$ as close to $1$ as desired.
A: As AgCl pointed out (and François G. Dorais, in the answer AgCl linked to), the statement is false. (And Nate also pointed out, as I was writing my answer.)
I would even say that the point is that ${\mathbb R}\setminus{\mathbb Q}$ is homeomorphic to the set ${\mathbb N}^{\mathbb N}$ (with the product topology of discrete ${\mathbb N}$). 
But, obviously, ${\mathbb N}^{\mathbb N}$ contains lots of perfect subsets, such as $\{1,2\}^{\mathbb N}$.
[A really pretty example of such a homeomorphism in the first chapter of the book "Descriptive Set Theory and Forcing: how to prove theorems about Borel sets the hard way" by Arnold Miller (which you can download from his webpage). The usual construction (which is also pretty) is to note that a sequence $(a_1,a_2,\dots)$ can be identified with the continued fraction of an irrational in $(0,1)$, and a bit of thought shows that this map is indeed a homeomorphism.]
