The set of numbers whose decimal expansions contain only 4 and 7 Let $S$ be the set of numbers in $X=[0,1]$ that when expanded as a decimal form, the numbers are 4 or 7 only.
The following are the problems.
a), Is S countable ?
b), Is it dense in $X$ ?
c), Is it compact ?
d), Is it perfect ?
For a), I want to say that it is intuitively, but I have no idea how to prove this. I tried to come up with a bijection between $S$ and $\Bbb Z$ but I couldn't find one.
For b), my understanding of a set being "dense" means that all points in $X$ is either a limit point or a point in $S$. Am I right? Even if I were, I am not sure how to show this.
For c), my intuition tells me that it is because it is bounded. So if I could show that it is closed I will be done, I think.  But I am still iffy with the idea of limit points, and I am not sure what kind of limit points there are in $S$.
For d), Because I can't show that it's closed I am completely stuck.  
I am teaching myself analysis, and I only know up to abstract algebra. Since I never took topology, please give me an explanation that helps without knowledge of advanced math.

S is not countable because any countable subset of S must be a proper subset. To see this, suppose that A is contained in S and A is countable,  then its elements can be displayed  in some sequence x_1,x_2,...,x_n,.... We shall construct a number x of S that not belongs to A. 
If x_n have 4 at the nth decimal position, x have 7 at the nth position. And vice versa.
So x differs from any element of A,hence x does not belongs to A. This completes the proof.
