Relationship between Continuity and Countability This is a consequence of one of the problems in elementary real analysis that I am attempting to solve. I have this doubt.
Suppose $f$ is a continuous map from the reals to the reals. If the set $S=${$f(x)|x\in A$} is countable, is the set $S'=${$f(x+1)|x\in A$} countable? Here $A$ is just a subset of $R$. 
Well, the question may seem a bit artificial but let me also share the initial question which caused this doubt. 
The original question is: Suppose $f$ is continuous such that $f(x)$ is rational iff $f(x+1)$ is irrational. Show such an $f$ cannot exist.
My method:
Suppose such an $f$ exists. A quick thinking tells us that it suffices to consider a restricted domain [$0,1$] only. Consider the set of rationals, $Q$, in this domain. Clearly $f(Q)$ is countable. Consider those irrationals for which $f$ is rational. This particular range of $f$ also is countable as it is a subset of rationals. So we are left with those irrational points in this domain, at which $f$ is irrational. If we can somehow show that the range of $f$ on this particular domain is countable, then it follows that range of $f$ is countable for all real points and hence $f$ is not continuous by IVT. 
This is where I am facing a hurdle. I have to prove that the range of $f$ on the final particular domain i.e. set of all irrationals at which $f$ is irrational, is also countable. I figured out that considering any point $x$ in this domain, where $x$ is irrational, if $f(x)$ is irrational, then $f(x+1)$ is rational, so the set {$f(x+1)$} is countable. Does this imply that the range of $f$ when we are concentrating only those irrational points at which $f$ is irrational, is also countable?

Let $f$ be zero on the nonpositive numbers and then the identity on the nonnegative numbers.
If you take $A=[-1,0]$, $|f(A)|=|\{0\}|=1$, but $|f(A+1)|=|[0,1]|=|\Bbb R|$.
(Here I abuse notation a little by writing $A+1=\{x+1\mid x\in A\}$)
