Bijection function from R to R with discontinuous at uncountable number of points and continuous at uncountable number of points. Consider R with usual topology (R means set of all real numbers ). Can we construct a function f:R -->R ** such that **f is one-one and onto ** which has **discontinuous at uncountable number of points and f is continuous at uncountable number of points?
If i have a mistake suggest me.. Give some explanation about this question and hints. Thanks in advance.
 A: Consider the following function:
If $x>0$ and $x\in \mathbb{Q}$: $f(x)=\frac1x$ 
Otherwise, $f(x)=x$
This function is continuous on $]-\infty,0[ \cup \{1\}$ which is uncountable. Its complement is uncountable too.
A: The idea is to find a function which is (at the least) continuous everywhere in some $C \subset \Bbb{R}$ and discontinuous at an uncountable set of point in some  $D \subset \Bbb{R}$, with $D \cap C = \emptyset$.
We will do this using  $C = \Bbb{R}^- \cup 0$ and $D = \Bbb{R}^+$.
You then want to make the continuous map 1:1 onto some $C' \subset \Bbb{R}$ and it will simplify our efforts if we choose $C' = C = \Bbb{R}^-$.  For $x \in C$ we can then take $f(x) = x$, which is clearly a continuous bijection.
Now for the hard part:  We are from now on restricted, on $D$ to $f(x) > 0$ and we wish to make $f(x)$ discontinuous at an uncountably infinite number of points, yet no two values of $f(x)$ can be equal, and the range of $f(x)$ must be $D$.  Let's get greedy and try to make $f(x)$ discontinuous everywhere on $D$.
The usual way to construct a function that is discontinuous everywhere is to define it in a different way on the rationals than on the irrationals.  So let's 
try keeping $f(x) = x$ for $x$ irrational.  That leaves only rational values available for the rationals, and our first try might be to say that $f(q)=q+1$ 
for $q\in\Bbb{Q}$.  But that does not work; the interval between $0$ and $1$ is not in the range, so it is not a bijection.  Similarly $f(q) = q-1$ fails because it sprays values into the forbidden negative region.
The first solution I thought of that works is $f(q) = \frac1{q}$ which leads to discontinuities everywhere on the positive real line except at $x=1$.  But it is perhaps cooler to use $f(q)=2q$ which leads to discontinuities on all of $\Bbb{R}^+$. This exploits the fact that the doubling transformation is a bijection of the positive rationals to themselves.
So:
$$
f(x) =  \begin{cases} 2x & x \in \Bbb{R}^+ \cap \Bbb{Q} \\
x & x \not\in \Bbb{R}^+ \cap \Bbb{Q} \end{cases}
$$
is discontinuous on the positive real axis and continuous on the negative real axis and at zero.
