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.

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.

• Is this f bijection? – user322806 Mar 14 '16 at 20:16
• Can you explain, how this function f is continuos at a point "1"? – user322806 Mar 14 '16 at 20:17
• I think you can figure out by yourself why this function works for your example. – H. Potter Mar 14 '16 at 20:29
• Yes... i think this example is ok.. thank you..!! – user322806 Mar 15 '16 at 3:07

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.