MHB Working with Piecewise Functions

[Enzipino]

I'm given the following Piecewise function when $f:[0,1]\to[0,1]$:
$f(x) = x$ when $x\in\Bbb{Q}$
$f(x) = 1-x$ when $x\notin\Bbb{Q}$

I need to prove that $f$ is continuous only at the point $x=\frac{1}{2}$.

For this problem, I know I need to use the fact that a function $f$ is continuous at a point $x$ iff for every sequence ${{x}_{n}}$ that converges to a number $x$ as $n\to\infty$ we also have the sequence $f({x}_{n})$ converging to $f(x)$ as $n\to\infty$. So, my initial approach to this was to first assume that $x\ne\frac{1}{2}$. And then I let $x\in\Bbb{Q}$ and I choose a sequence of irrational numbers ${x}_{n}$ that converges to $x$ as $n\to\infty$. But then from here I don't know how to work with the functions of the sequences. Do I have the right idea for this?

I also wanted to know that if I could simply just say that $f(x)= f^{-1}(x)$ once I prove that the function is bijective? (This is a separate problem but involves using the same piecewise function.

 

[Fallen Angel]

Hi,

I don't think the fact that $f(x)=f^{-1}(x)$ is helpful there, but it's correct.

For the first part, notice that given a point $x\in [0,1]$ you can choose a sequence of rational numbers $(y_{n})$ converging to $x$ and you will get a sequence $(f(y_{n}))$ converging to $x$ and you can also choose a sequence of irrational ones $(z_{n})$ and you will get a sequence $(f(z_{n}))$ converging to $1-x$.