Question

Let I = [0,1] and let f : I to R be defined by f(x) = x if x is rational and f(x) = 1 - x for x irrational. Show that f is injective on I and that f(f(x))=x and that f is only continuous at the point x = 1/2 Let I = [0,1] and let f : I to R be defined by f(x) = x if x is rational and f(x) = 1 - x for x irrational. Show that f is injective on I and that f(f(x))=x and that f is only continuous at the point x = 1/2 @Mathematics

Answer 1

First, I'd like to say this is an interesting exercise.
Let's work together... uhm.
First above all some notation.
\[\large\begin{array}{l}
I = [0,1]\\
f:I \to \mathbb{R} \\
\left\{ \begin{array}{l}
f(x) = x\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; if \;\;\; x \in \mathbb{Q} \\
f(x) = 1 - x \;\;\;\;\;\;\;\;\;\;\;\; if \;\;\;x \in \mathbb{R}
\end{array} \right\}
\end{array}\]

Then we know that a function to be continuous...

\[\large \mathop {\lim }\limits_{x \to f({x_0})} f(x) = f({x_0})\]

If we think about it:

\[\large \begin{array}{l}
\mathop {\lim }\limits_{x \to {1^ + }} x \ne 1\\
\mathop {\lim }\limits_{x \to {1^ - }} x \ne 1
\end{array}\]

Because in the very moment we do 1+ or 1- we are using an irrational number so of course we have to use the "second part" of the function.

But, it is continuos in the 1/2 because:

\[\begin{array}{l}
\mathop {\lim }\limits_{x \to {{\frac{1}{2}}^ + }} x = 1/2\\
\mathop {\lim }\limits_{x \to {{\frac{1}{2}}^ - }} x = 1/2
\end{array}\]

As a matter of fact 1/2 is the only number that subtracted to one gives 1/2.

Answer 2

Now, about injectivity, we have to think about the definition, it means that...

\[\large a \ne b \to f(a) \ne f(b)\]

We have no kind of problem if they are both rational, because it'd reproduce:
\[\large a \ne b\]

If just one of them is integer, we find out that :
\[\large a \ne 1-a\]
and of course:
\[\large b \ne 1-b\]

What if they are both irrational? Well..

\[\large 1 - a \ne 1 - b \to - a \ne - b \to a \ne b\]

Answer 3

Hope I helped! :) Good luck!

Answer 4

thnx. but if you approach 1 from the right since I [ 0,1] isn't the right side limit equal to 1?