I have given an task to proof or refute that there NOT exists a function $f:[0,1]\rightarrow \left( 0,1\right)$ with $f(\frac{1}{2})=\frac{1}{2}$ which is bijective and also continuous.
What i was thinking about is that two sets have equal size when there exists a bijection between them. But i guess this wont work the other way around. Can anyone help me on this?
You can prove even more: there is no continous bijection $f:[0, 1]\to (0, 1)$.
If $f:[0, 1]\to (0, 1)$ is continous and bijective then $f$ has to be monotonic. This follows from the fact that every continous function $[0, 1]\to(0, 1)$ has intermediate value property. So assume that $f$ is not monotonic, i.e. it reverses order at some point. Well, such function has to move either from "increasing" to "decreasing" or vice versa. Both cases are analogous so without a loss of generality assume that there are $x,y,z\in[0, 1]$ such that $x < y < z$ and $f(x) < f(y)$ and $f(z) < f(y)$. Now pick a value $v\in(0, 1)$ such that
$$\max(f(x), f(z)) < v < f(y)$$
By intermediate value property there is $a_1\in (x, y)$ and $a_2\in(y, z)$ such that $f(a_1)=v=f(a_2)$. Contradicts with bijectivness of $f$.
Now without a loss of generality assume that $f$ is increasing. Then obviously there is no argument $x\in[0,1]$ such that $f(x)\in (0, f(0))$ (due to the increasing nature of $f$). Contradiction. $\Box$
