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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?

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    Possible duplicate of [Continuous bijection from $(0,1)$ to $[0,1]$](http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1) – 2017-02-21
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    @B.Mehta Not quite. Non-existence of a continous bijection $(0, 1)\to [0, 1]$ doesn't imply non-existence of a continous bijection $[0,1]\to (0,1)$. – 2017-02-21
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    Do you know compactness property? – 2017-02-21
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    well i should do – 2017-02-21
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    We have a very nice result then. That a continuous image of a compact set is compact. So $[0,1]$ is compact in $\Bbb R$. Which should get mapped to a compact set under continuous map. But $(0,1)$ is not compact. Hence such $f$ can't be continuous. Note that even injectivity wasn't required here. – 2017-02-21
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    @freakish Ah yes, my mistake – 2017-02-21

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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. $\Box$