real analysis - Continuous, bijective function from $f:[0,1)to mathbb{R}.$

Prove that there does not exist a continuous, bijective function $f:[0,1)\to \mathbb{R}.$

By contradiction I can assume a function exists, so that function is surjective, onto and continuous. And I know I need to use the intermediate value theorem but I can't create such a contradiction.

Answer

From the interaction in comments I think OP needs bit more elaboration on the hint by Asaf. I will however refrain from providing a complete solution.

Assume that there is a function $f$ defined on $[0, 1)$ which is continuous and a bijection from $[0, 1)$ to $\mathbb{R}$. It means that $f$ is one-one function in particular i.e $f(a) = f(b)$ implies $a = b$.

Next let's consider $f(0)$ and $f(1/2)$. Because $f$ is one-one we must have $f(0) \neq f(1/2)$.

Let's assume that $f(0) < f(1/2)$ (the case $f(0) > f(1/2)$ can be handled similarly). Now we need to prove that if $x \in (0, 1)$ then $f(0) < f(x)$. This is where you need to use IVT for continuous functions. You should be able to do this by following Asaf's comments. And then we know that $f(0)$ is the minimum value of $f(x)$ and hence the part $(-\infty, f(0))$ of $\mathbb{R}$ is not mapped by this function $f$ and thus $f$ is not onto $\mathbb{R}$.

The proof of $f(0) < f(x)$ for all $x \in (0, 1)$ proceeds as follows. Clearly $f$ is one-one so $f(0) \neq f(x)$. If $x = 1/2$ then we already know that $f(0) < f(1/2) = f(x)$. So let $x \neq 1/2$. If $f(0) > f(x)$ then $f(x) < f(0) < f(1/2)$ so that $f(0)$ lies between $f(x)$ and $f(1/2)$ and hence by IVT we have ..... (I hope OP will be able to complete the dots)

If $f(0) > f(1/2)$ then we can show that $f(x) < f(0)$ for all $x \in (0, 1)$ so that $f(0)$ is the maximum value of $f$ and again $f$ is not onto $\mathbb{R}$.

Another thing to note. There is nothing special in $1/2$ we have chosen above. It can be replaced by any number lying in $(0, 1)$.