TITLE: Continuous, bijective function from $f:[0,1)\to \mathbb{R}.$
QUESTION [8 upvotes]: 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.

REPLY [7 votes]: Let $f : [0,1) \rightarrow \mathbb{R}$ be a bijection. Then the restriction of $f$ to $(0,1)$ is a continuous bijection with range $\mathbb{R} - {f(0)}$ which is clearly disconnected but (0,1) is connected.

REPLY [4 votes]: Since there are continuous bijections (even homeomorphisms) from $(0,1)$ to $\mathbb R$, it somehow must have to do with that $0$. Now, how do we use it?
One observation would be that such a map $f$ restricts to a continuous bijection from $(0,1)$ to $\mathbb R\setminus\{f(0)\}$, the real line without one point. Why can such a map not exist?
A different version of the same argument would be that for such a map $f$ there would have to be an $a\in(0,1)$ with $f(a)<f(0)$ and a $b\in(0,1)$ with $f(b)>f(0)$. Now, what would the intermediate value theorem tell us then, and how would that lead to a contradiction?