Q:

Continuous bijection between $[0, 1)$ and $\mathbb{R}$

I have to prove the non-existence of a continuous bijection between $[0,1)$ and $\mathbb{R}$.
My attempt:
Since $\mathbb{R}$ is homeomorphic to $(0, 1)$, I tried to prove the non-existence of a continuous bijection $g: [0, 1) \rightarrow (0, 1)$. So, assume by contradiction that such a map exists. I wanted to find a contradiction using the Intermediate Value Theorem, but without success.
Any suggestions? Thanks in advance!

A:

Suppose $f:[0,1) \to \Bbb R$ is a continuous bijection. 
Then $f[(0,1)]=\Bbb R \setminus \{f(0)\}$ which is a contradiction, as $(0,1)$ is connected and so its continuous image is too, while $\Bbb R\setminus \{p\}$ is disconnected for any $p \in \Bbb R$.

A:

If a map $f:[0,1)\to (0,1)$ satifies the intermediate value property, then $f\left[(0,1)\right]=I$ is a subinterval of $(0,1)$. If $f$ is bijective, then $I=(0,1)\setminus\{f(1)\}$. However, $(0,1)\setminus \{c\}$ is never an interval forany $c\in (0,1)$.

