real analysis - Bijective function need not implies continious

Is there exist a bijective function from $[0, 1)$ to $\mathbb{R}$?

I think it will not possible because $[0, 1)$ is not isomorphic $\mathbb{R}$

Any hints/solution will be appreciated

thanks u

Answer

Start with the identity on $[0,1)$. Then send $0$ to $1/2,$ $1/2$ to $1/3,$ etc. In the end you get a bijection from $[0,1)$ to $(0,1)$. To get a bijection, in fact, a homeomorphism from $(0,1)$ to $\mathbb R$, you can use eg. $\tan(\pi x - \pi /2)$.

You cannot get a continuous bijection of $[0,1)$ to $\mathbb R$ though because then the image of $(0,1)$ has to be an interval. But you get $\mathbb R \setminus\{\text{image of 0}\}.$