Cardinality of a subset of R to R

Question

To Check whether the cardinality is same I was solving cardinality of finite set.But couldn't find a way to mapp a bounded subset of $ \mathbb{ R } $ to $\mathbb{R}$ For example How to prove that cardinality of [0,1] and $ \mathbb{R} $ are same

Answer 1

Hint: Remember that you're not limited to continuous functions. Also, you can build a bijection in several steps, for instance $$[0, 1]\to [0,1)\to (0,1)\to(-1, 1)\to \Bbb R$$

Answer 2

$\star$ $[0,1]$ and $(0,1)$ are equivalent

To see this, let $A=[0,1] \setminus \{0,1,\frac{1}{2},\frac{1}{3},...\}$. Define $f:[0,1] \rightarrow (0,1)$ by

$$f(x)= \left\{ \begin{array}{lcc} \frac{1}{2} & x=0 \\ \\ \frac{1}{n+2}, & x=\frac{1}{n} \\ \\ x & x\in A \end{array} \right.$$ Then $f$ is a bijection.

$\star$ $(0,1)$ and $(-\frac{\pi}{2},\frac{\pi}{2})$ are equivalent.

Consider $g:(0,1) \rightarrow (-\frac{\pi}{2},\frac{\pi}{2})$ defined by $$g(x)=-\frac{\pi}{2}+\pi x$$

$\star$ $(-\frac{\pi}{2},\frac{\pi}{2})$ and $\Bbb{R}$ are equivalent.

Consider $h:(-\frac{\pi}{2},\frac{\pi}{2}) \rightarrow \Bbb{R}$ by $$h(x)=\arctan x$$

Finally, we conclude $$[0,1] \sim (0,1) \sim (-\frac{\pi}{2},\frac{\pi}{2}) \sim \Bbb{R}$$

To explicitly find the map from $[0,1]$ to $\Bbb{R}$, take the composition of the above maps!