bijection between closed and open interval

For mapping the end points of the closed unit interval  $[0,\mathrm{\hspace{0.17em}1}]$  and its inner points bijectively onto the corresponding open unit interval  $(0,\mathrm{\hspace{0.17em}1})$,  one has to discern suitable denumerable subsets in both sets:

$[0,\mathrm{\hspace{0.17em}1}]=\{0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}1}/2,\mathrm{\hspace{0.17em}1}/3,\mathrm{\hspace{0.17em}1}/4,\mathrm{\dots }\}\cup S,$

$(0,\mathrm{\hspace{0.17em}1})=\{1/2,\mathrm{\hspace{0.17em}1}/3,\mathrm{\hspace{0.17em}1}/4,\mathrm{\dots }\}\cup S,$

where

$$S:=[0,\mathrm{\hspace{0.17em}1}]\setminus \{0,\mathrm{\hspace{0.17em}1},\mathrm{\hspace{0.17em}1}/2,\mathrm{\hspace{0.17em}1}/3,\mathrm{\hspace{0.17em}1}/4,\mathrm{\dots }\}.$$

Then the mapping $f$ from  $[0,\mathrm{\hspace{0.17em}1}]$  to  $(0,\mathrm{\hspace{0.17em}1})$  defined by

$$f(x):=\{\begin{array}{cc}1/2\hspace{1em}\text{for}\hspace{1em}x=0,\hfill & \\ 1/(n+2)\hspace{1em}\text{for}\hspace{1em}x=1/n\hspace{1em}(n=1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}3},\mathrm{\dots }),\hfill & \\ x\mathit{\hspace{1em}\hspace{1em}}\text{for}\mathit{\hspace{1em}}x\in S\hfill & \end{array}$$

is apparently a bijection.  This means the equicardinality of both intervals.

Note that the bijection is neither monotonic (e.g. $0\mapsto \frac{1}{2}$,  $\frac{1}{2}\mapsto \frac{1}{4}$, $1\mapsto \frac{1}{3}$) nor continuous.  Generally, there does not exist any continuous surjective mapping $[0,\mathrm{\hspace{0.17em}1}]\to (0,\mathrm{\hspace{0.17em}1})$,  since by the intermediate value theorem a continuous function maps a closed interval to a closed interval.