Question:
Prove that the interval $[0,2)$ and $[5,6) \cup [7,8)$ have the same cardinality by constructing a bijection between the two sets. You can add a graph to support your argument.
I tried graphing a function with domain $[0,2)$ and target space $[5,6) \cup [7,8)$ but I can't come up with an explicit formula. Will the function be linear?
Essentially we want to map part of $[0,2)$ injectively onto $[5,6)$ and similarly the rest onto $[7,8)$. We will map $[0,1)$ onto $[5,6)$ and $[1,2)$ onto $[7,8)$ in the natural way; namely, we define $f:[0,2)\to[5,6)\cup[7,8)$ by $$ f(x) = \begin{cases} 5+x & \text{if}\ x\in[0,1), \\ 6+x & \text{if}\ x\in[1,2). \end{cases} $$ It remains to show that this function is both one-to-one and onto.
To map the interval $[0;2)$ to the union of two disjoint intervals $[5;6)\cup[7;8)$ you will clearly need a piecewise function. To be bijective, this function must have an inverse.
$$\newcommand{\hide}[1]{\underline{\phantom{#1}}}\begin{align*} f(x) =\begin{cases} \hide{x+5} & :& x\in [\hide{0;1}) \\[1ex] \hide{x+6} &:& x\in[\hide{1;2}) \\[1ex] \text{undefined} &:& \text{elsewhere}\end{cases} \\[2ex] f^{-1}(y) =\begin{cases} \hide{y-1} & :& y\in [5;6) \\[1ex] \hide{y-6} &:& y\in[7;8) \\[1ex] \text{undefined} &:& \text{elsewhere}\end{cases}\end{align*}$$
Can you find such a function and show $\forall x{\in}[0;2):\Big(\big(f(x){\in}[0;5){\cup}[6;7)\big)\wedge\big( f^{-1}\circ f(x)=x\big)\Big)$.
