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I need to create a bijection between $[4,5)$ and $(9,12)$. My strategy was to do it in two parts. First I created a set $X=\{x_n \in [4,5) | n \in \mathbb N \cup \{0\}\}$ Such that $x_n=5-\frac {4}{2^{n+2}}$. Then I defined $h(x_n)=x_{n+1}$ and $p(x)=x$. Then $f(x)=h(x)$, if $x \in X$ and $f(x)=p(x)$ otherwise. Then $f(x)$ takes the interval $[4,5)$ to $(4,5)$ Next I defined $g(x)=3x-3$ to take the open interval $(4,5)$ to $(9,12)$. Then I composed the two functions and got $F(x)=g(x_{n+1})$, if $x \in X$ and $F(x)=g(x)$ otherwise. Is this correct or have I made a mistake? How to I prove if this is a bijection?

Markus Punnar
  • 565
  • This is correct. As for how to prove that it is in fact a bijection, you can prove that each individually is a bijection and prove that the composition of bijections is a bijection. Proving the second part is a bijection should be trivial. Proving the first, show that it is both surjective and injective. I.e. show that each element in the codomain both has a preimage and has only one preimage. See this related question – JMoravitz Dec 12 '17 at 17:46
  • That's correct. Another way could be defining a bijection from $[4,5]$ onto $(4,5)$, then a straightforward bijection from $(4,5)$ onto $(9,12)$..For the first step, choose a sequence $(x_n)$ such that all terms belong to $(4,5)$ and define $f$ such that $f(4)=x_1, f(5)=x_2, f(x_n)=x_{n+2}$ and for all $x$ not belonging to the sequence $f(x)=x$. – CiaPan Dec 12 '17 at 17:56

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