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A bijection is a function that is both one-to-one and onto.

I need to find such a function that maps $\{ -2, -1 \} \cup [0, 1] \rightarrow (0, 1)$ or equivalently, $ (0, 1) \rightarrow \{-2, -1 \} \cup [0, 1]$.

This problem is very confusing because if I exclude any function output for $x \in [0, 1]$ (output occupied by $\{ -1, -2 \}$) I have to find another function output in $(0, 1)$ that associates with $x$, even though all numbers in $(0, 1)$ have already been covered.

Help needed.

Alexander Bullen
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  • Do you know how to map $[0,1] \rightarrow (0,1)$ bijectively? The trick will be to map $[0,1]$ to $(0,1/2]$ bijectively and then map $[0,1]$ to $(1/2,1)$ bijectively. See https://math.stackexchange.com/questions/213391/how-to-construct-a-bijection-from-0-1-to-0-1 and https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1?rq=1 for ideas – User203940 Nov 05 '20 at 19:43

1 Answers1

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We can compose bijections. First let's get a bijection from $[0,1]\to(0,1)$.

To do it, we can consider the following function (see this answer): $f(x) = \left\{ \begin{array}{1 1} \frac{1}{2} & \mbox{if } x = 0\\ \frac{1}{2^{n+2}} & \mbox{if } x = \frac{1}{2^n}\\ x & \mbox{otherwise} \end{array} \right.$

So we have the bijection $f:[0,1]\to(0,1)$. Now consider $g:[0,1]\cup\{-1,-2\}\to[0,1]$ defined by $g(x) = \left\{ \begin{array}{1 1} f(x) & \mbox{if } x \in[0,1]\\ 0 & \mbox{if } x = -1\\ 1 & \mbox{if } x=-2 \end{array} \right.$

Then $h=f\circ g:[0,1]\cup\{-1,-2\}\to(0,1)$ is a bijection.

Darsen
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