Wednesday, May 2, 2018

functions - Concerning bijections between intervals of real numbers.




To construct a bijection between, say, $A = (0,1) \subset \mathbb{R}$ and $B = [0,1] \subset \mathbb{R}$, the following function is often used:



$$ f(a) =
\begin{cases}
0, & \text{if $a = \frac{1}{2}$} \\
1, & \text{if $a= \frac{1}{3}$} \\
\frac{1}{n-2}, & \text{if $a = \frac{1}{n}$, for $n \ge 4$}
\end{cases}$$



where $a \in A$, and $b \in B$.




But, what about all of the irrationals in $A$ and $B$? Clearly, $f$ only accounts for rational numbers, does it not? Perhaps it's trivial, as one could simply add the case that $f(a) = a$ if $a \in \mathbb{R} \setminus \mathbb{Q}$ (although, I'm not sure if that's actually even well-defined); but since this bijection is often presented without any qualification, I assume I'm missing something obvious here...


Answer



As you suggest, making $f$ the identity on the irrationals will do just fine. You also have to define $f$ on the rationals whose numerator is greater than $1$. Easy question: what do you suggest there?



This is perfectly well defined. It's even continuous at most points.


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