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.