 Indeed, a classical counterexample to this theorem, the nowhere continuous function   \[  f(x) =    \begin{cases}     0 \text{ if \(x\) is a rational number } \\     1 \text{ if \(x\) is an irrational number}    \end{cases} \]   is not a legitimate function from the intuitionistic point of view since the property of being rational is not decidable on the real numbers. The theorem above implies that the continuum is not decomposable, and in van Dalen 1997, it is shown that this even holds for the set of irrational numbers.