no continuous function switches the rational and the irrational numbers

Let 𝕁= denote the irrationals. There is no continuous functionMathworldPlanetmathPlanetmath f: such that f()𝕁 and f(𝕁).


Suppose there is such a function f.

First, =i{qi} implies


This is because functions preserve unions (see properties of functionsPlanetmathPlanetmath).

And then, f() is first category, because every singleton in is nowhere dense (because with the Euclidean metric has no isolated points, so the interior of a singleton is empty).

But f(𝕁), so f(𝕁) is first category too. Therefore f() is first category, as f()=f()f(𝕁). Consequently, we have f()=i{zi}.

But functions preserve unions in both ways, so

=f-1(i{zi})=if-1({zi}). (1)

Now, f is continuous, and as {zi} is closed for every i, so is f-1({zi}). This means that f-1({zi})¯=f-1({zi}). If int(f-1({zi})), we have that there is an open interval (ai,bi)f-1({zi}), and this implies that there is an irrational number xi and a rational numberPlanetmathPlanetmathPlanetmath yi such that both lie in f-1({zi}), which is not possible because this would imply that f(xi)=f(yi)=zi, and then f would map an irrational and a rational number to the same elementMathworldMathworld, but by hypothesisMathworldPlanetmathPlanetmath f()𝕁 and f(𝕁).

Then, it must be int(f-1({zi}))= for every i, and this implies that is first category (by (1)). This is absurd, by the Baire Category Theorem.

Title no continuous function switches the rational and the irrational numbers
Canonical name NoContinuousFunctionSwitchesTheRationalAndTheIrrationalNumbers
Date of creation 2013-03-22 14:59:15
Last modified on 2013-03-22 14:59:15
Owner yark (2760)
Last modified by yark (2760)
Numerical id 13
Author yark (2760)
Entry type Result
Classification msc 54E52