No swapping

There is no continuous function f:\Bbb{R}\to\Bbb{R} such that f(\Bbb{Q})\subseteq \Bbb{R}\setminus\Bbb{Q} and f(\Bbb{R}\setminus\Bbb{Q}) \subseteq \Bbb{Q}.

No swapping

2 thoughts on “No swapping

  1. gciruelos says:

    Suppose such f exists.
    Then f(\mathbb{R}) is connected, because \mathbb{R} is connected and f is continuous. But f(\mathbb{R}) = f(\mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})) = f(\mathbb{Q}) \cup f(\mathbb{R} \setminus \mathbb{Q}) \subseteq f(\mathbb{Q}) \cup \mathbb{Q}, which is countable.
    So f(\mathbb{R}) is at most countable. But then f(\mathbb{R}) has only one element (because if C is a connected metric space then \# C = 1 or \# C \geq c). Then f is constant.

    A contradiction follows easily from this fact

    Liked by 2 people

  2. A second solution:

    Suppose f were such a function. Then f(x) \neq qx for all x\in \Bbb{R} and q\in\Bbb{Q}. Even more, by the mean value theorem we either have f(x) > qx for all x\in\Bbb{R} or f(x)<qx for all x\in\Bbb{R}.

    Let m=f(1). Then, if q,q'\in\Bbb{Q} are such that q<m<q', then qx<f(x)<q'x for all x\in\Bbb{R}, since the last set of inequalities holds for x=1. Letting q,q' approach m we get f(x)=mx for all x\in\Bbb{R}. Since obviously m can not be zero, f is bijective; in particular, it induces a bijection between the rational and the irrational numbers.

    Liked by 1 person

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