No continuous function switches Q\mathbb{Q} and the irrationals

Is there a way to prove the following result using connectedness?

Result:

Let J=RQ denote the set of irrational numbers. There is no continuous map f:RR such that f(Q)J and f(J)Q.

http://planetmath.org/encyclopedia/ThereIsNoContinuousFunctionThatSwitchesTheRationalNumbersWithTheIrrationalNumbers.html

Answer

Here’s a way to use connectedness, really amounting to using the intermediate value theorem.

If f(Q)RQ and f(RQ)Q, then f(0)f(2). Because intervals are connected in R and f is continuous, f[0,2] is connected. Because connected subsets of R are intervals, f[0,2] contains the interval [min{f(0),f(2)},max{f(0),f(2)}]. The set of irrational numbers in this interval is uncountable, yet contained in the countable set f(Q), a contradiction.

A slightly briefer outline: The hypothesis implies that f is nonconstant with range contained in the countable set Qf(Q), whereas the intermediate value theorem and uncountability of R imply that a nonconstant continuous function f:RR has uncountable range.

Attribution
Source : Link , Question Author : user10 , Answer Author : Jonas Meyer

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