Bdtyktl

Let $f colon mathbb{R} to mathbb{R} $ be a continuous function so that $ f(x) in mathbb I = mathbb Rsetminus mathbb Q, forall x in mathbb{R}$. Prove that $f$ is constant.

I tried assuming it's not a constant but I can't get to a contradiction with continuity.

Assuming we know 1. The Intermediate Value Theorem and 2. that between any two irrationals there is a rational, then the proof by contradiction should work fine.

If $f(a)<f(b)$ then there has to be a rational number $r$ with $f(a)<r<f(b).$ By IVT, there is a $c$ such that $f(c)=r$, contradiction.

The image $f(mathbb{R})$ is path-connected and the only non-empty path-connected subspaces of $mathbb{I}$ are the points. So $f$ is constant.

The continuous image of a connected set is connected.

The only connected sets in $mathbb Rsetminus mathbb Q$ are single points.

Done.

Suppose that $f$ is not constant. Then there are $i_1, i_2 in mathbb I$ such that $i_1 ne i_2$ and $i_1,i_2 in f(mathbb R)$. WLOG we can assume that $i_1 <i_2$. Now pick some rational number in $(i_1, i_2)$. The intermediate value theorem gives $r in f(mathbb R)$, a contradiction.