# Problem: proving that function is constant [closed]

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

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

## closed as off-topic by user21820, Xander Henderson, Did, TheSimpliFire, Chinnapparaj RNov 20 '18 at 2:58

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• What's $\mathbb{I}$? – Dante Grevino Nov 19 '18 at 12:26
• @DanteGrevino it's set of irrational numbers – user560461 Nov 19 '18 at 12:27
• Hint: assume it is not and then use the Intermediate Value Theorem to show that $f$ must take on a rational value. – John Douma Nov 19 '18 at 12:29
• Basically, rationals are everywhere. You cannot move an inch, a millimeter or an angstrom without stepping on a rational, or, for that matter, an infinity of rationals. – Eric Duminil Nov 19 '18 at 15:50

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 $$f(a) then there has to be a rational number $$r$$ with $$f(a) 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 $$\mathbb R\setminus \mathbb Q$$ 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 $$i_1 . Now pick some rational number in $$(i_1, i_2)$$. The intermediate value theorem gives $$r \in f(\mathbb R)$$, a contradiction.