# Proving that there is no continuous function $f:\Bbb R\to\Bbb R$ satisfying $f(\Bbb Q)\subset\Bbb R-\Bbb Q$ and $f(\Bbb R-\Bbb Q) \subset\Bbb Q$. [duplicate]

How can I prove that there is no continuous function $f:\mathbb{R}\to \mathbb{R}$ satisfying $f(\mathbb{Q}) \subset \mathbb{R}\backslash \mathbb{Q}$ and $f(\mathbb{R}\backslash \mathbb{Q} ) \subset \mathbb{Q}$?

• The answers to this question are an impenetrable mess because the first version was flawed. You can't tell which version of the question is being answered! How can we fix this?? – TonyK Jan 29 '16 at 17:20
• Sorry, man. =/ Arpit Kansal answered me. – Filburt Jan 29 '16 at 17:24

Hint: Since $\bf R$ is connected and $f$ is non constant hence$f(\bf R)$ is connected and hence uncountable.

• I changed the question. Because Q is countable and R-Q is not, the other question was obvious. Your hint works for this case? – Filburt Jan 29 '16 at 17:07
• Note that because $Q$ is countable hence $f(Q)$ is countable,also $f(R\Q)$ is countable because contained in $Q$.Hence $f(R)$ is countable,contradiction! – Arpit Kansal Jan 29 '16 at 17:11
• what's the problem with $f(R)$ be countable? – Filburt Jan 29 '16 at 17:16
• @FernandoVieiraCostaJúnior Do you know only connected set in $R$ are intervals? – Arpit Kansal Jan 29 '16 at 17:17
• oh, yes. Sorry, man. Since f is continuous and non constant, f(R) is connected, and so uncountable. You made it! – Filburt Jan 29 '16 at 17:20

Hint: $f(\Bbb Q)$ is countable.

New hint given the edit: $f(\Bbb R)$ must be connected, and hence an interval, and hence must contain uncountably many irrationals.

• Sorry, I saw this problem. So I changed the question. See it again. – Filburt Jan 29 '16 at 17:11
• @FernandoVieiraCostaJúnior: The hint is still good. Note that $[a,b]\setminus \mathbb Q$ is not countable. – hmakholm left over Monica Jan 29 '16 at 17:18

HINT: Use the fact that

Between any two rational numbers, we have infinite irrational numbers

and similarly,

Between any two irrational numbers, we have infinite rational numbers.