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# Understand the problem

True or false: There exists a 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}\$$
##### Source of the problem
TIFR 2019 GS Part B, Problem 1
Real analysis
Easy
##### Suggested Book
Real Analysis; Bartle and Sherbert

Do you really need a hint? Try it first!

Can you use connectedness and intermediate value theorem?

If $f(\mathbb{Q})\subseteq \mathbb R\setminus\mathbb Q$ and $f(\mathbb R\setminus \mathbb Q)\subseteq\mathbb Q$, then $f(0)\neq f(\sqrt 2)$. Because intervals are connected in $\mathbb R$ and $f$ is continuous, $f[0,\sqrt 2]$ is connected. Now, what are the connected sets in $\Bbb R$?

Because connected subsets of $\mathbb R$ are intervals, $f[0,\sqrt 2]$ contains the interval $\left[\min\{f(0),f(\sqrt 2)\},\max\{f(0),f(\sqrt 2)\}\right]$. Now, what is the cardinality of irrational numbers in $\left[\min\{f(0),f(\sqrt 2)\},\max\{f(0),f(\sqrt 2)\}\right]$?
The set of irrational numbers in this interval is uncountable, yet contained in the countable set $f(\mathbb Q)$, a contradiction.

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

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