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This question might turn out to be really trivial.

$f$ is a one-to-one function from the interval $[0,1]$ to $\mathbb{R}$. Is it necessary that $\exists q \in \mathbb{Q}$ such that $f(x) = q$ for some $x \in [0,1]$ i.e. is it necessary that the image of $f$ contains a rational number?

I came across this question when I was browsing through some website.

I think this is false. But I am unable to come up with a counter example.

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    The cardinality of the rationals is countable. The cardinality of the reals is a continuum. So the irrationals also have the cardinality of the continuum. So compose any 1-1 function $[0,1] \to \mathbb R$ with a bijection $\mathbb R \to \mathbb R \setminus \mathbb Q$ and you have an answer to your question.2010-11-22
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    It would be nice if the title reflected the question a little more closely. Right now, it reads as if the question is about whether there *exists* a one-to-one function from $[0,1]$ to $\mathbb{R}$.2010-11-22
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    @Rahul: I edited the title. @Sivaram: I hope it is OK that I changed your title along the lines of Rahul's suggestion.2010-11-22
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    @Rahul, Jonas: Yes. The title now makes better sense. Thanks Jonas.2010-11-22

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If you want an explicit counterexample, define $f:[0,1]\to\mathbb{R}$ by $f(x)=x$ if $x$ is irrational, $f(x)=\sqrt2+x$ if $x$ is rational.

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    Thanks Jonas. I realized there must some trivial example like this, just that I didn't give it enough thought.2010-11-22
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    Ah, my mistake.2010-11-22
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It is obviously false since $[0,1]$ and $\mathbb{R}\setminus\mathbb{Q}$ have the same cardinality, so there exists a function $f: [0,1] \to \mathbb{R}\setminus\mathbb{Q}$ that is bijective. Do codomain expansion and you get an injective map from the unit interval to $\mathbb{R}$ that has no rationals in its range.