My intuiton says that the answer is no, but I don't know how to prove it. I think the problem is in the cardinality of $\mathbb R$ and $\mathbb Q$... Any hint?
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8hint: connectedness – Mr. Cooperman Sep 29 '21 at 01:09
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2@Timkinsella : Connectedness is not normally introduced until well after the intermediate value theorem. And the latter is plenty. – Michael Hardy Sep 29 '21 at 03:26
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4Does this answer your question? Continuous Functions from $\mathbb{R}$ to $\mathbb{Q}$ – TryingHardToBecomeAGoodPrSlvr Sep 29 '21 at 04:24
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@MichaelHardy, how would you use IVT if the codomain is $\mathbb{Q}$? I thought the generalization of IVT to other codomains was the concept of connectedness. – Joe Sep 29 '21 at 11:13
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@Ethan, is this really a *calculus* question?! Or is it a topology question? – Joe Sep 29 '21 at 11:15
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Since f is continuous and takes only rational value , so the function must be a constant one : If not constant then obviously there will exist two real numbers a and b so that f(a) and f(b) are unequal. Since f is continuous then by intermediate value property f must take on all the values lying between f(a) and f(b) , which contradicts that f takes only rational values . Since f is a constant function, it can't be a surjective one.
Grets Etya
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