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Is there a continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for any two real numbers $a>b$, $f(x)=0$ has exactly a countable infinite many solutions with $a>x>b$?

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    @KaratugOzanBircan Why is `\rightarrow` better than `\mapsto`, the latter is more appropriate IMO2011-11-05
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    @Sasha: $\mapsto$ is usually used when explicitly saying where $x$ is mapped to, e.g. $$x\mapsto x^2$$ When the function is just from one set to another, $\to$ is more common.2011-11-05
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    @Sasha: It is appropriate to use \mapsto when we deal with the elements. For example, we define a function $f(x): \mathbb{R} \rightarrow \mathbb{R}$ such that $x \mapsto x^2.$2011-11-05
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    Thanks for the clarification, good to know.2011-11-05
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    @Karatung: I don't remember I've ever seen that written this way. Usually people write $f: \mathbb{R} \to \mathbb{R}, x \mapsto x^2$.2011-11-05

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No. If that were the case, then for every $a\in \mathbb R$ and every $n\in\mathbb N$, there would be an $x_n$ such that $a

More briefly: The zero set of a continuous function is closed, so if it is also dense, it must be all of $\mathbb R$. This makes the existence of such $f$ impossible, because intervals in $\mathbb R$ are uncountable.

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    Welcome back! It's a pleasure to see you active again.2011-11-05
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    @t.b.: Thanks! My participation will likely continue only sporadically for a while, but I at least visit now and then (sometimes just reading and maybe voting). It's good to see you here, too.2011-11-05