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 $a>x>b$?
Continuous $f: \mathbb R \to \mathbb R$ with infinitely many zeroes in every interval
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calculus
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3@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
1 Answers
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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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0@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