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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0@KaratugOzanBircan Why is `\rightarrow` better than `\mapsto`, the latter is more appropriate IMO – 2011-11-05
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5@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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3@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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0Thanks for the clarification, good to know. – 2011-11-05
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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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0Welcome back! It's a pleasure to see you active again. – 2011-11-05
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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