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Let $(P,<)$,$|P|>1$, be a densely totally ordered set, i.e. $(P,<)$ is totally ordered and for all $x, y\in P$, $x<y$ implies there is $z\in P$ such that $x<z<y$.

Can there exist a strictly positive map $f:X\rightarrow \mathbb{R}$ that satisfies that $f$ converges to zero at every point in $P$? Namely for every $\varepsilon>0$ and every $x\in P$, there are $x_1, x_2 \in P$, $x_1<x<x_2$, such that for all $y\in (x_1,x_2)\setminus \{x\}$, $f(y)<\varepsilon$?

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On $P = (0,1) \cap \mathbb Q$, define $f(p/q) = 1/q$, where $p/q$ is in simplest terms. This is known as Thomae's function or the popcorn function, except restricted to the rationals. For each $\epsilon > 0$, there are only finitely many points in $P$ that get mapped to a value $\geq \epsilon$, so it is easy to find an interval around any point that avoids them all.

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