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Given infinitely many (different) points $x_1,x_2,\ldots$ in some bounded domain $U\subset \mathbb{R^n}$ and two arbitrary points $P,Q\in U$ (with $P\neq Q$). I would like to show that there exist an analytic curve $\gamma:[0,1]\rightarrow U$, with $\gamma(0)=P$, $\gamma(1)=Q$, and such that infinitely many of the points $\lbrace x_i \rbrace$ are in the image of $\gamma$. Is that possible?

For all the examples I can think of it is indeed possible. But I think there should be an example of wildly distributed points such that it is not possible anymore to find a nice analytic curve going through infinitely many points. Can somebody help me with my problem?

Best regards

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Such a curve does not exist in general. Let $n=2$ and $x_k = (1/k, 0)$. Also pick $P$ and $Q$ anywhere not on the $x$-axis.

Let $t\mapsto (u(t), v(t))$, $t\in [0, 1]$ be a parametric equation of the curve. By construction, the function $v$ is nonzero at $0, 1$ but has infinitely many zeros on the interval $(0, 1)$. Therefore, these zeros have an accumulation point $t_0\in (0, 1)$. Since $v$ is analytic, this implies $v\equiv 0$, a contradiction.

Note this only requires the parametrized curve to be continuous on $[0, 1]$ and analytic on the open interval $(0, 1)$.

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