A smooth path passing through infinitely many points of a given sequence While doing my research I came across an interesting question. 
Let $\{\bf{x}_n \}\subset R^m$ be any sequence of points with $\bf{x}_n\rightarrow \bf{0}$, as $n \rightarrow \infty$. Is there a smooth path $c: [0,a) \rightarrow \mathbf{R}^m$, $a>0$, $c(0)=\bf{0}$, a (decreasing) sequence $\{t_k\}\subset (0,a)$, an (infinite) subsequence $\{\bf{x}_{n_k} \}\subset \{\bf{x}_n \}$ so that $c(t_k)=\bf{x}_{n_k}$? Can we make the path analytic at zero? 
Any help, or suggestion would be appreciated.
 A: I'll just tackle the analytic case: Assume that you have a sequence 
$y_k = \phi(t_k)$ where $\phi$ analytic at $0$, $\phi$ has $0$ of order $d$ at $0$. Then we have $\frac{\log |y_k|}{\log |t_k|}\to d$. Now if you have a sequence $(x'_n, x''_n)\in \mathbb{R}^2$ approaching $(0,0)$ such that $\frac{\log|x'_n|}{\log|x''_n|}\to \sqrt{2}$, then no subsequence can be interpolated by an analytic function. 
$\bf{Added:}$ Consider the $C^{\infty}$ case now.  We'll show the following: if the sequence $a_n$ decreases fast enough to $0$ then there exists a smooth function $f$ defined around $0$ so that $f(\frac{1}{n}) = a_n$. Indeed, consider $\phi_n$ smooth bump function such that $\phi_n(\frac{1}{m}) =\delta_{mn}$. Now the take $a_n$ so that $|a_n| \le \frac{1}{2^n \|\phi_n\|_n}$, where the norm $\|\psi\|_n = \sup_{t\in [-1,1], 1\le k \le n} |\psi^{(k)}(t)|$. Then the function $f$ equals the series $\sum a_n \phi_n$, which converges to a smooth function.
Now it's enough to choose a subsequence of $x_n$ that converges to $0$ fast enough and thus can be interpolated at $\frac{1}{n}$. I think it can be arranged that the whole sequence can be interpolated by a smooth function at a sequence of values ( we have to retard enough the convergence to $0$ of this). 
