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No. If $\varphi = (\sqrt{5}+1)/2$ (which is a root of $x^2 - x - 1$) and $F_n$ are the Fibonacci numbers, $|F_n - \varphi F_{n-1}|$ goes to $0$ exponentially. Take $a_n = F_{n+k}$ for sufficiently large $k$. EDIT: For the revised problem, it's still no. Consider the polynomial $P(x) = x^2 - m x - 1$ where $m$ is a positive integer. The roots of $P$ are ...


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If a countable open cover $\{U_j\}$ of a complete metric space is everywhere not locally finite, it can't be point finite. This is because of the Baire Category Theorem: the sets $G_n = \bigcup_{j > n} U_j$ are open and dense, so their intersection must be dense.


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The following is an open cover of the reals: $\mathcal{U} = \{ (\leftarrow, \frac{1}{2}), (\frac{1}{2},\rightarrow)\} \cup \{(0,\frac{1}{n}): n \in \mathbb{N}^+\}$. This is an open cover, that is point-finite but not locally finite at $0$. Your example looks correct to me at first blush, but seems quite complicated for a topology problem. It is nice in ...



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