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My idea: I got the inspiration from Diophantine approximation. According to Thue–Siegel–Roth theorem, we have $$\left|\alpha-\dfrac{p}{q}\right|>\dfrac{1}{q^{2+\epsilon}}$$ for irrational $\alpha$ for all except finitely many $\dfrac{p}{q}$s; And for rational $\dfrac{a}{b}$, $$\left|\dfrac{a}{b}-\dfrac{p}{q}\right|\ge\dfrac{1}{bq}$$ . So you can assign a $B_{r(q)}(\dfrac{p}{q})$ to a $\dfrac{p}{q}$, with $r(q)$ sufficiently fast approaching $0$(e.g. $\tfrac{1}{500q!}$), so that if $\alpha$ is irrational, it is contained in only finitely many such balls(intervals), and if $\dfrac{a}{b}$ is rational, $q$ satisfying $r(q)>\dfrac{1}{bq}$ is finite, so $\dfrac{a}{b}$ is also contained in only a finite number of such balls.

These balls, combined with $\mathbb{R}$, is an open cover. As demonstrated above, it is point finite. But as every rational number, except $0$, provides a unique ball, and any interval contains infinitely many rationals, it is not locally finite.

Is my argument correct? Are there simple examples?(It is just an exercise of general topology, and I don't think it need to be so involving.)

UPDATE: I misinterpreted the simple exercise, which asked for a non-locally finite cover. But the problem remains, so I have edited the title to the one I actually have in mind.

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    $\begingroup$ You're misinterpreting the Thue-Siegel-Roth theorem. That applies only to irrational algebraic $\alpha$. Transcendental numbers can have much better Diophantine approximations. $\endgroup$ – Robert Israel Jul 8 '15 at 15:40
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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 the sense that it is "everywhere non locally finite" in a way, not just at one point, which is easy enough, as shown by the first example.

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  • $\begingroup$ Right. I am realizing that I might be misinterpreting the problem. It asked for a cover that is not locally finite, not everywhere non-locally finite. $\endgroup$ – Colliot Jul 8 '15 at 15:19
  • $\begingroup$ It is a good example, but somewhat overkill. To get a cover that is not locally finite, to fail it at one point is enough. $\endgroup$ – Henno Brandsma Jul 8 '15 at 15:21
  • $\begingroup$ yes :-). Now I have solved the mystery of the simplicity of an exercise. But the problem I have in mind is actually the complex one, so I will edit the post. $\endgroup$ – Colliot Jul 8 '15 at 15:23

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