Give an example of a cover for $\Bbb R$ that isn't locally finite, but has the property that every $x \in \Bbb R$ belongs to only finitely many covers 
Give an example of a cover for $\Bbb R$ that isn't locally finite, but has the property that every $x \in \Bbb R$ belongs to only finitely many covers.

We may assume that $\Bbb R$ has the standard topology. A class $\{U_i\}_{i \in I}$ of subsets of a topological space $X$ is locally finite if for every $x \in X$, there exists $O_x$ such that $O_x \cap U_i \ne \emptyset$ for finitely many $i$.
Is the question poorly stated or am I not understanding it properly? What do they mean by "finitely many covers"? As stated it makes very little sense to me. By covers do they mean the elements of the collection that covers $\Bbb R$ or what?
 A: The question is worded a little strangely. I think that by "belongs to only finitely many covers" they mean that for the open cover $\{U_i\}_{i\in I}$, for each $x$ there are only finitely many $i$ such that $x \in U_i$. By "covers", I think they mean elements of the set $\{U_i\}_{i\in I}$.
My idea for a cover that satisfies the given constraints is as follows. For $i\in \mathbb{Z}_{\geq 0}$, let $$U_i = (\sum_{j=1}^i2^{-j}-\delta_i, \sum_{j=1}^{i+1}2^{-j}-\delta_i)$$
Where $\delta_i$ is a very small number such that each interval overlaps the intervals directly adjacent to it (something like $\delta_i=100^{-i}$).
Then, let $U_{-1}=(1,\infty)$ and $U_{-2}=(-\infty,0)$. The collection $\{U_i\}_{i\geq-2}$ forms an open cover of $\mathbb{R}$. Furthermore, each $x$ lies in at most 2 of the intervals. However, any neighborhood of $x=1$ intersects with infinitely many intervals in the open cover. So, $\{U_i\}_{i\geq -2}$ is not locally finite.
Let me know your thoughts or if anyone can think of a simpler cover that does the trick!
