Find an example of non-locally finite collection I got stuck on this problem and can't find any hint to solve this. Hope some one can help me. I really appreciate.


Give an example of a collection of sets $A$ that is not locally finite, such that the collection $B = \{\bar{X} | X \in A\}$ is locally finite.


Note: Every element in $B$ must be unique, so maybe there exist 2 distinct sets $A_{1}, A_{2} \in A$, but have $\bar{A_{1}} = \bar{A_{2}}$. So that's why this problem would be right even though $A \subset \bar{A}$.
Thanks everybody.
 A: Just take some non-empty open set $U$ such that $\overline{U} \setminus U$ is infinite. Actually $U$ does not have to be open, just non-empty and such that $\overline{U} \setminus U$ is infinite. This includes the correct example of dense subset of $\mathbb{R}$ other people mentioned.
A: Sorry, typo, now corrected. All elements of A are dense in R.
A: Let Q be the rationals in R, and A be the collection of sets of the form {a+√p, a ∈ Q}, p ∈ N.
A: If someone wants to check my easy proof for metric spaces that would be good.
Let the topological space be a metric space. Here I will use X = $(0,1)$ with the basis of (0,1) being the standard basis of R consisting of open balls, but those balls not contained in (0,1) are removed. Use the absolute value function as the metric for simplicity. (|x-y| = d(x,y) for x,y $\in$ R).
Then let $A_{N}$ = $\bigcup_{i=0}^{N-1}(\frac{i}{N}, \frac{i+1}{N})$ and A = $\{A_{N}\}_{N=1}^{N=\infty}$
It should be easy to see that $A_{N} = (0, \frac{1}{N})\bigcup(\frac{1}{N}, \frac{2}{N})\bigcup...\bigcup(\frac{N-1}{N}, 1)$
1.) Choose $x_{0} \in (0,1)$. Then a neighborhood U containing $x_{0}$ intersects (0,1), in which case U will intersect infinitely many $A_{N} \in A$.
This is because an open set U containing $x_{0}$ will be the union of open balls B contained in U. Some ball B in U contains $x_{0}$. This ball has a radius $r > 0$.
The existence of a ball B in U containing $x_{0}$ with positive radius r implies that some $N_{0} \in \mathbb{Z^{+}}$ satisfies $\frac{1}{N_{0}} < r$.
Thus B intersects at least 2 balls of radius $\frac{1}{N}$ in $A_{N}$ for all $N > N_{0}$ no matter where x lies in (0,1). Thus B intersects all $A_{N}$ which have $N > N_{0}$. Since B is contained within U, U must intersect the same two balls. As N > $N_{0}$ goes to infinity, any neighborhood U of $x_{0}$ intersects 2 balls of infinitely many $A_{N}$.
Here we note that A is not locally finite since no open set U of X intersects only finitely many $A_{N}$.
2.) Now take the closure of each $A_{N}$.
3.) Then $\overline{A_{N}}$ = (0,1) for each N. (Note 0 and 1 are not elements of X = (0,1)).
Hence the neighborhood U containing $x_{0}$ intersects the only member of $\overline{A} = (0,1)$

Now, to expand this to any metric space X: first expand this proof to a bounded subset of the metric space X. Then use a limiting process to expand to the entire space. Should work great.
Ill do R^n here for example: Just use a non-empty bounded open ball B of radius R' in R^n, instead of (0,1) in R. Approximate the set from the inside with almost disjoint closed cubes of side length r' = R'/N. (r = R/2, R/3, R/4, etc, etc...) Now take the interiors of these cubes to get a collection of disjoint open cubes. Call the collection $C_{N}$. The interior of $\overline{\bigcup{C_{N}}}$ can now be exhausted with almost disjoint closed cubes $C_{N+1}$ such that the $\overline{\bigcup{C_{N}}} = \overline{\bigcup{C_{N+1}}}$ = V. Then as above, any ball B containing x in U will intersect 2 cubes of each $C_{N}$ as long as N is large enough. Such as when R'/N = r' is smaller than 2 times the radius r of B containing x. Then do some sort of limiting process on R' to exhaust the entire set $R^{n}$; $\lim_{R' \to \infty} B \supset X$.
