Let's call the following conditions (1): $X$ is a complete metric space with metric $d$, $X = \cup\_{n=1}^\infty A\_n$. Let $\bar{A}$ denote the closure of $A$.


Let's call the following statement (2): at least one of the $\bar{A}\_n$ contains a ball.


Baire category theorem gives: 


Fact1: 
(1) $\Rightarrow$ (2)


Now take any $x\in X$ and consider the closed ball $B = B(x,\delta)=\{y: d(x,y)\le \delta\}$. This is a complete metric space itself and it is covered by $ A\_n \cap B$. Thus these sets satisfy (1). Fact1 gives us an $n$ such that the closure of $A\_n \cap B$ contains a ball in $B$. Thus, we have strengthened Fact1 to


Fact1': (1) $\Rightarrow$ (2')


where (2') is: For every $x\in X$, every neighborhood of $x$ contains a ball that is contained in one of the $\bar{A}\_n$.


Question: Can (2') be strengthened further? Here are some example statements, both of which are too strong:


* For every $x$, there is a $\delta$ such that $B(x,\delta)$ is contained in one of the $\bar{A}\_n$
* For every $x$, there is an $A\_n$ such that $\bar{A}\_n$ contains an open set $G$ with $d(x,G)=0$.




---


Many thanks for the responses. The motivation for this question was as follows.


1) What does it mean for a set $A$ to have a closure with empty interior? Take an element
 $a \in \bar{A}$. This means that $a$ is either in $A$ or there is a sequence in $A$ that converges to $a$. This can be rephrased as $a$ is a point that can be approximated with infinite precision by $A$.' If $\bar{A}$ has no interior, perturbing $a$ by a small amount will give an $a'$, such that $a'$ is a finite distance away from $\bar{A}$. Thus, $a'$ will be a point that can only be approximated by $A$ with finite precision. Then, one can think of $A$ as a multi resolution grid with discontinuous approximation capability: you perturb any point that can be approximated infinitely well by it and you get a point that can only be approximated with finite precision.


2) $X$ itself can be thought of as the perfect multi resolution grid for itself: every point $x\in X$ is well approximated by $X$, and perturbing $x$ will not change this. The way I wanted to think of $X= \cup\_{n=1}^\infty A\_n$ was this: for every $x \in X$, one of the $A\_n$ provides a multiresolution grid around $x$ that has continuous approximation capability. I wanted to think, similar to Leonid's response, that the set of $X$ where this is not possible was to be in some sense to be exceptional.



Baire category theorem
Maybe if you allow an exceptional set, as in "For every $x$ outside of some meagre set, there is a $\delta>0$ such that $B(x,\delta)$ is contained in one of the $\overline{A}\_n$" 


(Indeed, the set of all $x$ where the above fails is a closed subset of $X$ with empty interior.)


