fa.functional analysis - Baire category theorem

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 $xin X$ 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 $xin X$, 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 $d(x,G)=0$.

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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 $xin X$ 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.

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