Baire category theorem

Question

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 $d(x,G)=0$.

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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 $x\in X$ 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.

Answer 1

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.)

Answer 2

The following example should place limits on any strengthening of that kind. Define a sequence of sets by taking the first one to be the Cantor set, the second to be the Cantor set with the middle third filled in, the third to be the Cantor set with the middle third and two remaining middle ninths filled in, and so on. The union of all these (closed) sets is [0,1]. Now if I take a number with a ternary representation such as 0.202020202... then it will not belong even to the boundary of the middle third, or the middle ninth on the right, etc. etc. In the light of this, and many similar examples, I'm not sure what more one can ask for than what the Baire category theorem gives.

Answer 3

You need to be more careful. "A ball" in (2') means a ball of the neighbourhood, not a ball of the original space. The first answer above demonstrates this difference.