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To say that $p$ is a condensation point of a subset $S$ of a topological space $X$ is to say that any open neighbourhood of $p$ contains uncountably many points of $S$.

Let us suppose we have written $\mathbb{R} = \bigcup_{i=1}^\infty A_i$. I would like to know whether the condensation points of some $A_i$ must form a set with nonempty interior.

My thoughts: First of all, some of the $A_i$ may be countable. But, the union of all such $A_i$ is 1st category in the Baire space $\mathbb{R}$, so the union of all the uncountable $A_i$ is 2nd category in $\mathbb{R}$. Hence there is some uncountable $A_n$ which is not nowhere dense. The difficulty is that we could have $A_n$ equal to, say, the union of the rationals with a Cantor set, in which case the condensation points of $A_n$ are exactly the Cantor set (which has empty interior).

It is my feeling, although I can't seem to prove it, is that there must be at least one $A_i$ out there which makes use of its uncountability to fail to be nowhere dense.


I think the answer to this question is "yes". The key thing I hadn't noticed which makes this easy, pointed by Bryan Scott below, is that if $A \subset \mathbb{R}$ and $C$ is the set of condensation points of $A$, then $A - C$ is countable. As I see it, this is because of the following

Fact: If $S$ is an uncountable subset of a second-countable topological space $X$, then $S$ has a condensation point.

Proof. Otherwise, there is a cover $(U_s)_{s \in S}$ of $S$ such that each $U_s$ contains only countably many points of $S$. Since $X$ is second-countable, $S$ is second-countable, hence Lindelof in the subspace topology. Thus we can extract a countable subcover from $(U_s)$ and this shows, contrary to assumption, that $S$ is countable. QED.

Continuing, if $A-C$ were uncountable, it would have a condensation point which would also be a condensation point of $A$ so $A-C$ must be countable. Since countable implies 1st category, it follows from $\mathbb{R} = \bigcup_{i=1}^n A_i = \bigcup_{i=1}^n (A_i-C_i) \cup C_i$ that for some $i$ the condensation points $C_i$ of $A_i$ do not form a nowhere dense set. Since the condensation points of a subset of a topological space always form a closed set (clearly their complement is open), this $C_i$ is closed and not nowhere dense, or equivalently, is closed with nonempty interior.

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Are you assuming that the $A_i$ are nested? –  JSchlather Aug 4 '11 at 22:43
    
@Jacob: Despite the title, I wasn't assuming anything in particular about the $A_i$, save that their union is all of $\mathbb{R}$. –  Mike Aug 4 '11 at 22:45
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For each $i \in \mathbb{Z}^+$ let $C_i$ be the set of condensation points of $A_i$; then $A_i \setminus C_i$ is countable. Let $C_0 = \bigcup\limits_{i \in \mathbb{Z}^+} (A_i \setminus C_i)$; $C_0$ is countable, and $\bigcup\limits_{i\in\omega}C_i = \mathbb{R}$. Clearly some $C_i$ with $i>0$ must be somewhere dense in $\mathbb{R}$. However, it’s entirely possible for all of the $A_i$ themselves to have empty interior, even if each consists entirely of condensation points.

Enumerate $2^\omega \times \omega \times \omega$ as $\{\langle \alpha_\xi,m_\xi,n_\xi \rangle:\xi \in 2^\omega\}$, let $\mathbb{R} = \{x_\xi:\xi < 2^\omega\}$, and let $\{I_n:n \in \omega\}$ be an enumeration of the open intervals with rational endpoints. Recursively construct sets $A_n$ for $n\in\omega$ as follows. Suppose that at stage $\xi < 2^\omega$ the approximations $A_n(\eta)$, each of cardinality $|\eta|$, have been constructed for each $n\in\omega$ and $\eta<\xi$. For $k\in\omega\setminus\{m_\xi\}$ let $A_k(\xi)=\bigcup\limits_{\eta<\xi}A_k(\eta)$. Let $\zeta$ be minimal such that $x_\zeta \in I_{n_\xi}\setminus \left(\bigcup\limits_{k\in\omega\setminus\{m_\xi\}}A_k(\xi)\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)\right)$, and let $A_{m_\xi} = \{x_\zeta\}\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)$. This is always possible, since $|I_{n_\xi}|=2^\omega>\left|\bigcup\limits_{k\in\omega\setminus\{m_\xi\}}A_k(\xi)\cup\bigcup\limits_{\eta<\xi}A_{m_\xi}(\eta)\right|$. The construction goes through to $2^\omega$, and for each $n\in\omega$ we let $A_n = \bigcup\limits_{\xi<2^\omega}A_n(\xi)$. The sets $A_n$ partition $\mathbb{R}$, and since each of them intersects every non-empty open interval with rational endpoints (in a set of power $2^\omega$, in fact), all of them have empty interior.

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Thanks! I understand why your first paragraph is true. I'll try to work through your counterexample and then accept this answer. –  Mike Aug 4 '11 at 21:32
    
Something which confuses me is this: presumably each of the $C_i$ is supposed to be the set of condensation points of some "$A_i$". Are you claiming that each $C_i$ is supposed to be equal to its own set of condensation points? I guess not since it seems like the condensation points of any $C_i$ are all of $\mathbb{R}$. –  Mike Aug 4 '11 at 21:42
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Your example is interesting, but now that I think about it the answer to this question is "yes". The set of condensation points of a subset of a topological space is closed (since its complement is clearly open). As you say in your first paragraph, the set of condensation points of one of my $A_i$s is nowhere dense. Since closed sets are nowhere dense precisely when they have nonempty interiors, we are done. –  Mike Aug 4 '11 at 22:11
    
@Mike: No, a closed set is nowhere dense if it has empty interior. However, I did slip up, because I was using a different (and non-standard) definition of condensation point of $A_i$, namely, the set of points in $A_i$ whose nbhds all contain uncountably many points of $A_i$. That’s why some of what I wrote was confusing, and why the recursive construction was unnecessary (though it does show that it’s possible for each of the sets to have empty interior and consist entirely of condensation points). I’ll edit the answer to clear up that confusion. –  Brian M. Scott Aug 5 '11 at 1:18
    
@Bryan: Sorry that was a typo in my last comment. I meant to say "closed sets fail to be nowhere dense precisely when they have nonempty interiors". –  Mike Aug 5 '11 at 2:12
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