Question

Let $X$ be a topological space. When I call a set nowhere dense, meagre or similar without qualification, I mean that it has this property as a subset of $X$. Call a subset of $X$ weager (for weakly meagre) if it is the union of a chain (wrt containment) of nowhere dense sets. Using that finite unions of nowhere dense sets are nowhere dense, it is easy to see that meagre implies weager. Call $X$ an Astaire space (for a stronger Baire space) if every weagre subset of $X$ has empty interior. Obviously Astaire implies Baire. Two natural, but rather silly (not just because of the terminology) questions are:

Does weagre imply meagre? If not, does Baire imply Astaire?

Unsurprisingly, the 2nd (and hence also the 1st) question has a negative answer. Let $X$ be uncountable. In fact, for convenience, take $X$ to be the well-ordered set of all countable ordinals. Topologize $X$ by putting open all sets which are either empty or have countable complement. Then $X$ is a Baire space - in fact the notions countable; closed and not $X$; nowhere dense; and meagre all coincide for subsets of $X$. However, $X$ is the union of the chain of all its countable initial segments so $X$ is not an Astaire space.

The above example is somewhat unsatisfactory since the space is far from Hausdorff, but the ease with which it arose made me wonder whether my question had a positive answer even when $X = \mathbb{R}$. Adapting my example, it is at least possible to express an uncountable subset of $\mathbb{R}$ as the union of a chain of countable subsets of $\mathbb{R}$ but this is quite unhelpful because, in this context, there is no guarantee that countable implies nowhere dense, or that uncountable implies nonempty interior (or even nonmeagre for that matter). So that I don't spend too much more time today thinking about things I know nothing about and/or dreaming up silly names for concepts that probably already have much more respectable names - I pose to you the following question:

Is the real line an Astaire space? If not, are there at least weagre subsets of $\mathbb{R}$ which are not meagre?

Or, in plain English for those of you who only skimmed this nonsense:

Does there exist a chain of nowhere dense subsets of $\mathbb{R}$ whose union has nonempty interior? If not, is there such a chain whose union is not meagre?

Thank you, Michael.

Answer 1

Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union contains an interval.

Proof. Suppose there was such a chain $\{\ B_i \mid i\in I\ \}$, where $\langle I,\lt\rangle$ is a linear order and $i\lt j$ implies $B_i\subset B_j$. First, I claim that this chain cannot have countable cofinality, since then we could find a countable cofinal subset of $I$, and the union of the $B_j$ from this cofinal subset would also contain an interval, violating the Baire category theorem. So every countable subset of $I$ is bounded. In this case, consider the set $Q$ of rational numbers $q$ in the interval from the union $\bigcup_i B_i$. Each of them appears in some $B_{i_q}$, and the set of all $i_q$ for $q\in Q$ is a countable set and hence bounded in $I$. Thus, there is some $j\in I$ beyond all $i_q$. So $B_j$ contains all those $q$ and thus is not nowhere dense. QED

Edit. As George pointed out in the comments, essentially the same argument establishes the full property:

Theorem. There is no chain of nowhere dense subsets of $\mathbb{R}$ whose union is non-meager.

Proof. Suppose that $B_i$ for $i\in I$ is a chain of nowhere dense sets whose union $\bigcup_i B_i$ is non-meager. Again, we see that every countable subset of $I$ must be bounded, for otherwise the union would be a countable union of nowhere dense sets and hence meager. Since $\bigcup_i B_i$ is non-meager, it must be dense on an interval, and so it must have a countable subset $Q$ that is also dense on an interval. By the argument above, since $I$ has uncountable cofinality, this set $Q$ must be in some $B_j$ for large enough $j\in I$, contradicting that $B_j$ is nowhere dense. QED

Answer 2

(Joel's answer appeared as I was typing this.)

I think the answer is no.

Suppose to the contrary there exists a nonmeager set $A \subset \mathbb{R}$ which is the union of some chain $\{K_i\}_{i \in I}$ of nowhere dense sets. $A$ is separable, so we may enumerate a countable dense set $\{x_n\} \subset A$. Then we can find an increasing sequence $\{K_{i_n}\}$ with $x_n \in K_{i_n}$. Set $K = \bigcup_n K_{i_n}$. Since $K$ is meager $K \ne A$, so there exists $x \in A \backslash K$. Now there must be some $K_j$ with $x \in K_j$. Now for each $n$ we certainly don't have $K_j \subset K_{i_n}$, so we must have $K_{i_n} \subset K_j$ since the $K_i$ are a chain. Thus $K \subset K_j$, but then $K_j$ contains all the $x_n$ and so is not nowhere dense.

Added: This indeed shows that a second countable Baire space cannot be the union of a chain of nowhere dense subsets. Here is a stab at a counterexample in the non-second countable case.

Consider the non-separable complete metric space $\ell^\infty = \ell^\infty(\mathbb{N})$. I claim its Hamel dimension $\dim \ell^\infty$ is $\mathfrak{c}$. First, we have the natural inclusion $\ell^1 \subset \ell^\infty$; $\ell^1$ is a separable Banach space, so it is known that $\ell^1$ has Hamel dimension $\mathfrak{c}$, and thus $\dim \ell^\infty \ge \mathfrak{c}$. On the other hand, $\ell^\infty$ is the continuous dual of $\ell^1$, and thus is naturally included into the algebraic dual of $\ell^1$, which must also have Hamel dimension $\mathfrak{c}$; thus $\dim \ell^\infty \le \mathfrak{c}$. By Schroeder-Bernstein, $\dim \ell^\infty = \mathfrak{c}$.

Now suppose we believe the continuum hypothesis $\mathfrak{c} = \aleph_1$. Pick a Hamel basis $B$ for $\ell^\infty$; since it is in bijection with the least uncountable ordinal, we can well-order it in such a way that for any $x \in B$, $B_x = \{y \in B : y < x\}$ is countable. Note $B$ has no greatest element, so $\bigcup_{x \in B} B_x = B$. Let $E_x = \mathrm{span } B_x$; clearly ${E_x}$ is a chain, and $\bigcup_{x \in B} E_x = \ell^\infty$. But each $E_x$ has countable Hamel dimension and therefore is separable, so it must be nowhere dense in $\ell^\infty$.