Timeline for Wanted: chain of nowhere dense subsets of the real line whose union is nonmeagre, or even contains intervals
Current License: CC BY-SA 2.5
14 events
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| Jan 18, 2011 at 22:03 | comment | added | George Lowther | @Nate: For $X=\ell^\infty(\mathbb{N})$, consider a dense subset $A$ of the unit ball. For each $S\subseteq\mathbb{N}$, let $x_S\in X$ be defined by $x_S(i)=1_{\{i\in S\}}$. Then, there exists $y_S\in A$ with $\Vert x_S-y_S\Vert < 1/2$. Then $S\mapsto y_S$ gives an injection from $2^{\mathbb{N}}$ to $A$, so $A$ has cardinality at least (in fact, equal to) $\mathfrak{c}$. A subset $A$ of infinite cardinality $\kappa <\mathfrak{c}$ will be nowhere dense. The rational linear combinations of $A$ also has cardinality $\kappa$, so is nowhere dense, as is its closure. | |
| Jan 18, 2011 at 21:54 | comment | added | George Lowther | @Nate: For $\ell^2(\omega_1)$, let $\{e_x\}$ be the obvious orthonormal basis, and $E_x$ be the closed subspace generated by $\{e_y\}\_{y\le x}$ which, being the closure of a countable set, is nowhere dense. Then, $E\equiv\bigcup E_x$ is dense. It is also closed, as any convergent sequence in $E$ lies in a countable collection of the $E_x$ and, as $\omega_1$ does not have countable cofinality, lies in $E_y$ for some $y$. So, its limit is in $E_y\subseteq E$. So $\bigcup E_x=\ell^2(\omega_1)$. | |
| Jan 18, 2011 at 20:53 | comment | added | Nate Eldredge | @George Lowther: I thought about $\ell^2(\omega_1)$, but I didn't see how to make it work. If I replace $B$ by the obvious orthonormal basis ordered in the obvious way, then $\bigcup E_x$ is not all of $\ell^2(\omega_1)$. Regarding your second comment, how can one see that a subspace of Hamel dimension smaller than $\mathfrak{c}$ must be nowhere dense? | |
| Jan 18, 2011 at 17:56 | comment | added | George Lowther | In fact, I don't think you need CH for $\ell^\infty(\mathbb{N})$ either. Every subset of size smaller than $\mathfrak{c}$ is nowhere dense. So, use the first ordinal of size $\mathfrak{c}$ instead of the first countable ordinal. Think that works. | |
| Jan 18, 2011 at 17:42 | comment | added | George Lowther | @Nate: Nice counterexample. However, I don't think you even need the continuum hypothesis if you use $\ell^2(\omega_1)$. | |
| Jan 18, 2011 at 16:44 | history | edited | Nate Eldredge | CC BY-SA 2.5 |
Minor correction, A is assumed nonmeager so we don
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| Jan 18, 2011 at 15:31 | history | edited | Nate Eldredge | CC BY-SA 2.5 |
Add counterexample assuming CH
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| Jan 18, 2011 at 1:12 | comment | added | George Lowther | @Michael: I agree with that. In fact, a chain of nowhere dense sets either has countable cofinality or has a nowhere dense union. | |
| Jan 18, 2011 at 1:06 | comment | added | Michael | So, I suppose one concludes that, in a 2nd countable space, a set is meagre if and only if it is the union of some (possibly uncountable) chain of nowhere dense sets? | |
| Jan 18, 2011 at 1:00 | comment | added | Michael | I'll reiterate while convincing myself. If $A$ is nonmeagre then, by definition of meagre rather than the Baire Category theorem, any countable subchain is bounded. Certainly $A$ is not nowhere dense so its closure contains an interval. $\mathbb{R}$ is 2nd countable, so $A$ is 2nd countable and hence separable (I'm not sure that separability always passes to subspaces directly). Let $C \subset A$ be countable and dense in $A$. $C$ is in the union of a countable subchain, so one of our supposedly nowhere dense $K_i$ contains $C$ - contradiction since $C$ is not nowhere dense... Looks good! | |
| Jan 18, 2011 at 0:59 | comment | added | Nate Eldredge | Yeah, second countable looks like a better condition than separable, since a subset of a separable space need not in general be separable. | |
| Jan 18, 2011 at 0:47 | comment | added | George Lowther | ...or should that just be "any second countable space with the Baire property"? | |
| Jan 18, 2011 at 0:44 | comment | added | George Lowther | Nice answer too, along the same lines as Joel's, and works for any separable space with the Baire property. Interesting. | |
| Jan 18, 2011 at 0:20 | history | answered | Nate Eldredge | CC BY-SA 2.5 |