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I was thinking a bit more about the setting of my recent question about unions of chains of nowhere dense subsets of the reals and got stuck almost immediately on a follow-up question. I suspect that my difficulty is rooted in my profound ignorance of set theory, so I have abstracted the problem away from its original setting.

Suppose that $B$ is a family (set) of sets which is closed downward ($x \in B, y \subset x \Rightarrow y \in B$ - I don't expect this property to be essential) and closed under finite unions ($x,y \in B \Rightarrow x \cup y \in B$). Let $W$ be the family of all sets which are the union of some chain in $B$. I have two questions.

  1. If $x,y \in W$, does it follow that $x \cup y \in W$?
  2. If $C \subset W$ is a chain of sets, does it follow that $\bigcup C \in W$?

Some optional motivation: $B$ is intended to model the family of nowhere dense subsets of some topological space. If $M$ is the set of countable unions of sets from $B$ (ie. models the meagre subsets) then it is easy to see that $M$ inherits both of the abstracted closure properties of $B$ and is furthermore closed under countable unions. If we instead form the (larger) family $W$, I was wondering if something analogous holds. That is, I wondered if $W$ inherits both of the closure properties of $B$ (hence my 1st question - $W$ is obviously closed downward) and if it also picks up a new closure property reminiscent of the way in which we constructed it (hence my 2nd question).

I do not embarrass easily and am fully prepared to embrace the possibility that one or both of my queries has a laughably obvious answer! Thanks.

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@Andre: Thanks, I can't say I surprised to see a negative answer for 2. This example seems optimally nice. –  Michael Jan 18 '11 at 6:16
I think the first question also has a negative answer. Let $X$ be a copy of omega, and $Y$ a disjoint copy of omega_1, and let $B$ consist of all sets which intersect $X$ in a finite set and intersect $Y$ in a countable set. Both $X$ and $Y$ are in $W$, but $X\cup Y$ is not. Any chain whose union was $X\cup Y$ would have to be of uncountable cofinality; but then all of $X$ is included in some initial segment, impossible. –  Justin Palumbo Jan 18 '11 at 6:42
@Justin: Hehe, I just posted the same construction. –  Andres Caicedo Jan 18 '11 at 6:49
@Justin, thanks you nailed it. –  Michael Jan 18 '11 at 7:30
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1 Answer

up vote 5 down vote accepted

Let $X$ be the disjoint union of a countable set $A$ and $\omega_1$, the first uncountable ordinal. Let $B$ be the family of sets of the form $a\cup c$ where $a$ is a finite subset of $A$ and $c$ is a countable subset of $\omega_1$. $B$ is closed under subsets and finite unions. The union of a chain in $B$ has the form $d\cup e$ where $d$ is a subset of $A$, $e$ is a subset of $\omega_1$, and (if $d$ is infinite then $e$ is countable).

Note that $A$ and $\omega_1$ belong to $W$, but their union $X$ does not. This proves that the answer to the first question is no.

(In general, we can get a counterexample by looking at the disjoint union of sets whose cardinalities have different cofinalities.)

Now, let $B$ be the family of finite subsets of $X$, where $X$ is uncountable. Then $B$ is closed downwards, and under finite unions. If $A$ is a chain in $B$, then the union of $A$ is countable. Conversely, any countable subset of $X$ is in $W$. Now, there is a chain of countable subsets of $X$ with uncountable union. This shows that the second question also has negative answer.

A natural question that remains is to see what happens if we assume now that the axiom of choice fails and all well-ordered cardinals have cofinality $\omega$, so the counterexample above to the first question does not work.

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Thanks. In retrospect, when my original naive attempt at a proof of 2 hit a stumbling point, it was because I had tripped over precisely this flavour of counterexample. Unfortunately I wasn't savvy enough to connect the dots. –  Michael Jan 18 '11 at 7:42
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