Let $u$ be a free ultrafilter on $\omega$.

I am interested in constructing an ultrafilters on the closed subsets of $\mathbb R$ which contain the collection $\mathcal C$$=${$\bigcup_{n\in A} [n,n+1]:A\in u$}.

If $(x_n)\in \mathbb R ^\omega$ is a sequence of reals with the property $x_n\in (n,n+1)$ for each $n\in\omega$, then I can show the filter generated by $\mathcal C\cup${$x_n:n\in \omega$} is an ultrafilter.

Question: It is not true that every ultrafilter of the type I want can be constructed in this manner, but I am having some difficulty seeing why. Could someone describe an ultrafilter containing $\mathcal C$ which can not be generated from $\mathcal C$ and a sequence (of the type I used)?

EDIT: What about this. For each $A\in u$, let $(s^A_n)_{n\in A}$ be a real-valued sequence on $A$. Then consider a collection $\mathcal S$$=${$(s^A_n):A\in u$} that has the finite intersection property. It looks like $\mathcal C\cup \mathcal S$ generates an ultrafilter?


My original answer overlooked, as David pointed out in a comment, that he wants an ultrafilter in the lattice of closed sets, not in the Boolean algebra of all subsets of $\mathbb R$. I'll append a correction below.

First, here's the original, wrong answer:

Let $\mathcal B$ be the collection of those subsets of $\mathbb R$ whose intersections with the intervals $[n,n+1]$ have at most $1$ element (for each $n$). Then let $$ \mathcal D=\mathcal C\cup\{\mathbb R-X:X\in\mathcal B\}. $$ This family $\mathcal D$ has the finite intersection property, so it can be extended to an ultrafilter on $\mathbb R$. And this ultrafilter cannot contain any set of the form $\{x_n:n\in\omega\}$ as in your question, because sets of this form are in $\mathcal B$, so their complements are in $\mathcal D$.

Now here's the correction:

As David said, $\mathcal D$ should consist of closed sets, so $\mathcal B$ should consist of open sets, so I'll define a corrected version of $\mathcal B$. Put an open subset $X$ of $\mathcal R$ into $\mathcal B$ if, for each $n$, the Lebesgue measure of $X\cap[n,n+1]$ is at most $2^{-n}$.

Thie resulting corrected $\mathcal D$ still has the finite intersection property, because any set in $\mathcal C$ contains intervals $[n,n+1]$ for arbitrarily large $n$, and finitely many sets from $\mathcal B$ ccan't cover all such intervals because their total measure will be too small when $n$ is large. So there is an ultrafilter of closed sets extending $\mathcal D$.

This ultrafilter cannot contain any set $\{x_n:n\in\omega\}$ as in the question because any such set is covered by a set in $\mathcal B$.

  • $\begingroup$ But $\mathcal B$ would need to be a collection of open sets, so that $\mathcal D$ is a collection of closed sets. But no open sets can intersect an interval at only one point... $\endgroup$ Jul 31 '13 at 21:26
  • $\begingroup$ My point is that you are finding ultrafilter on $\mathcal P (\mathbb R)$ instead of on the closed subsets of $\mathbb R$. $\endgroup$ Jul 31 '13 at 21:36
  • $\begingroup$ Oops, I misread the question and overlooked "closed. I've edited the answer to correct this error. $\endgroup$ Jul 31 '13 at 21:40
  • $\begingroup$ 1) Is the filter generated by $\mathcal D$ automatically an ultrafilter? 2)I wonder if one could characterize all ultrafilters containing $\mathcal C$ by (i) those of the type I gave, and (ii) those collections whose complements approach sets of measure 0. $\endgroup$ Jul 31 '13 at 22:25
  • $\begingroup$ @David The answer to (1) is no. For example, each finite intersection of sets from $\mathcal D$ meets both $\bigcup_n[n,n+\frac12]$ and $\bigcup_n[n+\frac12,n+1]$. As for (2), I would be very surprised if there were a characterization along these lines. I would guess that there are lots more sources of complexity of ultrafilters of closed sets. $\endgroup$ Jul 31 '13 at 23:19

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