Complete subspaces of a metric space

$\DeclareMathOperator{\add}{add}\DeclareMathOperator{\cof}{cof}\DeclareMathOperator{\cov}{cov}\DeclareMathOperator{\non}{non}\newcommand{\T}{\mathcal{T}}\newcommand{\M}{\mathcal{M}}$
Let $X$ be an arbitrary metric space. What are the possible structural properties for the ideal $\T(X)$ generated by the complete subspaces of $X$? How does the topology of $X$ affect $\cof(\T(X))$? $\cov(\T(X))$? What about the $\sigma$-ideal $\T_\sigma(X)$ generated by the complete subspaces of $X$?

These questions are superficially very similar to questions asked about ideals and $\sigma$-ideals on Polish spaces. However, there is one crucial difference that the most interesting cases are when $X$ is not a Polish space. For that reason, it looks like answers to the above questions may need different tools than the ones traditionally used in the study of ideals on Polish spaces.

These questions originate from BEST 2010, where Frank Tall asked Andreas Blass what possible coinitialities the neighborhood filter of a subset $X$ of $\R$ could have. Blass figured out that the neighborhood filter of $\Z$ has coinitiality $\mathfrak{d}$ and that the coinitialility of the neighborhood filter of a Bernstein set is $\mathfrak{c}$. Upon his return to Ann Arbor, Blass related Tall’s question to me. After some thought, I realized that Tall’s question was really asking about the possible values of $\cof(\T(Y))$, where $Y$ is a metric subspace of $\R$. Indeed, the complements of the neighborhoods of a subset $X$ of $\R$ are precisely the complete subspaces of the complement $Y = \R – X$ with respect to the metric inherited from $\R$. So the coinitiality of the neighborhood filter of $X$ is the cofinality of the ideal $\T(Y)$.

From this point of view, the fact that the coinitiality of the neighborhood filter of Bernstein set is $\mathfrak{c}$ is more simply explained by the fact that the complement of a Bernstein set, which is also a Bernstein set, has no perfect subset. Indeed, since an uncountable separable complete metric space always contains a perfect set, the ideal of complete subspaces of a Bernstein set must consist only of countable subspaces. The fact that the coinitiality of the neighborhood filter of $\Z$ is $\mathfrak{d}$ is explained by the fact that complete subspaces of $\R-\Z$ must stay outside of some open ball around each integer. Considering the inclusion relations between the various ways to choose an open ball around each integer immediately leads to the number $\mathfrak{d}$.

To partly answer Tall’s question, I proved that $\cof(\T(X))$ cannot be strictly between $\aleph_0$ and $\mathfrak{d}$. Furthermore, there is a simple topological criterion to determine when $\cof(\T(X)) \geq \mathfrak{d}$. Beyond that, I haven’t had a chance to investigate much about these ideals $\T(X)$. So there could be plenty of low-hanging fruit to pick here…

A dichotomy for $\cof(\T(X))$

To state my answer to Tall’s question, I need to introduce a new term: the metric remainder $X^*$ of a metric space $X$ is the complement of $X$ in its completion $\widehat{X}$ [2]. Rather than simple cardinal inequalities, I will state the result in terms of Galois–Tukey connections comparing the poset $(\T(X),{\subseteq})$ with the posets $(\omega,{\leq})$ and $(\omega^\omega,{\leq})$. Galois–Tukey connections originate from Peter Vojtáš [3], but I will follow the terminology introduced by Andreas Blass [1].

Theorem. Let $X$ be an arbitrary metric space and let $X^*$ be its metric remainder.

  1. If $X^*$ is compact then there is a morphism $\phi:(\omega,{\leq})\to(\T(X),{\subseteq})$. Consequently, $$\cof(\T(X)) \leq \aleph_0.$$
  2. If $X^*$ is not compact then there is a morphism $\phi:(\T(X),{\subseteq})\to(\omega^\omega,{\leq})$. Consequently, $$\mathfrak{d} \leq \cof(\T(X)).$$

In case 1, the dual inequality is $\aleph_0 \leq \add(\T(X))$, which is not so interesting. In case 2, the dual inequality is $\add(\T(X)) \leq \aleph_0$. However, apart from the case where $X$ is complete, we must have $\add(\T(X)) = \aleph_0$. So neither of the dual results is of much interest.

Proof of 1. Assume further that $X^* \neq \varnothing$, i.e., that $X$ is not complete in which case the result can be strengthened to the existence of a morphism from $\phi:(1,\leq)\to(\T(X),{\subseteq})$.

The forward part $\phi_{+}:\omega\to\T(X)$ of the morphism $\phi$ is defined by $$\phi_{+}(n) = \set{ x \in X : d(X^*,x) \geq 2^{-n}}.$$ Since $\phi_{+}(n)$ is closed in $\widehat{X}$ and disjoint from $X^*$, $\phi_{+}(n)$ is always an element of $\T(X)$.

The backward part $\phi_{-}:\T(X)\to\omega$ of the morphism $\phi$ is defined by $$\phi_{-}(C) = \min\set{n \in \omega: 2^{-n} \leq d(C,X^*)}.$$ Since the map $x \mapsto d(C,x)$ is continuous, it achieves a minimum value on the compact set $X^*$. This minimum value must be positive since $\overline{C} \cap X^* = \emptyset$. Thus $\phi_{-}(C)$ is a well-defined element of $\omega$.

The fact that $\phi_{-}(C) \leq n \THEN C \subseteq \phi_{+}(n)$ is immediate from the definitions. QED

Proof of 2. Since $X^*$ is not compact, there is a sequence $(y_i)_{i\lt\omega}$ in $X^*$ which has no accumulation point in $X^*$. By passing to a subsequence if necessary, we may assume that there are $\delta_i \gt 0$ such that $\lim_{i\to\infty} \delta_i = 0$ and $d(y_i,y_j) \geq \delta_i+\delta_j$ when $i \neq j$. Since $X$ is dense in $\widehat{X}$, for each $n$ we can find a sequence $(x_{i,j})_{j\lt\omega}$ in $X \cap B(y_i,\delta_i)$ such that $y_i = \lim_{j\to\infty} x_{i,j}$.

The forward part $\phi_{+}:\T(X)\to\omega^\omega$ of the morphism $\phi$ is defined as follows. Given $C \in \T(X)$, the function $\phi_{+}(C)$ is defined by $\phi_{+}(C)(i) = \max\set{j : x_{i,j} \in C}.$ This is always well-defined since $\lim_{j\to\infty} x_{i,j} = y_i$ lies outside the closure $\overline{C}$.

The backward part $\phi_{-}:\omega^\omega\to\T(X)$ of the morphism $\phi$ is defined by $$\phi_{-}(f) = \set{x_{i,j} : j \leq f(i)}.$$ To see that this is an element of $\T(X)$, consider the closure $D_f$ of $\phi_{-}(f)$ in $\widehat{X}$. If $z \in D_f$ then we can find $x_{i(n),j(n)}$ such that $j(n) \leq f(i(n))$ and $z = \lim_{n\to\infty} x_{i(n),j(n)}$. We may assume that $i(n)$ is either constant or strictly increasing. If $i(n)$ is constant with value $i$ then the limit $j = \lim_{n\to\infty} j(n) \leq f(i)$ must exist and hence $z = x_{i,j} \in X$. If $i(n)$ is strictly increasing, then $$d(z,y_{i(n)}) \leq d(z,x_{i(n),j(n)}) + \delta_{i(n)} \to 0,$$ which means that $z$ is an accumulation point of the sequence $(y_i)_{i\lt\omega}$. Since the accumulation points of $(y_i)_{i\lt\omega}$ are all in $X$, it follows again that $z \in X$. It follows that $D_f \subseteq X$ and hence that $\phi_{-}(f) \in \T(X)$ since $D_f$ is complete.

The fact that $\phi_{-}(f) \subseteq C \THEN f \leq \phi_{+}(C)$ follows immediately from the definitions. QED

References

[1] [doi] A. Blass, “Combinatorial cardinal characteristics of the continuum,” in Handbook of set theory. Vols. 1, 2, 3, Dordrecht: Springer, 2010, pp. 395-489.
[Bibtex]
@incollection {Blass10,
AUTHOR = {Blass, Andreas},
TITLE = {Combinatorial cardinal characteristics of the continuum},
BOOKTITLE = {Handbook of set theory. {V}ols. 1, 2, 3},
PAGES = {395--489},
PUBLISHER = {Springer},
ADDRESS = {Dordrecht},
YEAR = {2010},
MRCLASS = {03E17},
MRNUMBER = {2768685},
DOI = {10.1007/978-1-4020-5764-9_7},
URL = {http://dx.doi.org/10.1007/978-1-4020-5764-9_7},
}
[2] F. G. Dorais, Is there a common name for the complement of a metric space in its completion?.
[Bibtex]
@misc {MO22050,
TITLE = {Is there a common name for the complement of a metric space in its completion?},
AUTHOR = {Fran{\c{c}}ois G. Dorais},
HOWPUBLISHED = {MathOverflow},
NOTE = {\url{http://mathoverflow.net/questions/22050} (version: 2010-04-21)},
URL = {http://mathoverflow.net/questions/22050},
}
[3] P. Vojtáš, “Generalized Galois-Tukey-connections between explicit relations on classical objects of real analysis,” in Set theory of the reals (Ramat Gan, 1991), Ramat Gan: Bar-Ilan Univ., 1993, vol. 6, pp. 619-643.
[Bibtex]
@incollection {Vojtas93,
AUTHOR = {Vojt{\'a}{\v{s}}, Peter},
TITLE = {Generalized {G}alois-{T}ukey-connections between explicit relations on classical objects of real analysis},
BOOKTITLE = {Set theory of the reals ({R}amat {G}an, 1991)},
SERIES = {Israel Math. Conf. Proc.},
VOLUME = {6},
PAGES = {619--643},
PUBLISHER = {Bar-Ilan Univ.},
ADDRESS = {Ramat Gan},
YEAR = {1993},
MRCLASS = {03E15 (18B99 28A99 54A25)},
MRNUMBER = {1234291 (95e:03139)},
}
 

Stationary strategies in Choquet games

The (strong) Choquet game on a topological space $X$ is played as follows. There are two players, Empty and Nonempty, who alternate turns for infinitely many rounds. On round $i$, Empty moves first, choosing a point $x_i$ and an open neighborhood $U_i$ of $x_i$ and, if $i \geq 1$, such that $U_i \subseteq V_{i-1}$ (the open set that Nonempty played on the previous round). Then, Nonempty responds with an open neighborhood $V_i$ of the same point $x_i$ such that $V_i \subseteq U_i$. After all the rounds have been played, we obtain a descending sequence of open sets $$U_0 \supseteq V_0 \supseteq U_1 \supseteq V_1 \supseteq U_2 \supseteq V_2 \supseteq \cdots$$ together with a sequence of points $x_0,x_1,x_2,\dots$ Empty wins this play if $\bigcap_{i=0}^\infty V_i = \emptyset$; Nonempty wins if $\bigcap_{i=0}^\infty V_i \neq \emptyset$.

A Choquet space is a topological space $X$ such that Nonempty has a winning strategy in the Choquet game played on the topological space $X$. The Choquet game was originally designed by Choquet to give a topological characterization of which metrizable spaces admit a complete metric. However, not all Choquet spaces are metrizable. In general, the Choquet game turns out to be a good measure of completeness for topological spaces.

In the case of complete metric spaces $X$, Nonempty has a relatively simple winning strategy in the Choquet game on $X$. Once Empty has played the point-neighborhood pair $x_i \in U_i$, Nonempty responds by picking an open ball $V_i$ around $x_i$ that fits inside $U_i$ and has radius no larger than $1/2^i$. This forces Empty to play a Cauchy sequence of points $x_0,x_1,x_2,\dots$ whose limit witnesses that $\bigcap_{i=0}^\infty V_i \neq \emptyset$. Note that to carry out this strategy, Nonempty only needs to know the last move played by Empty and to remember which round is currently being played. In fact, with just a small change in strategy, Nonempty doesn’t even need to remember which round is being played: Nonempty simply needs to pick an open ball $U_i$ around $x_i$ whose radius is no larger than a quarter of the diameter of $V_i$ since that ensures that the radius of each open ball played by Nonempty decreases by at least one half at each step.

A strategy for Nonempty that only uses the last move played by Empty to decide what to play next is called a stationary strategy. Thus, we see that for a metrizable space $X$, the following are equivalent:

  1. $X$ admits a complete metric.
  2. Nonempty has a winning strategy in the Choquet game played on $X$.
  3. Nonempty has a stationary winning strategy in the Choquet game played on $X$.

Since the Choquet game makes sense for arbitrary topological spaces, it makes sense to ask whether items 2 and 3 are equivalent in the general case. This is not the case, but it is known that the equivalence holds for classes of spaces much broader than metrizable spaces.

In our paper [1], Carl Mummert and I show that the equivalence between the existence of general and stationary strategies for Nonempty in the Choquet game holds for an interesting class of spaces, which includes all second-countable T1 spaces. To state our main result, I must introduce an unusual property of topological bases. A base $\mathcal{B}$ for a topology is said to be open-finite if every open set has only finitely many supersets in $\mathcal{B}$. While it is unusual for a base to have this property, it turns out that many spaces happen to have such a base. For example, all second-countable T1 spaces have such a base. The main result of our paper is the following.

Theorem (Dorais–Mummert). Let $X$ be a topological space with an open-finite base. If Nonempty has a winning strategy in the Choquet game on $X$, then Nonempty has a stationary winning strategy in the Choquet game on $X$.

The method for proving this is new and interesting, but you will have to read our paper to find out…

The Choquet game appears to be tied with certain types of representability of topological spaces. Representability issues are very important in the context of reverse mathematics since second-order arithmetic offers very limited resources to talk about large multi-layered objects like topological spaces. In [2], Carl Mummert introduced a broad class of topological spaces that can be represented in second-order arithmetic: countably based maximal filter (MF) spaces. The basic datum for these spaces consists of a countable partial order $(\mathcal{P},{\leq})$, the points of the space is the class $MF(\mathcal{P},{\leq})$ of maximal filters on $(\mathcal{P},{\leq})$, and the basic open sets consist of all classes $U_p = \set{F \in MF(\mathcal{P},{\leq}) : p \in F}$. It is not hard to see that these second-countable spaces are all T1 and Choquet.

A topological characterization of countably based MF spaces was obtained by Carl Mummert and Frank Stephan [3], who established that the countably based MF spaces are precisely the second-countable T1 Choquet spaces. The original proof of this result is long and intricate. The existence of stationary winning strategies for Nonempty in such spaces leads to a much easier proof of this representation theorem. This proof was not included in our paper since it was too far from the main topic and not short enough to include in passing. Therefore, I am recording this proof here for prosperity.

Theorem (Mummert–Stephan). Every second-countable T1 Choquet space is homeomorphic to a countably based MF space.

Proof. Suppose that $X$ is a second-countable T1 Choquet space. Let $\mathcal{B}$ be a countable open-finite base for $X$, and let $\mathfrak{S}$ be a stationary strategy for Nonempty in the Choquet game on $X$. We will define a transitive relation ${\prec}$ on $\mathcal{B}$ such that $X$ is homeomorphic to $MF(\mathcal{B},{\preceq})$. (Although the notation suggests otherwise, the relation ${\prec}$ is not necessarily irreflexive.)

A natural choice for ${\prec}$ would be to define $V \prec U$ to hold if and only if $V \subseteq \mathfrak{S}(x,U)$ for some $x \in V$. However, this relation is not necessarily transitive. To remedy this, we define $V \prec U$ to hold if and only if there is a point $x \in V$ such that $V \subseteq \mathfrak{S}(x,W)$ for every $W \in \mathcal{B}$ such that $U \subseteq W$. This relation is clearly transitive. Moreover, since $\mathcal{B}$ is open-finite, there are only finitely many such $W$, so the intersection of all corresponding $\mathfrak{S}(x,W)$ is an open neighborhood of $x$. This guarantees that for every $x \in U$, there is a $V \in \mathcal{B}$ such that $x \in V$ and $V \prec U$.

We begin by recording a lemma that will be used repeatedly in this proof.

Lemma. For every maximal filter $F$ on $(\mathcal{B},{\preceq})$ there is a descending sequence $$\cdots \prec U_2 \prec U_1 \prec U_0$$ such that $F = \set{W \in \mathcal{B} : (\exists i)(U_i \preceq W)}$.

Proof. Since $F$ is countable and downward directed in $(\mathcal{B},{\preceq})$, it is easy to get a sequence $$\cdots \preceq V_2 \preceq V_1 \preceq V_0$$ such that $F = \set{W \in \mathcal{B} : (\exists i)(V_i \preceq W)}$. If this sequence is not eventually constant, we can eliminate repeated elements to obtain the a sequence as required by the lemma. Otherwise, we may assume that $V_0 = V_i$ for every $i$. As observed above, there must be some $U \in \mathcal{B}$ such that $U \prec V$. Since $F$ is a maximal filter in $(\mathcal{B},{\preceq})$, we must have $U = V$, which means that the constant sequence $U_i = U$ is as required by the lemma. QED

The first step of the proof is to define the map $h:MF(\mathcal{B},{\preceq}) \to X$ that will witness that the two spaces are homeomorphic. Fix $F \in MF(\mathcal{B},{\preceq})$, we will show that $\bigcap F$ is always a singleton, so that we may define $h(F)$ to be the unique point of $X$ that belongs to every element of $F$.

First find a descending sequence $$\cdots \prec U_2 \prec U_1 \prec U_0$$ that generates $F$ as in the above lemma. By definition of $\prec$, we can find corresponding points $x_1,x_2,\dots$ such that $x_i \in U_{i+1}$ and $U_{i+1} \subseteq \mathfrak{S}(x_i,U_i)$. This defines a valid sequence of moves for Empty against Nonempty’s stationary strategy $\mathfrak{S}$ in the Choquet game on $X$. Since $\mathfrak{S}$ is a winning strategy for Nonempty, it follows that $\bigcap_{i=0}^\infty U_i = \bigcap F$ is nonempty.

To see that $\bigcap F$ has only one point, suppose for the sake of contradiction that $\bigcap_{i=0}^\infty U_i$ contains two distinct points $x$ and $y$. Because $X$ is T1, we can find a neighborhood $V_0$ of $x$ in $\mathcal{B}$ that does not contain $y$. Define the descending sequence $$\cdots \prec V_2 \prec V_1 \prec V_0$$ so that $x \in V_{i+1}$ and $V_{i+1} \subseteq \mathfrak{S}(x,W)$ for every $W \in \mathcal{B}$ such that $V_i \cap U_i \subseteq W$. The filter $$G = \set{W \in \mathcal{B} : (\exists i)(V_i \preceq W)}$$ extends $F$ since $V_i \preceq U_i$ for each $i$. Since $V_0 \in G$ but $V_0 \notin F$, this contradicts the maximality of $F$.

Now that $h:MF(\mathcal{B},{\preceq}) \to X$ is properly defined, it remains to show that it is a homeomorhism. We first show that $h$ is a bijection, which we break into two facts:

  • $h$ is injective. Suppose that $F_0$ and $F_1$ are maximal filters that map to the same point $x$. By the lemma, we can find two sequences $$\cdots \prec U_2^d \prec U_1^d \prec U_0^d \qquad (d \in \set{0,1})$$ that generate these two filters. Since $x \in U_i^0 \cap U_i^1$ for each $i$, we can find another sequence $$\cdots V_2 \prec V_1 \prec V_0$$ of neighborhoods of $x$ in $\mathcal{B}$ such that $V_{i+1} \subseteq \mathfrak{S}(x,W)$ for every $W \in \mathcal{B}$ such that $U_i^0 \cap U_i^1 \cap V_i \subseteq W$. Then the filter $$G = \set{W \in \mathcal{B} : (\exists i)(V_i \preceq W)}$$ extends both $F_0$ and $F_1$, which means that $F_0 = G = F_1$.
  • $h$ is surjective. Let $U_0,U_1,U_2,\dots$ be an enumeration of $\mathcal{B}$ (possibly with repetitions). Given $x \in X$, define the descending sequence $$\cdots \preceq V_2 \preceq V_1 \preceq V_0$$ of neighborhoods of $x$ in $\mathcal{B}$ as follows. Pick $V_0 \in \mathcal{B}$ so that $x \in V_0$. If $x \in U_i$ then pick $V_{i+1}$ in such a way that $V_{i+1} \subseteq \mathfrak{S}(x,W)$ for every $W \in \mathcal{B}$ such that $V_i \cap U_i \subseteq W$; if $x \notin U_i$ then simply set $V_{i+1} = V_i$. Since $X$ is T1, we immediately see that $\bigcap_{i=0}^\infty V_i = \set{x}$. Therefore, any maximal filter extending $$F = \set{W \in \mathcal{B} : (\exists i)(V_i \preceq W)}$$ will map to $x$. (In fact, $F$ is already maximal.)

Since the choice of $V_0$ was essentially arbitrary in the process we just used to show that $h$ is surjective, for every $V_0 \in \mathcal{B}$ and every $x \in V_0$ we can find some $F \in MF(\mathcal{B},{\preceq})$ such that $V_0 \in F$ and $h(F) = x$. It follows that $$h(F) \in V_0 \quad\IFF\quad V_0 \in F,$$ which shows that $h$ is a homeomorphism. QED

References

[1] [doi] F. G. Dorais and C. Mummert, “Stationary and convergent strategies in Choquet games,” Fund. math., vol. 209, iss. 1, pp. 59-79, 2010.
[Bibtex]
@article {DoraisMummert10,
AUTHOR = {Dorais, Fran{\c{c}}ois G. and Mummert, Carl},
TITLE = {Stationary and convergent strategies in {C}hoquet games},
JOURNAL = {Fund. Math.},
FJOURNAL = {Fundamenta Mathematicae},
VOLUME = {209},
YEAR = {2010},
NUMBER = {1},
PAGES = {59--79},
ISSN = {0016-2736},
MRCLASS = {91A44 (06B35 54D20 54D70 91A24)},
MRNUMBER = {2652592 (2011h:91054)},
MRREVIEWER = {L{\'a}szl{\'o} Zsilinszky},
DOI = {10.4064/fm209-1-5},
URL = {http://dx.doi.org/10.4064/fm209-1-5},
EPRINT = {0907.4126}
}
[2] [doi] C. Mummert, “Reverse mathematics of MF spaces,” J. math. log., vol. 6, iss. 2, pp. 203-232, 2006.
[Bibtex]
@article {Mummert06,
AUTHOR = {Mummert, Carl},
TITLE = {Reverse mathematics of {MF} spaces},
JOURNAL = {J. Math. Log.},
FJOURNAL = {Journal of Mathematical Logic},
VOLUME = {6},
YEAR = {2006},
NUMBER = {2},
PAGES = {203--232},
ISSN = {0219-0613},
MRCLASS = {03B30 (03D45 03F35 06A06 06B35 54E50)},
MRNUMBER = {2317427 (2008d:03011)},
MRREVIEWER = {Christian Bennet},
DOI = {10.1142/S0219061306000578},
URL = {http://dx.doi.org/10.1142/S0219061306000578},
}
[3] [doi] C. Mummert and F. Stephan, “Topological aspects of poset spaces,” Michigan math. j., vol. 59, iss. 1, pp. 3-24, 2010.
[Bibtex]
@article {MummertStephan10,
AUTHOR = {Mummert, Carl and Stephan, Frank},
TITLE = {Topological aspects of poset spaces},
JOURNAL = {Michigan Math. J.},
FJOURNAL = {Michigan Mathematical Journal},
VOLUME = {59},
YEAR = {2010},
NUMBER = {1},
PAGES = {3--24},
ISSN = {0026-2285},
MRCLASS = {06E15 (03B30 03E15 54E52 91A44)},
MRNUMBER = {2654139 (2011k:06026)},
MRREVIEWER = {Jimmie D. Lawson},
DOI = {10.1307/mmj/1272376025},
URL = {http://dx.doi.org/10.1307/mmj/1272376025},
EPRINT = {0912.3191}
}