$\DeclareMathOperator{\add}{add}\DeclareMathOperator{\cof}{cof}\DeclareMathOperator{\cov}{cov}\DeclareMathOperator{\non}{non}\newcommand{\T}{\mathcal{T}}\newcommand{\M}{\mathcal{M}}$
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] 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},
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.},
YEAR = {1993},
MRCLASS = {03E15 (18B99 28A99 54A25)},
MRNUMBER = {1234291 (95e:03139)},
}

I just read an interesting paper by Thomas Forster: The Paris–Harrington Theorem in an NF context [1]. This paper is a progress report on the status of the Paris–Harrington Theorem in Quine’s New Foundations (NF). Forster shows that the Paris–Harrington Theorem is at least consistent with NF, but leaves open the question whether it is provable in NF. Recall that the Paris–Harrington Theorem is a variation on Ramsey’s Theorem, where the homogeneous set $x$ is required to be relatively large: $\min(x) < |x|$. This result is not provable in Peano Arithmetic (PA). In fact, it is logically equivalent to $\Sigma_1$-reflection for PA.

The main issue with the Paris–Harrington Theorem in NF is that the notion of relative largeness is ill-typed. In addition to Forster, Harvey Friedman had also observed this [2]:

In ‘relatively large’, an integer is used both as an element of a finite set and as a cardinality (of that same set).

This is a serious problem in NF where cardinal numbers are defined as equivalence classes of equinumerous sets. Thus, the relative largeness condition $\min(x) < |x|$ entails a direct comparison between an element of the finite set $x$ and the cardinal number $|x|$ to which $x$ belongs. Needless to say that this is difficult to formulate in a stratified manner, as required for set comprehension in NF. Forster manages to work around this, but the winding road has so many twists that he only manages to get a consistency result rather than an actual proof.

While Forster does not directly discuss the issue of the strength of the Paris–Harrington Theorem in the paper, he does raise an interesting question whether the added strength of the theorem has any link to this failure of typing. I used to believe that while the naturalness of the Paris–Harrington Theorem could honestly be questioned, there was one real advantage of this example over the non-provable statements of Gödel and Rosser in that the Paris–Harrington Theorem did not appear to directly depend on self-reference. This observation by Forster and Friedman casts some serious doubts on this…

• Is the notion of relative largeness really self-referential?
• Is the fact that relative largeness is ill-typed the key to the strength of the Paris–Harrington Theorem?

#### References

[1] T. E. Forster, “The Paris-Harrington theorem in an NF context,” in One hundred years of axiomatic set theory, Acad.-Bruylant, Louvain-la-Neuve, 2010, vol. 17, pp. 97-109.
[Bibtex]
@incollection{Forster10,
AUTHOR = {Forster, Thomas Edward},
TITLE = {The {P}aris-{H}arrington theorem in an {NF} context},
BOOKTITLE = {One hundred years of axiomatic set theory},
SERIES = {Cahiers Centre Logique},
VOLUME = {17},
PAGES = {97--109},
YEAR = {2010},
MRCLASS = {03E70 (03E35)},
MRNUMBER = {2796910},
}
[2] H. Friedman, [FOM] PA incompleteness, 2007.
[Bibtex]
@misc{Friedman07,
AUTHOR = {Friedman, Harvey},
TITLE = {{[FOM]} {PA} incompleteness},
HOWPUBLISHED = {FOM},
NOTE = {\url{http://www.cs.nyu.edu/pipermail/fom/2007-October/012046.html}},
YEAR = {2007},
URL = {http://www.cs.nyu.edu/pipermail/fom/2007-October/012046.html},
}