# François G. Dorais

## Complete subspaces of a metric space

Let $X$ be an arbitrary metric space. What are the possible structural properties for the ideal $\newcommand{\T}{\mathcal{T}}\T(X)$ generated by the complete subspaces of $X$? How does the topology of $X$ affect $\DeclareMathOperator{\cof}{cof}\cof(\T(X))$? $\DeclareMathOperator{\cov}{cov}\cov(\T(X))$? 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 $\widehat{X}$. 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 Vojáš (1993), but I will follow the terminology introduced by Blass (2010).

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 $\aleph_0 \leq \DeclareMathOperator{\add}{add}\add(\T(X))$, 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

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

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

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

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, 2010: Combinatorial cardinal characteristics of the continuum, Handbook of set theory, Vols. 1, 2, 3, Springer, Dordrecht, 395–489.

2. P. Vojáš, 1993: Generalized Galois-Tukey-connections between explicit relations on classical objects of real analysis, Set theory of the reals (Ramat Gan, 1991), Israel Math. Conf. Proc., 6, Bar–Ilan Univ., Ramat Gan, 619–643.

Originally posted on by François G. Dorais. To the extent possible under law, François G. Dorais has waived all copyright and neigboring rights to this work.

Here are some additional observations on the other cardinal characteristics of $\T(X)$.
As observed in the post, if $X$ is not complete then $\operatorname{add}(\T(X)) = \aleph_0$. For essentially the same reason, $\operatorname{non}(\T(X)) = \aleph_0$. Let $% $ be a Cauchy sequence in $X$ that does not converge. Then $\set{x_0,x_1,\dots} \notin \T(X)$, which shows that $\operatorname{add}(\T(X)) = \non(\T(X)) = \aleph_0$.
The characteristic $\operatorname{cov}(\T(X))$ is more mysterious. If $X$ is open in $\widehat{X}$, then $\operatorname{cov}(\T(X)) \leq \aleph_0$. Indeed, the metric remainder $X^*$ is then closed, which means that $% $. Beyond that, I only know the trivial upper bound $\operatorname{cov}(\T(X)) \leq \card{X}$.