
Topics
 Topology
 Set Theory
Complete subspaces of a metric space
Let be an arbitrary metric space. What are the possible structural properties for the ideal generated by the complete subspaces of ? How does the topology of affect ? ? What about the ideal generated by the complete subspaces of ?
These questions are superficially very similar to questions asked about ideals and ideals on Polish spaces. However, there is one crucial difference that the most interesting cases are when 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 of could have. Blass figured out that the neighborhood filter of has coinitiality and that the coinitialility of the neighborhood filter of a Bernstein set is . 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 , where is a metric subspace of . Indeed, the complements of the neighborhoods of a subset of are precisely the complete subspaces of the complement with respect to the metric inherited from . So the coinitiality of the neighborhood filter of is the cofinality of the ideal .
From this point of view, the fact that the coinitiality of the neighborhood filter of Bernstein set is 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 is is explained by the fact that complete subspaces of 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 .
To partly answer Tall’s question, I proved that cannot be strictly between and . Furthermore, there is a simple topological criterion to determine when . Beyond that, I haven’t had a chance to investigate much about these ideals . So there could be plenty of lowhanging fruit to pick here…
A dichotomy for
To state my answer to Tall’s question, I need to introduce a new term: the metric remainder of a metric space is the complement of in its completion . Rather than simple cardinal inequalities, I will state the result in terms of Galois–Tukey connections comparing the poset with the posets and . Galois–Tukey connections originate from Vojáš (1993), but I will follow the terminology introduced by Blass (2010).
Theorem. Let be an arbitrary metric space and let be its metric remainder.
 If is compact then there is a morphism . Consequently,
 If is not compact then there is a morphism . Consequently,
In case 1, the dual inequality is , which is not so interesting. In case 2, the dual inequality is . However, apart from the case where is complete, we must have . So neither of the dual results is of much interest.
Proof of 1. Assume further that , i.e., that is not complete in which case the result can be strengthened to the existence of a morphism from .
The forward part of the morphism is defined by
Since is closed in and disjoint from , is always an element of .
The backward part of the morphism is defined by
Since the map is continuous, it achieves a minimum value on the compact set . This minimum value must be positive since . Thus is a welldefined element of .
The fact that is immediate from the definitions. QED
Proof of 2. Since is not compact, there is a sequence in which has no accumulation point in . By passing to a subsequence if necessary, we may assume that there are such that and when . Since is dense in , for each we can find a sequence in such that .
The forward part of the morphism is defined as follows. Given , the function is defined by This is always welldefined since lies outside the closure .
The backward part of the morphism is defined by
To see that this is an element of , consider the closure of in . If then we can find such that and . We may assume that is either constant or strictly increasing. If is constant with value then the limit must exist and hence . If is strictly increasing, then
which means that is an accumulation point of the sequence . Since the accumulation points of are all in , it follows again that . It follows that and hence that since is complete.
The fact that follows immediately from the definitions. QED
References

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

mr: 2768685

zbl: 1198.03058


P. Vojáš, 1993: Generalized GaloisTukeyconnections 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.

mr: 1234291

zbl: 0829.03027

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.
Comments

François G. Dorais wrote
Here are some additional observations on the other cardinal characteristics of .
As observed in the post, if is not complete then . For essentially the same reason, . Let be a Cauchy sequence in that does not converge. Then , which shows that .
The characteristic is more mysterious. If is open in , then . Indeed, the metric remainder is then closed, which means that . Beyond that, I only know the trivial upper bound .