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On the Borel complexity and the complete metrizability of spaces of metrics

  • Katsuhisa Koshino EMAIL logo
Published/Copyright: November 11, 2024
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Analysis and Geometry in Metric Spaces
From the journal Analysis and Geometry in Metric Spaces Volume 12 Issue 1

Abstract

Given a metrizable space X , let A M ( X ) be the space of continuous bounded admissible metrics on X , which is endowed with the sup-metric. In this article, we shall investigate the Borel complexity and the complete metrizability of A M ( X ) and show that a separable metrizable space X is σ -compact if and only if A M ( X ) is completely metrizable.

Keywords: admissible metric; sup-metric; absolute Borel class; completely metrizable
MSC 2010: Primary 54C35; Secondary 54E35; 54H05

Recognizing topologies on function spaces is one of the most important issues in infinite-dimensional topology, and finding their Borel complexity plays a significant role in it. In this article, we shall study the Borel complexity and the complete metrizability of spaces of metrics. For a metrizable space X , let C ( X 2 ) be the space of continuous bounded real-valued functions on X 2 , whose topology induced by the sup-metric D : for any f , g ∈ C ( X 2 ) ,

D ( f , g ) = sup { ∣ f ( x , y ) − g ( x , y ) ∣ : ( x , y ) ∈ X 2 } .

Denote the subspace consisting of continuous bounded pseudometrics on X by P M ( X ) and the subspace consisting of bounded admissible metrics on X by A M ( X ) . Recall that the sup-metric D is complete on C ( X 2 ) and P M ( X ) , which is closed in C ( X 2 ) . Let A M ∗ ( X ) be the space of admissible metrics on X with the topology induced by D (namely, the topology of uniform convergence). Ishiki [6] showed that A M ∗ ( X ) is a Baire space, which implies that the open subspace A M ( X ) ⊂ A M ∗ ( X ) is also Baire. Koshino [7] proved that if X is σ -compact, then A M ( X ) is a G δ set in C ( X 2 ) , and hence, it is completely metrizable, see also [5]. It is well-known that completely metrizable spaces are Baire. However, the inverse does not hold. For example, let

Y = R × ( 0 , ∞ ) ∪ Q × { 0 }

and endow it with the usual topology, where R is the real line and Q is the set of rationals. Then, Y is Baire but not completely metrizable. Indeed, since the Baire space R × ( 0 , ∞ ) is an open dense subspace of Y , Y is also Baire. Conversely, the closed subspace Q × { 0 } is not completely metrizable, and hence, Y is also not completely metrizable. It is natural to ask whether A M ( X ) is always completely metrizable or not. We will show that even if X is separable, the space A M ( X ) is not necessarily completely metrizable.

Given a metrizable space Y , let A 0 ( Y ) be the family of open sets in Y , and let M 0 ( Y ) be the one of closed sets in Y . For a natural number n ≥ 1 , define inductively the collections

A n ( Y ) = { ⋃ k ∈ ω A k : A k ∈ ⋃ m < n M m ( Y ) } and M n ( Y ) = { ⋂ k ∈ ω A k : A k ∈ ⋃ m < n A m ( Y ) } .

Denote by A n (respectively, M n ) a class of spaces X satisfying that X is belonging to A n ( Y ) (respectively, M n ( Y ) ) for every metrizable space Y which contains X as a subspace. We call A n and M n to be the absolute Borel classes. Recall that A 0 = { ∅ } and M 0 is the class of compact metrizable spaces. Moreover, the class M 1 is consisting of completely metrizable spaces, and the class A 1 is consisting of σ -locally compact metrizable spaces ( σ -compact metrizable spaces in the separable case), refer to [10]. We can establish the following.

Theorem 1.1

Suppose that X is a metrizable space and n ≥ 1 is a natural number. If A M ( X ) is in A n (respectively, M n ), then X is in M n (respectively, A n ).

As a corollary, we can give a necessary and sufficient condition on the complete metrizability of A M ( X ) as follows.

Corollary 1.2

Let X be a separable metrizable space. Then, X is σ -compact if and only if A M ( X ) is completely metrizable.

Proof

The “only if” part follows from [7, Proposition 3] and the “if” part follows from Theorem 1.1 in the case where n = 1 .□

First, we give the following example, which contains a good technique to prove Theorem 1.1.

Example 1.3

Equipping the real line with the usual metric, let Q = [ 0 , 1 ] ∩ Q and P = [ 0 , 1 ] ∩ P , where P is the set of irrationals. Suppose that X is the topological sum P ⊕ ( 0 , 1 ] . Then, A M ( X ) is not completely metrizable. Since Q is not completely metrizable, it is sufficient to show that A M ( X ) contains a closed topological copy of Q . For each q ∈ Q , define d q ∈ A M ( X ) by

d q ( x , y ) = ∣ x − y ∣ if ( x , y ) ∈ P 2 or ( 0 , 1 ] 2 , ∣ x − q ∣ + y if x ∈ P and y ∈ ( 0 , 1 ] , x + ∣ y − q ∣ if x ∈ ( 0 , 1 ] and y ∈ P .

Put S = { d q ∈ A M ( X ) : p ∈ Q } , so it is a closed subset of A M ( X ) and is isometric to Q .

First, we shall prove that S is closed in A M ( X ) . Take any sequence { d q n } ⊂ S that is converging to some d ∈ A M ( X ) . It remains to show that d ∈ S . When ( x , y ) is in P 2 or ( 0 , 1 ] 2 ,

∣ d ( x , y ) − ∣ x − y ∣ ∣ = ∣ d ( x , y ) − d q n ( x , y ) ∣ ≤ D ( d , d q n ) → 0

as n → ∞ , which means that d ( x , y ) = ∣ x − y ∣ . Replacing { q n } with a subsequence converging in [ 0 , 1 ] , we can find a point z ∈ [ 0 , 1 ] so that q n → z as n tends to ∞ . For every x ∈ P and every y ∈ ( 0 , 1 ] , since

∣ d ( x , y ) − ( ∣ x − q n ∣ + y ) ∣ = ∣ d ( x , y ) − d q n ( x , y ) ∣ ≤ D ( d , d q n ) ,

d ( x , y ) = ∣ x − z ∣ + y . In the case that x ∈ ( 0 , 1 ] and y ∈ P , we have that d ( x , y ) = x + ∣ y − z ∣ similarly. Then, z ∈ Q . Otherwise, z ∈ P and

d z , 1 n = ∣ z − z ∣ + 1 n = 1 n → 0

for the sequence { 1 ⁄ n } ⊂ ( 0 , 1 ] , that is not converging in X . Therefore d ∉ A M ( X ) , which is a contradiction. It follows that z ∈ Q , and hence, d = d z ∈ S .

Next, we will show that S is isometric to Q . To prove it, let i : S → Q be a map defined by i ( d q ) = q and fix any q , p ∈ Q . If ( x , y ) is belonging to P 2 or ( 0 , 1 ] 2 , then

∣ d q ( x , y ) − d p ( x , y ) ∣ = ∣ ∣ x − y ∣ − ∣ x − y ∣ ∣ = 0 .

Moreover, if x ∈ P and y ∈ ( 0 , 1 ] , then

∣ d q ( x , y ) − d p ( x , y ) ∣ = ∣ ( ∣ x − q ∣ + y ) − ( ∣ x − p ∣ + y ) ∣ ≤ ∣ q − p ∣ .

Similarly, for each x ∈ ( 0 , 1 ] and each y ∈ P ,

∣ d q ( x , y ) − d p ( x , y ) ∣ ≤ ∣ q − p ∣ .

It follows that D ( d q , d p ) ≤ ∣ q − p ∣ . We may assume that q ≤ p . In the case where 0 < q , taking any x ∈ [ 0 , q ] ∩ P and any y ∈ ( 0 , 1 ] , we have that

∣ d q ( x , y ) − d p ( x , y ) ∣ = ∣ ( ∣ x − q ∣ + y ) − ( ∣ x − p ∣ + y ) ∣ = ∣ q − p ∣ .

In the case that p < 1 , fix any x ∈ [ p , 1 ] ∩ P and any y ∈ ( 0 , 1 ] , so

∣ d q ( x , y ) − d p ( x , y ) ∣ = ∣ ( ∣ x − q ∣ + y ) − ( ∣ x − p ∣ + y ) ∣ = ∣ q − p ∣ .

When q = 0 and p = 1 , for every number ε ∈ ( 0 , 1 ⁄ 2 ) , letting any x ∈ [ 0 , ε ] ∩ P and any y ∈ ( 0 , 1 ] , observe that

∣ d q ( x , y ) − d p ( x , y ) ∣ = ∣ ( ∣ x − q ∣ + y ) − ( ∣ x − p ∣ + y ) ∣ ≥ ∣ q − p ∣ − 2 ε .

Consequently, D ( d q , d p ) = ∣ q − p ∣ , and hence the map i is isometry. It conclude that A M ( X ) is not completely metrizable.

Due to the efforts of Bessaga [2], Banakh [1], Pikhurko [8], and Zarichnyi [11] (see also [3, Theorem 2]), we can establish the following lemma.

Lemma 1.4

Let Y be a metrizable space and A ⊂ Y be a closed subset. Then, there exists a continuous function e : A M ( A ) → A M ( Y ) such that e ( d ) ∣ A 2 = d for all d ∈ A M ( A ) .

In Example 1.3, the subset P is closed in X and is homeomorphic to P . More generally, by virtue of the similar argument, we have the following.

Proposition 1.5

If a metrizable space X contains a closed subset homeomorphic to P , then A M ( X ) is not completely metrizable.

Proof

By the assumption, there exists a closed embedding h : P → X , and then let A 1 = h ( [ 0 , 1 ⁄ 3 ] ∩ P ) and A 2 = h ( [ 2 ⁄ 3 , 1 ] ∩ P ) . Set Q ′ = [ 2 ⁄ 3 , 1 ] ∩ Q and define a function i : Q ′ → A M ( A 1 ⊕ A 2 ) by for every q ∈ Q ′ ,

i ( q ) ( x , y ) = ∣ h − 1 ( x ) − h − 1 ( y ) ∣ if ( x , y ) ∈ A 1 2 or A 2 2 , h − 1 ( x ) + ∣ h − 1 ( y ) − q ∣ if x ∈ A 1 and y ∈ A 2 , ∣ h − 1 ( x ) − q ∣ + h − 1 ( y ) if x ∈ A 2 and y ∈ A 1 .

It follows from the same argument as Example 1.3 that i is an isometric embedding.

According to Lemma 1, we can obtain a continuous map e : A M ( A 1 ⊕ A 2 ) → A M ( X ) such that e ( d ) ∣ ( A 1 ⊕ A 2 ) 2 = d for every d ∈ A M ( A 1 ⊕ A 2 ) . Then, the composition e ∘ i is a closed embedding. For any q , p ∈ Q ′ with q < p and any x ∈ A 1 , letting y = h ( r ) with r ∈ [ q , ( 2 q + p ) ⁄ 3 ] ∩ P , observe that

∣ e ∘ i ( q ) ( x , y ) − e ∘ i ( p ) ( x , y ) ∣ = ∣ i ( q ) ( x , y ) − i ( p ) ( x , y ) ∣ = ∣ ( h − 1 ( x ) + ∣ h − 1 ( y ) − q ∣ ) − ( h − 1 ( x ) + ∣ h − 1 ( y ) − p ∣ ) ∣ = ∣ ( h − 1 ( x ) + r − q ) − ( h − 1 ( x ) + p − r ) ∣ ≥ p + q − 2 r ≥ ( p − q ) ⁄ 3 > 0 ,

which implies that e ∘ i is injective. It remains to prove that e ∘ i is a closed map. Suppose that { q n } is a sequence in Q ′ such that e ∘ i ( q n ) is converging to some metric d ∈ A M ( X ) . Then, there exists a subsequence of { q n } that converges to some point z ∈ Q ′ . Indeed, replace { q n } with a converging subsequence to some z ∈ [ 2 ⁄ 3 , 1 ] . By the similar way as Example 1.3, for every x ∈ A 1 and every y ∈ A 2 ,

d ( x , y ) = h − 1 ( x ) + ∣ h − 1 ( y ) − z ∣ .

Assume that z ∈ P . Take any sequence { p n } consisting irrational numbers so that 1 ⁄ 3 ≥ p n → 0 as n → ∞ , so

d ( h ( z ) , h ( p n ) ) = ∣ z − z ∣ + p n = p n → 0 .

Conversely, the sequence { h ( p n ) } is not converging in A 1 ⊕ A 2 , which is a contradiction. Hence z ∈ Q ′ and e ∘ i is a closed embedding. Since A M ( X ) contains a closed subset homeomorphic to Q ′ , which is not completely metrizable, A M ( X ) is also not completely metrizable.□

Every space homeomorphic to some member of an absolute Borel class also belongs to it. Moreover, to prove Theorem 1, we will use the following basic properties on absolute Borel classes.

Lemma 1.6

The following are true.

  1. If a space is a union of a locally finite family consisting of subspaces belonging to an absolute Borel class, then it is also in the same class (cf. [4, 4.5.8 (a)]).

  2. For a metrizable space X , X ∈ A n , n ≥ 2 , (respectively, X ∈ M n , n ≥ 1 ), if and only if X ∈ A n ( Y ) (respectively, M n ( Y ) ) for some completely metrizable space Y (cf. [9, Theorem 5.11.2]).

  3. If X ∉ A 1 , then there exists a completely metrizable space Y such that X ∉ A 1 ( Y ) .

  4. All closed subspaces of some space in an absolute Borel class also belong to it.

  5. For all 0 ≤ m < n , A m ∪ M m ⊂ A n ∩ M n .

For spaces A ⊂ Y , denote the interior and closure of A in Y by int Y A and cl Y A , respectively. Now, we will show Theorem 1.1.

Proof of Theorem 1.1

We only prove it in the case that A M ( X ) ∈ M n . Assume that X is not belonging to A n and take a bounded complete metric space Y = ( Y , ρ ) such that X ∉ A n ( Y ) by (ii) and (iii) of Lemma 1.6. Put B = X \ int Y X . Then, B ∉ A n ( Y ) . We also let Z = cl Y B \ X , so Z ∉ M n . Remark that B and Z are non-degenerate, since A 0 ∪ M 0 ⊂ A n ∩ M n due to (v) of Lemma 1.6. Note that the subset cl Y B = B ∪ Z is complete and the diameter of Z is positive. Let O be the family of all open balls in cl Y B with radii one eighth of the diameter of Z and take a locally finite open cover U of cl Y B refining O . According to the local finiteness of U and (i) of Lemma 1.6, there is U ∈ U such that Z ∩ cl Y U ∉ M n . Since U is a refinement of O , there exists U ′ ∈ U such that Z ∩ U ′ ≠ ∅ but cl Y U ∩ cl Y U ′ = ∅ . Fix any point a ∈ Z ∩ U ′ and find a sequence { a n } ⊂ X ∩ U ′ converging to the point a . Let A = { a n } , B ′ = B ∩ cl Y U and Z ′ = Z ∩ cl Y U . Remark that A and B ′ are closed in X . Define a continuous function i : Z ′ → A M ( A ⊕ B ′ ) by

i ( z ) ( x , y ) = ρ ( x , y ) if ( x , y ) ∈ A 2 or B ′ 2 , ρ ( x , a ) + ρ ( y , z ) if x ∈ A and y ∈ B ′ , ρ ( x , z ) + ρ ( y , a ) if x ∈ B ′ and y ∈ A .

Choosing an extending map e : A M ( A ⊕ B ′ ) → A M ( X ) by Lemma 1.4, we can obtain a closed embedding e ∘ i . Indeed, let any distinct points z 1 , z 2 ∈ Z ′ . Since B is dense in cl Y ( X \ int Y X ) , find a point x ∈ B ′ with ρ ( x , z 1 ) ≤ ρ ( z 1 , z 2 ) ⁄ 3 . For every y ∈ A ,

∣ e ∘ i ( z 1 ) ( x , y ) − e ∘ i ( z 2 ) ( x , y ) ∣ = ∣ i ( z 1 ) ( x , y ) − i ( z 2 ) ( x , y ) ∣ = ∣ ( ρ ( x , z 1 ) + ρ ( y , a ) ) − ( ρ ( x , z 2 ) + ρ ( y , a ) ) ∣ = ρ ( x , z 2 ) − ρ ( x , z 1 ) ≥ ρ ( z 1 , z 2 ) − 2 ρ ( x , z 1 ) ≥ ρ ( z 1 , z 2 ) ⁄ 3 > 0 ,

and hence e ∘ i is an injection. To prove that e ∘ i is a closed map, take any sequence { z n } in Z ′ so that e ∘ i ( z n ) is converging to some d ∈ A M ( X ) . As is observed in the above inequality, for all natural numbers n and m ,

ρ ( z n , z m ) ≤ 3 D ( e ∘ i ( z n ) , e ∘ i ( z m ) ) .

Thus, { z n } is a Cauchy sequence in the complete metric space cl Y ( X \ int Y X ) ∩ cl Y U , so it is converging to some z ∈ cl Y ( X \ int Y X ) ∩ cl Y U . Supposing that z ∈ B ′ , we have that

d ( z , a n ) = ρ ( z , z ) + ρ ( a n , a ) → 0 ,

but { a n } is not converging in A ⊕ B ′ . This is a contradiction. Hence, the point z ∈ Z ′ and the composition e ∘ i is a closed embedding. Therefore, A M ( X ) admits a closed embedding from Z ′ , that is not belonging to M n . Consequently, the space A M ( X ) ∉ M n by (iv) of Lemma 1.6. The proof is completed.□

Remark 1.7

Theorem 1 holds on A M ∗ ( X ) because A M ( X ) is open in A M ∗ ( X ) and any open subspace of a member in A n (respectively, M n ), where n ≥ 1 , is also belonging to A n (respectively, M n ).

Acknowledgements

This study is motivated by some works of Yoshito Ishiki, and the author would like to thank him for his helpful advice on the non-separable case of Theorem 1.1 and on the unbounded case in Remark 1.7. The author should like to also express his gratitude to the reviewers for their helpful suggestions.

  1. Funding information: The author states no funding involved.

  2. Author contributions: The author confirms the sole responsibility for the conception of the study, presented results, and manuscript preparation.

  3. Conflict of interest: The author states no conflict of interest.

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Received: 2024-05-01
Revised: 2024-07-18
Accepted: 2024-09-18
Published Online: 2024-11-11

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/agms-2024-0014
Keywords for this article
admissible metric; sup-metric; absolute Borel class; completely metrizable
Creative Commons
BY 4.0

Articles in the same Issue

  1. Research Articles
  2. C1,α-rectifiability in low codimension in Heisenberg groups
  3. (In)dependence of the axioms of Λ-trees
  4. Metric quasiconformality and Sobolev regularity in non-Ahlfors regular spaces
  5. An approach to metric space-valued Sobolev maps via weak* derivatives
  6. Gromov-Hausdorff limits of closed surfaces
  7. Curvature exponent and geodesic dimension on Sard-regular Carnot groups
  8. Qualitative Lipschitz to bi-Lipschitz decomposition
  9. On the Borel complexity and the complete metrizability of spaces of metrics
  10. Metric lines in the jet space
  11. Contractibility of boundaries of cocompact convex sets and embeddings of limit sets
  12. Lipschitz extension theorems with explicit constants
  13. Special Issue: Second Order Subelliptic PDEs - Part I
  14. On the heat kernel of the Rumin complex and Calderón reproducing formula
  15. On a critical Choquard-Kirchhoff p-sub-Laplacian equation in ℍn
  16. A view on Liouville theorems in PDEs
  17. Schauder estimates on bounded domains for KFP operators with coefficients measurable in time and Hölder continuous in space
  18. Liouville's type results for singular anisotropic operators
  19. On the role of embeddability in conformal pseudo-hermitian geometry
  20. On an evolution equation in sub-Finsler geometry
  21. One-side Liouville theorems under an exponential growth condition for Kolmogorov operators
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