Approximability, compactness and random dense sequences Robert - PDF document

Approximability, compactness and random dense sequences Robert Kenny

Notation X separable metric space R X := { ρ ρ : ⊆ N N → X onto } I d := { ν ν : ⊆ N → X dense } Fix open cover ( U i ) r 1 , [ r ] := { 1 , . . . , r } R : X ⇒ [ r ] , x �→ { i U i ∋ x } S : [ r ] ⇒ X, i �→ U i T ǫ : X ⇒ X, x �→ { y d ( x, y ) < ǫ } ( ǫ > 0) δ ≤ a ρ : ⇐ ⇒ ( ∀ ǫ > 0) ( T ǫ is ( δ, ρ )-computable ) δ ≤ ρ ⇐ ⇒ (id X : X → X is ( δ, ρ )-computable ) δ ∈ R X compact if for each finite open cover 1 , R : X ⇒ [ r ] is ( δ, δ N | [ r ] )-computable ( U i ) r

Example: If X compact and ν ∈ I d , the stan- dard representation (equivalent to Cauchy rep- resentation in case of a computable metric space) defined by p ∈ δ − 1 { x } : ⇐ ⇒ { p i − 1 i ∈ N ∧ p i ≥ 1 } = { k α ( k ) ∋ x } , α : N → T X , � i, j � �→ B d ( ν ( i ); ν Q + ( j )). Lemma 1. Suppose X compact, δ, ρ ∈ R X , δ compact. If ρ -computable points are dense in X then δ ≤ a ρ . Proof. Given ǫ > 0, pick open cover ( U i ) r 1 with max i diam U i < ǫ . Then ( δ, δ N | [ r ] )-computability of R and ( δ N | [ r ] , ρ )-computability of S implies ( δ, ρ )-computability of T ǫ . Some generalisation is possible. Consider e.g. Lemma 2. If ( X, d ) a totally bounded metric space and ν ∈ I d then for any r ∈ Q + there ex- ists finite A ⊆ dom ν with X = ∪ a ∈ A B d ( ν ( a ); r ) .

For ν 0 , ν 1 , ν, λ ∈ I d , write ⇒ ( ∀ ǫ > 0)( ∃ h ∈ P (1) )( ∀ c ) � ⇒ c ∈ ( λ ◦ h ) − 1 B ( ν ( c ); ǫ ) � ν ≤ a λ : ⇐ c ∈ dom ν = . . dom( ν 0 ⊕ ν 1 ) := ∪ i (2 dom ν i + i ) , ν 0 ⊕ ν 1 (2 a + i ) := ν i ( a ) ν ∈ I d compact if any finite open cover ( U i ) r 1 admits some f ∈ P (1) with dom ν ⊆ f − 1 [ r ] ∧ ( ∀ a ∈ dom ν ) � � ν ( a ) ∈ U f ( a ) (1) Proposition 3. Let X be a separable metric space, ρ ν the Cauchy representation for ν ∈ I d . For any ν, λ ∈ I d and δ, ρ ∈ R X , 1. ν ≤ a λ ⇐ ⇒ ρ ν ≤ a ρ λ 2. δ ≤ ρ = ⇒ δ ≤ a ρ 3. δ ⊔ ρ is a least upper bound of { δ, ρ } w.r.t. ≤ a 4. ν ⊕ λ is a least upper bound of { ν, λ } w.r.t. ≤ a Proof of (3): First apply (2) in δ i ≤ δ 0 ⊔ δ 1 ( i < 2). If also δ i ≤ a ρ , say via F i at precision ǫ ( i < 2), then δ 0 ⊔ δ 1 ≤ a ρ at precision ǫ via F : ⊆ N N → N N , i.p �→ F i ( p ).

Each ≤ a also reflexive and transitive. Thus above op- erations give rise to upper semilattice structures, on R X / ≡ a , I a := I d / ≡ a and ( I d ∩ X N ) / ≡ a . If X compact, analogue of Lemma 1 for dense partial sequences im- plies compact ν ∈ I d form a least element of I a (later we show there exists compact ν ∈ I d ). Proposition 4. 1. ρ ≤ �≤ a ρ > 2. ρ ≤ a ρ Cf 3. ρ < ≤ a ρ ≤ and symmetrically (replace <, ≤ by >, ≥ ). Hence ρ ≤ a ρ Cf , ρ ≤ , ρ ≥ and ρ b ,n ≤ a ρ Cf , ρ ≥ , ρ ≤ and (3) are the only ≤ a -reductions not shown in the left figure. Proof. (1): Let F realise ρ ≤ ≤ a ρ > to precision ǫ ; where p ∈ ρ − 1 ≤ { x } enumerates all rationals ≤ x , q = F ( p ) enu- merates strict right Dedekind cut of some y s.t. | x − y | < ǫ . q 0 is output after finitely many steps, with only finite prefix p N of p read from input. Let p ′ ∈ ρ − 1 ≤ { z } for some z ≥ ν Q ( q 0 ) + ǫ ( > y + ǫ > x ) with ( p ′ ) N = p N . Then F ( p ′ ) 0 = q 0 , so ( ρ < ◦ F )( p ′ ) < ν Q ( q 0 ) ≤ z − ǫ , contradiction. (3): Recall the T 0 -topology τ < = { ( x, ∞ ) x ∈ R } ∪ {∅ , R } on R (with respect to which ρ < is admissible) and the following Lemma 5. For any D ⊆ R , a function f : ⊆ R → R with dom f = D is ( τ < , τ < ) -continuous iff it is left-continuous and nondecreasing.

We specify f which is ( ρ < , ρ ≤ )-computable and lies in the open ǫ -envelope of id R . First, consider f : R → R as in Lemma 5 which is also piecewise constant: take f ( t ) = c i for t i < t ≤ t i +1 where strictly increasing ( c i ) i ∈ Z , ( t i ) i ∈ Z have inf t i = −∞ , sup t i = ∞ , c i − ǫ ≤ t i ∧ t i +1 < c i + ǫ ( i ∈ Z ). If ( t i ) i are uniformly right-computable and ( c i ) i uniformly left- computable then f is ( ρ < , ρ < )-computable. If ( c i ) i ∈ Z ⊆ R \ Q then any ( ρ < , ρ < )-realiser of f also ( ρ < , ρ ≤ )-realises For instance, if ǫ = 2 − j + 1 2 we can take t n := nǫ , it. c n := t n + ǫ 2 = 2 n +1 ǫ ( �∈ Q ), n ∈ Z . 2 Note now that (1) (with transitivity of ≤ a ) implies δ 0 �≤ a δ 1 for all ( δ 0 , δ 1 ) ∈ { ρ ≤ , ρ < , ρ Cn } × { ρ Cf , ρ b ,n , ρ ≥ , ρ, ρ > } . With sym- metric version (exchanging <, ≤ with >, ≥ ), shows ρ Cn �≤ a δ for all δ ∈ S \{ ρ Cn } , where S := { ρ Cf , ρ ≤ , ρ ≥ , ρ b ,n , ρ, ρ < , ρ > , ρ Cn } (for convenience we fix n ), while δ ≤ ρ Cn for all δ ∈ S . For δ 0 ∈ { ρ ≤ , ρ ≥ } again (1) and figure completely deter- mine either δ 0 �≤ a δ 1 or δ 0 ≤ δ 1 as δ 1 ∈ S varies. Case δ 0 ∈ { ρ < , ρ > } is similar except where (3) and its sym- metric version apply. So assume δ 0 ∈ { ρ Cf , ρ b ,n , ρ } , and δ 1 ∈ S \ { ρ < , ρ > , ρ Cn } (otherwise δ 0 ≤ δ 1 ). By (2) we already have δ 0 ≤ a δ 1 , which completes the proof.

ρ Cn ρ Cn � � � ��������� � � � � � � � � � � � � � � � � � � � � � ρ < ρ > � � � � � � � � � � � ��������� ρ < ≡ a ρ ≤ ρ > ≡ a ρ ≥ � � � � � � � � � � ρ ≤ ρ ≥ � � ρ � � � � � � � � � � � � ���������������� � � � � � � � � � � � � � ρ ≡ a ρ b ,n ≡ a ρ Cf � ρ b ,n � � � � � � � � � ρ Cf From the figure ( ρ ≡ a ρ b ,n ≡ a ρ Cf ) and known facts it fol- lows ≤ a does not imply continuous reducibility ≤ t (nor does ≡ a guarantee the same final topology). The con- verse inclusion ( ≤ t ) ⊆ ( ≤ a ) also fails in general, using next construction from [ ? ] (with Proposition 3(1)). Definition 6. Let ( Y, ν ) be a numbered set with | Y | ≥ 2 and fix x, y ∈ Y with x � = y . For each A ⊆ N write  ν ( k ) , if i = 0 ∧ k ∈ dom ν ,   ν A (2 k + i ) = ν A x,y (2 k + i ) := x, if i = 1 ∧ k ∈ A ,  y, if i = 1 ∧ k ∈ N \ A  ( i < 2). Iterating, we can compare towers constructed this way. First consider a generalisation of compact ν ∈ I d .

Definition 7. Let X be a separable metric space. ν ∈ I d separating at finite precision if for any distinct x, y ∈ X there exist open cover ( U i ) r 1 , V, W ∈ T X , h ∈ P (1) with x ∈ V ∧ y ∈ W ∧ ( ∀ c ∈ ν − 1 ( V ∪ W ))( c ∈ h − 1 [ r ] ∧ ν ( c ) ∈ U h ( c ) ) ∧ { i U i ∩ V � = ∅ � = U i ∩ W } = ∅ . Theorem 8. Let X be a separable metric space, ν 0 , . . . , ν n , λ, λ ′ ∈ I d , A i , B ⊆ N ( i < n ∈ N ), E n := { x i , y i i < n } ⊆ X and x, y ∈ X distinct. Suppose ν 0 separat- x i ,y i for all i < n and λ ′ = λ B ing a.f.p., ν i +1 = ( ν i ) A i x,y . If E n ∩ { x, y } = ∅ and B is nonrecursive then λ ′ �≤ a ν n . Proof. Fix ( U i ) r 1 , V, W, h as in definition of sep- aration a.f.p., and pick ǫ > 0 suff. small that B ( x ; ǫ ) ⊆ V ∧ B ( y ; ǫ ) ⊆ W ∧ N ǫ ( E n ) ∩ { x, y } = ∅ . Also suppose λ ′ | 2 N +1 ≤ a ν n at precision ǫ via f ∈ P (1) . i =0 a i 2 i where Write k ∈ N , a = f (2 k + 1) = � n a n ∈ N , ( a i ) i<n ⊆ { 0 , 1 } . The last requirement on ǫ means λ ′ (2 k + 1) ∈ { x, y } = ⇒ ν n ( a ) ∈ im ν 0 \ E n and a i = 0 for i < n (inductively for m = n, . . . , 1, use m m � b j 2 j ) = ν A m − 1 � � b j 2 j ) = ν m − 1 ( b j +1 2 j ) im ν 0 \ E n ∋ ν m ( m − 1 ( j =0 j =1 j<m

0 , ( a i +1 ) n − 1 where ( b j ) m 0 = ( a i ) n , . . . , ( a i + n − 1 ) 1 0 ). So, we 0 get ν n ( a ) = ν n ( a n 2 n ) = · · · = ν 1 ( a n 2 1 ) = ν 0 ( a n ) � � 2 − n f (2 k + 1) while g : k �→ a n = is com- putable with im g ⊆ dom ν 0 . ( λ ′ (2 k + 1) = x g ( k ) ∈ h − 1 B 0 ) ∧ ( λ ′ (2 k + 1) = = ⇒ ⇒ g ( k ) ∈ h − 1 B 1 ) for all k ∈ N where B 0 := { i y = B ( x ; ǫ ) ∩ U i � = ∅} and B 1 := { i B ( y ; ǫ ) ∩ U i � = ∅} . In particular, B ≤ m B 0 via h ◦ g , which implies B recursive, a contradiction. So, λ ′ �≤ a ν n . ⇒ ν A ≤ ν B , Lemma 9. 1. A ≤ m B = 2. If x � = y ∧ { x i , y i i < n } ∩ { x, y } = ∅ ∧ ∅ � = A � = N ∧ ( ∀ i < n )( ν i +1 = ( ν i ) A i x i ,y i ) and λ B x,y ≤ a ( ν n ) A x,y where ν 0 separating a.f.p. then B ≤ m A . Proof. (1): Fix f ∈ R (1) such that A = f − 1 B  2 k, if i = 0,  and let g : N → N , 2 k + i �→ if i = 1 . 2 f ( k ) + 1 ,  Then one checks ν A ≤ ν B via g . (2): Fix ( U i ) r 1 , V, W, h as in definition of separa- tion a.f.p. (for x � = y ), ǫ > 0 such that B ( x ; ǫ ) ⊆