A unified approach to Gelfand and de Vries dualities

Lemma A.1.

Let α : ( S , ≪ ) → ( S ′ , ≪ ′ ) be a weak proximity morphism between proximity Baer–Specker algebras and a , b ∈ S .

1. α ⁢ ( a ) + α ⁢ ( b ) ≤ α ⁢ ( a + b ) .

2. If 0 ≤ a , b , then α ⁢ ( a ) ⁢ α ⁢ ( b ) ≤ α ⁢ ( a ⁢ b ) .

Proof.

The proofs of (i) and (ii) are similar, and we only prove (i). By [7, Proposition 5.1], the restrictions of ≪ and ≪ ′ to idempotents are de Vries proximities,

is a de Vries morphism, and we have the following commutative diagram by [6, Corollary 6.7], where the vertical maps are 𝒃 ⁢ 𝒂 ⁢ ℓ -isomorphisms:

It then suffices to show that the inequality in (i) holds for σ ♭ . For this, let a , b ∈ ℝ ⁢ [ Id ⁡ ( S ) ] ♭ . Recalling the operations on ℝ ⁢ [ Id ⁡ ( S ) ] ♭ given after Definition 6.12, if r ∈ ℝ , then

and

Thus,

for each r ∈ ℝ , and so we have that σ ♭ ⁢ ( a ) + σ ♭ ⁢ ( b ) ≤ σ ♭ ⁢ ( a + b ) . ∎

Lemma A.2.

Let σ : ( B , ≺ ) → ( B ′ , ≺ ′ ) be a de Vries morphism between de Vries algebras. If e , f , g ∈ B with f ≺ g , then σ ⁢ ( e ∨ f ) ≤ σ ⁢ ( e ) ∨ σ ⁢ ( g ) .

Proof.

Since f ≺ g , we have σ ⁢ ( f * ) * ≺ σ ⁢ ( g ) , so σ ⁢ ( f * ) * ≤ σ ⁢ ( g ) . This yields σ ⁢ ( g ) * ≤ σ ⁢ ( f * ) . Therefore,

From this it follows that σ ⁢ ( e ∨ f ) ≤ σ ⁢ ( e ) ∨ σ ⁢ ( g ) . ∎

Lemma A.3.

Let S be a Specker algebra.

1. ([ 7 , Lemma 4.9 (5)]) If 0 ≠ e ∈ Id ⁡ ( S ) and r ∈ ℝ with r ⁢ e ≥ 0 , then r ≥ 0 .

2. ([ 7 , Lemma 4.9 (6)]) Let 0 ≠ e , f ∈ Id ⁡ ( S ) and 0 < r , p ∈ ℝ . Then r ⁢ e ≤ p ⁢ f if and only if r ≤ p and e ≤ f .

3. ([ 6 , Lemma 5.4 (1)]) Let a ∈ S . If r , p ∈ ℝ with r ≤ p , then

   ( a ∧ p ) - ( a ∧ r ) = [ ( a - r ) ∧ ( p - r ) ] ∨ 0 .

4. ([ 6 , Lemma 5.4 (1)]) Let a ∈ S with a = r 0 + ∑ i = 1 n r i ⁢ e i in decreasing form. Set p i = r 0 + ⋯ + r i for 1 ≤ i ≤ n . Then

   [ ( a - p i - 1 ) ∧ r i ] ∨ 0 = r i ⁢ e i .

5. ([ 6 , Lemma 5.4 (2)]) Let a , b ∈ S . Then there exist p 0 < ⋯ < p n in ℝ with p 0 ≤ a , b ≤ p n such that a and b have decreasing decompositions

   a = p 0 + ∑ i = 1 n ( p i - p i - 1 ) ⁢ e i   𝑎𝑛𝑑   b = p 0 + ∑ i = 1 n ( p i - p i - 1 ) ⁢ f i .

   Moreover, if a , b ≥ 0 , then we may assume p 0 = 0 .

6. ([ 6 , Lemma 6.4 (2)]) Suppose α : ( S , ≪ ) → ( S ′ , ≪ ′ ) is a weak proximity morphism between proximity Baer–Specker algebras. If a = r 0 + ∑ i r i ⁢ e i is in decreasing form, then α ⁢ ( a ) = r 0 + ∑ i r i ⁢ α ⁢ ( e i ) .

Lemma A.4.

Suppose α : ( S , ≪ ) → ( S ′ , ≪ ′ ) is a weak proximity morphism between proximity Baer–Specker algebras. Let 0 ≤ c ∈ S .

1. c is invertible if and only if there is 0 < r ∈ ℝ with r ≤ c .

2. If 0 ≤ b ≪ c and b is invertible, then α ⁢ ( c ) is invertible and α ⁢ ( c ) - 1 ≤ α ⁢ ( b - 1 ) .

Proof.

(i) Suppose 0 < r ≤ c for some r ∈ ℝ . Write

in full orthogonal form. Then r ⁢ e i ≤ c ⁢ e i = r i ⁢ e i , which implies that r ≤ r i by Lemma A.3 (ii). Consequently, each r i ≠ 0 , and hence r 0 - 1 ⁢ e 0 + ⋯ + r n - 1 ⁢ e n is the multiplicative inverse of c.

Conversely, let c be invertible, and write c = r 0 ⁢ e 0 + ⋯ + r n ⁢ e n as above. Without loss of generality, suppose that r 0 ≤ r i for each i. Since 0 ≤ c , e i , we have 0 ≤ c ⁢ e i = r i ⁢ e i . So, r i ≥ 0 by Lemma A.3 (i). If r 0 = 0 , then c ⁢ e 0 = r 0 ⁢ e 0 = 0 , which is false since c is invertible and e 0 ≠ 0 . Therefore,

(ii) Write

in compatible decreasing form by Lemma A.3 (v). We may assume that 1 > e 1 , f 1 . Since e 1 ≥ e i , we have e 1 * ⁢ e i = 0 for each i. Therefore, if r 0 = 0 , then b ⁢ e 1 * = 0 . So, e 1 * = 0 as b is invertible. This forces e 1 = 1 , which is false by assumption. So, r 0 ≠ 0 . In addition, since b ≥ 0 , we have r 0 ⁢ e 1 * = b ⁢ e 1 * ≥ 0 , which implies that r 0 > 0 by Lemma A.3 (i). Set p 0 = r 0 and p i = r 0 + r 1 + ⋯ + r i for i ≥ 1 . As each r i ≥ 0 for i ≥ 1 , we obtain that all p i > 0 . Therefore,

is in full orthogonal form. Consequently, since b is invertible and all p i ≠ 0 ,

Because 0 < p 0 ≤ ⋯ ≤ p n , we have p n - 1 ≤ ⋯ ≤ p 0 - 1 . From this we may write b - 1 in decreasing form as

Thus, by Lemma A.3 (vi),

Since b is invertible, there is 0 < r ∈ ℝ with r ≤ b by (i). Because b ≪ c , (P2) implies that r ≤ c . Therefore, r ≤ α ⁢ ( c ) , and so α ⁢ ( c ) is invertible by (i). Since α ⁢ ( c ) = r 0 + ∑ i = 1 n r i ⁢ α ⁢ ( f i ) by Lemma A.3 (vi), a similar calculation applied to α ⁢ ( c ) yields

From b ≪ c , we get e i ≪ f i for each i by Claim 7.15. Therefore, - α ⁢ ( - e i ) ≪ α ⁢ ( f i ) , and so α ⁢ ( e i * ) * ≪ α ⁢ ( f i ) by equation (5.1). Taking complements gives α ⁢ ( f i ) * ≪ α ⁢ ( e i * ) , and so α ⁢ ( f i ) * ≤ α ⁢ ( e i * ) for each i. Thus, α ⁢ ( c ) - 1 ≤ α ⁢ ( b - 1 ) . ∎

Lemma A.5.

Let ( S , ≪ ) be a proximity Baer–Specker algebra and let a , b ∈ S . If

are in compatible decreasing form, then the following hold:

1. a ∨ b = r 0 + ∑ i r i ⁢ ( e i ∨ f i ) .

2. a ∧ b = r 0 + ∑ i r i ⁢ ( e i ∧ f i ) .

Proof.

We prove (i); the proof of (ii) is similar. For 0 ≤ i ≤ n , set p i = r 0 + ⋯ + r i . By Lemma A.3 (iv), we have

Therefore, by standard vector lattice identities, for i ≥ 1 we have

where the last equality holds since r i ≥ 0 . Because p 0 ≤ a , b ≤ p n , we have p 0 ≤ a ∨ b ≤ p n . Therefore,

By the calculation above and Lemma A.3 (iii) and (iv),

Since r 0 = p 0 , it follows that a ∨ b = r 0 + ∑ i r i ⁢ ( e i ∨ f i ) . ∎

Lemma A.6.

Let α : ( S , ≪ ) → ( S ′ , ≪ ′ ) be a weak proximity morphism between proximity Baer–Specker algebras. Suppose that a , b , c ∈ S with b ≪ c .

1. α ⁢ ( a ∨ b ) ≤ α ⁢ ( a ) ∨ α ⁢ ( c ) .

2. α ⁢ ( a + b ) ≤ α ⁢ ( a ) + α ⁢ ( c ) .

3. If 0 ≤ a , b , then α ⁢ ( a ⁢ b ) ≤ α ⁢ ( a ) ⁢ α ⁢ ( c ) .

Proof.

(i) By Lemma A.3 (v), we may write

in compatible decreasing form. Then a ∨ b = r 0 + ∑ i = 1 n r i ⁢ ( e i ∨ f i ) by Lemma A.5 (i). Therefore,

by Lemma A.3 (vi). We have f i ≺ g i for each i by Claim 7.15. Thus, since the restriction of α to Id ⁡ ( S ) is a de Vries morphism,

by Lemma A.2. Consequently,

where the last equality follows from Lemma A.5 (i).

(ii) Since b ≪ c , we have - α ⁢ ( - b ) ≤ α ⁢ ( c ) by (PM3) and (P2), so - α ⁢ ( c ) ≤ α ⁢ ( - b ) . Therefore, by Lemma A.1 (i),

This yields α ⁢ ( a + b ) ≤ α ⁢ ( a ) + α ⁢ ( c ) .

(iii) First, suppose that b is invertible. Since 0 ≤ b , Lemma A.4 (i) shows that 0 < r ≤ b for some r ∈ ℝ . Because b ≪ c , by Lemma A.4 (ii) we have α ⁢ ( c ) - 1 ≤ α ⁢ ( b - 1 ) . Consequently,

by Lemma A.1 (ii). Multiplying by α ⁢ ( c ) yields α ⁢ ( a ⁢ b ) ≤ α ⁢ ( a ) ⁢ α ⁢ ( c ) .

For an arbitrary b ≥ 0 , by Lemma A.4 (i), 1 + b is invertible, and 1 + b ≪ 1 + c . Therefore, by the previous case, α ⁢ ( a ⁢ ( 1 + b ) ) ≤ α ⁢ ( a ) ⁢ α ⁢ ( 1 + c ) . Since α is a weak proximity morphism,

By Lemma A.1 (i),

Subtracting α ⁢ ( a ) yields (iii). ∎

Remark A.7.

Lemma A.6 (ii) follows from [7, Lemma 7.1 (2)], the proof of which is not choice-free.

Theorem A.8.

Let ( S , ≪ ) and ( S ′ , ≪ ′ ) be proximity Baer–Specker algebras. A map α : S → S ′ is a proximity morphism if and only if it is a weak proximity morphism.

Proof.

Clearly, if α is a proximity morphism, then it is a weak proximity morphism. For the converse, we only need to show that axioms (PM6)–(PM8) hold. Let a , c ∈ S with c ≪ c .

(PM6) The inequality α ⁢ ( a ∨ c ) ≥ α ⁢ ( a ) ∨ α ⁢ ( c ) holds since α is order preserving, and the reverse inequality holds by Lemma A.6 (i).

(PM7) The inequality α ⁢ ( a + c ) ≥ α ⁢ ( a ) + α ⁢ ( c ) holds by Lemma A.1 (i), and the reverse inequality holds by Lemma A.6 (ii).

(PM8) Let 0 ≤ c . First suppose that 0 ≤ a . The inequality α ⁢ ( a ⁢ c ) ≥ α ⁢ ( a ) ⁢ α ⁢ ( c ) holds by Lemma A.1 (ii), and the reverse inequality by Lemma A.6 (iii). Now apply the argument of [11, Remark 8.9] to conclude that the equality holds for all a. ∎